Mastering Physics Solutions Chapter 26 Geometrical Optics

Mastering Physics Solutions Chapter 26 Geometrical Optics

Mastering Physics Solutions

Chapter 26 Geometrical Optics Q.1CQ
Two plane mirrors meet at right angles at the origin, as indicated in Figure. Suppose an L-shaped object has the position and orientation labeled A. Draw the location and orientation of all the images of object A formed by the two mirrors.
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Solution:
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Chapter 26 Geometrical Optics Q.1P
Alaserbeam is reflected by a plane mirror. Itis observed that the angle between the incident and reflected beams is 28°. If the mirror is now rotated so that the angle of incidence increases by 5.0°, what is the new angle between the incident and reflected beams?
Solution:
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Chapter 26 Geometrical Optics Q.2CQ
Two plane mirrors meet at right angles at the origin, as indicated in Figure. Suppose an L-shaped object has the position and orientation labeled B. Draw the location and orientation of all the images of object B formed by the two mirrors.
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Solution:
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Chapter 26 Geometrical Optics Q.2P
The reflecting surfaces of two mirrors form a vertex with an angle of 120°. If a ray of light strikes mirror 1 with an angle of incidence of 55°, find the angle of reflection of the ray when it leaves mirror 2.
Solution:
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Chapter 26 Geometrical Optics Q.3CQ
What is the radius of curvature of a plane mirror? What is its focal length? Explain.
Solution:
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Chapter 26 Geometrical Optics Q.3P
A ray of light reflects from a plane mirror with an angle of incidence of 37°. If the mirror is rotated by an angle θ, through what angle is the reflected ray rotated?
Solution:
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Chapter 26 Geometrical Optics Q.4CQ
Dish receivers for satellite TV always use the concave side of the dish, never the convex side. Explain.
Solution:
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Chapter 26 Geometrical Optics Q.4P
IP Asmall vertical mirror hangs on the wall, 1.40 m above the floor. Sunlight strikes the mirror, and the reflected beam forms a spot on the floor 2.S0 m from the wall. Later in the day, you notice that the spot has moved to a point 3.75 m from the wall. (a) Were your two observations made in the morning or in the afternoon? Explain. (b) What was the change in the Sun’s angle of elevation between your two observations?
Solution:
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Chapter 26 Geometrical Optics Q.5CQ
Suppose you would like to start a fire by focusing sunlight onto a piece of paper. In Conceptual Checkpoint 26-2 we saw that a concave mirror would be better than a convex mirror for this purpose. At what distance from the mirror should the paper be held for best results?
Solution:
A fire can be started by focusing sunlight onto a piece of paper. For this purpose, a concave mirror is preferred over a convex mirror, because, when the light rays strike a concave mirror, it gets reflected and meets at the focal point of the mirror. Whereas, the reflected rays from a convex mirror diverge from the mirror, and appear as they come from the focal point behind the mirror.
The parallel rays from the sun are focused on a very small area of the paper. If a concave mirror is used, the light rays will be converged in front of the mirror. Whereas, when rays are focused on a convex mirror, the light rays appear to come from the focal point behind the mirror. So, a concave mirror focuses the light rays more strongly on the paper piece, when compared to a convex mirror.
For best results, the paper should be held at the focal length of the concave mirror, such that, the light rays converge on the small area of the paper and the heating would be the greatest.

Chapter 26 Geometrical Optics Q.5P
Sunlight enters a room at an angle of 32° above the horizontal and reflects from a small mirror lying flat on the floor. The reflected light forms a spot on a wall that is 2.0 m behind the mirror, as shown in Figure If you now place a pencil under the
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Solution:
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Chapter 26 Geometrical Optics Q.6CQ
When light propagates from one medium to another, does it always bend toward the normal? Explain.
Solution:
No, the bending of light depends upon the speed of the medium.
The light bends towards the normal when the light enters a medium in which its speed of propagation is less than it was in the first medium.
The light bends away from the normal when the light enters a medium in which its speed of propagation is greater than the first medium.

Chapter 26 Geometrical Optics Q.6P
You stand 1.50 m m front of a wall and gaze downward at a small vertical mirror mounted on it. In this mirror you can see the reflection of your shoes. If your eyes are 1.85 m above your feet, through what angle should the mirror be tilted for you to see your eyes reflected in the mirror? (The location of the mirror remains the same, only its angle to the vertical is changed.)
Solution:
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Chapter 26 Geometrical Optics Q.7CQ
A swimmer at point B in Figure needs help. Two lifeguards depart simultaneously from their tower at point A, but they follow different paths. Although both lifeguards run with equal speed on the sand and swim with equal speed in the water, the lifeguard who follows the longer path, ACB, arrives at point B before the lifeguard who follows the shorter, straight-line path from A to B. Explain.
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Solution:
One can run on sand with speed more than the speed with which one swim in water.
Since the lifeguard who follows the path ACB run more distance on sand compared to other, hence that lifeguard takes less time to travel more distance.
Distance travelled in water by lifeguard of path ACB is less than other, hence it takes less time to reach at same point B while speed of both lifeguards are same.

Chapter 26 Geometrical Optics Q.7P
IP Standing 2.3 m in front of a small vertical mirror, you see the reflection of your belt buckle, which is 0.72 m below your eyes. (a) What is the vertical location of the mirror relative to the level of your eyes? (b) What angle do your eyes make with the horizontal when you look at the buckle? (c) If you now move backward until you are 6.0 m from the mirror, will you still see the buckle, or will you see a point on your body that is above or below the buckle? Explain.
Solution:
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Chapter 26 Geometrical Optics Q.8CQ
When you observe a mirage on a hot day, what are you actually seeing when you gaze at the “pool of water” in the distance?
Solution:
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Chapter 26 Geometrical Optics Q.8P
How many times does the light beam shown in Figure reflect from (a) the top and (b) the bottom mirror?
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Solution:
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Chapter 26 Geometrical Optics Q.9CQ
Explain the difference between a virtual and a real image.
Solution:
The main difference between real image and virtual image is
(i) Real image:– If a divergent beam of light from a point after reflection or refraction actually converges to a second point, then the second point is called the real image of first point. The real image can be caught on a screen.
(ii) Virtual Image:– If a divergent beam of light from a point , after reflection or refraction , appears to diverge from a second point, then the second point is called the virtual image of the first point. The virtual image cannot be caught on a screen. The virtual image can be photographed.

Chapter 26 Geometrical Optics Q.9P
CE If you view a clock in a mirror, do the hands rotate clockwise or counterclockwise?
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Solution:
From the figure we can say that, the hands on mirror – image clock rotate counter clock wise

Chapter 26 Geometrical Optics Q.10CQ
Sitting on a deserted beach one evening, you watch as the last bit of the Sun approaches the horizon. Just before the Sun disappears from sight, is the top of the Sim actually above or below the horizon? That is, if Earth’s atmosphere could be instantly removed just before the Sun disappeared, would the Sun still be visible, or would it be below the horizon? Explain.
Solution:
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Chapter 26 Geometrical Optics Q.10P
A 12.5-foot-long, nearsighted python is stretched out perpendicular to a plane mirror, admiring its reflected image. If the greatest distance to which the snake can see cleariy is 26.0 ft, how close must its head be to the mirror for it to see a clear image of its tail?
Solution:
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Chapter 26 Geometrical Optics Q.11CQ
A large, empty coffee mug sits on a table. From your vantage point the bottom of the mug is not visible. When the mug is filled with wa ter, however, you can see the bottom of the mug. Explain.
Solution:
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Chapter 26 Geometrical Optics Q.11P
(a) How rapidly does the distance between you and your mirror image decrease if you walk directly toward a mirror with a speed of 2.6 m/s? (b) Repeat part (a) for the case in which you walk toward a mirror but at an angle of 38° to its normal.
Solution:
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Chapter 26 Geometrical Optics Q.12CQ
The Disappearing Eyedropper The accompanying photograph shows eyedroppers partially immersed in oil (left) and water (right). Explain why the dropper is invisible in the oil.
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Solution:
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Chapter 26 Geometrical Optics Q.12P
You are 1.9 m tall and stand 3.2 m from a plane mirror that extends vertically upward from the floor. On the floor 1.5 m in front of the mirror is a small table, 0.80 m high. What is the minimum height the mirror must have for you to be able to see the top of the table in the mirror?
Solution:
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Chapter 26 Geometrical Optics Q.13CQ
The Invisible Man In the H. G. Wells novel The Invisible Man, a person becomes invisible by altering his index of refraction to match that of air. This is the idea behind the disappearing eye-dropper in Conceptual Question. If the invisible man could actually do this, would he be able to see? Explain.
Question
12. The Disappearing Eyedropper The accompanying photograph shows eyedroppers partially immersed in oil (left) and water (right). Explain why the dropper is invisible in the oil.
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Solution:
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Chapter 26 Geometrical Optics Q.13P
The rear window in a car is approximately a rectangle, 1.3 m wide and 0.30 m high. The inside rearview mirror is 0.50 m from the driver’s eyes, and 1.50 m from the rear window. What are the minimum dimensions for the rearview mirror if the driver is to be able to see the entire width and height of the rear window in the mirror without moving her head?
Solution:
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Chapter 26 Geometrical Optics Q.14CQ
What’s the Secret? The top of Figure shows the words SECRET CODE written in different colors. If you place a cylindrical rod of glass or plastic just above the words, you find that SECRET appears inverted, but CODE does not Explain.
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Solution:
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Chapter 26 Geometrical Optics Q.14P
IP You hold a small plane mirror 0.50 m in front of your eyes, as shown in Figure (not to scale). The mirror is 0.32 cm high, and in it you see the image of a tall building behind you. (a) If the building is 95 m behind you, what vertical height of the building, H,can be seen in the mirror at any one time? (b) If you move the mirror closer to your eyes, does your answer to part (a) increase, decrease, or stay the same? Explain.
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Solution:
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Chapter 26 Geometrical Optics Q.15P
Two rays of light converge toward each other, as sljown in Figure forming an angle of 27°. Before they intersect, however, they are reflected from a circular plane mirror with a diameter of 11 cm. If the mirror can be moved horizontally to
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Solution:
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Chapter 26 Geometrical Optics Q.16P
For a corner reflector to be effective, its surfaces must be pre cisely perpendicular. Suppose the surfaces of a comer reflector left on the Moon’s surface by the Apollo astronauts formed a 90.001° angle with each other. If a laser beam is bounced back to Earth from this reflector, how far (in kilometers) from its starting point will the reflected beam strike Earth? For simplicity, assume the beam reflects h-om only two sides of the reflector, and that it strikes the first surface at precisely 45°.
Solution:
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Chapter 26 Geometrical Optics Q.17P
CE Astronomers often use large mirrors in their telescopes to gather as much light as possible from faint distant objects. Should the mirror in their telescopes be concave or convex? Explain.
Solution:
The mirrors used by the Astronomers in their telescopes are always concave, because concave mirrors focus all parallel rays of light (as from the stars) to a point in front of the mirror. On the other hand convex mirror disperse parallel rays of light by sending them outward on divergent paths.

Chapter 26 Geometrical Optics Q.18P
A section of a sphere has a radius of curvature of 0.86 m. If this section is painted with a reflective coating on both sides, what is the focal length of (a) the convex side and (b) the concave side?
Solution:
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Chapter 26 Geometrical Optics Q.19P
Amirrored-giass gazing globe in a garden is 31.9 cm in diameter. What is the focal length of the globe?
Solution:
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Chapter 26 Geometrical Optics Q.20P
Sunlight reflects from a concave piece of broken glass, converging to a point 15 cm from the glass. What is the radius of curvature of the glass?
Solution:
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Chapter 26 Geometrical Optics Q.21P
CE You hold a shiny tablespoon at arm’s length and look at the back side of the spoon. (a) Is the image you see of yourself upright or inverted? (b) Is the image enlarged or reduced? (c) Is the image real or virtual?
Solution:
Here the back of the spoon behaves like a convex mirror. Therefore from the conditions of the convex mirror we can say that the image is
a) Upright
b) The image is reduced in size.
c) Behind the spoon no light passes through it. So, the image is a virtual image.

Chapter 26 Geometrical Optics Q.22P
CE You hold a shiny tablespoon at arm’s length and look at the front side of the spoon. (a) Is the image you see of yourself upright or inverted? (b) Is the image enlarged or reduced? (c) Is the image real or virtual?
Solution:
Due to the silver coating, the spoon acts as a mirror and the front side of the spoon means we are looking at a concave mirror. In addition, holding the spoon at arms length means that we are outside the focal point of the mirror – clearly the focal length of the front side of a spoon is only a few centimeters.
If follows that our image is reduced, real and inverted.

Chapter 26 Geometrical Optics Q.23P
CE An object is placed in front of a convex mirror whose radius of curvature is R. What is the greatest distance behind the mirror that the image can be formed?
Solution:
We know that if the object is in front of the mirror then image produced by a convex mirror will be always behind the mirror. The greatest image distance occurs when the object is infinitely far from the mirror. In the case when the image is at the focal point, the greatest distance the image can be behind the mirror is \( f\quad =\cfrac { R }{ 2 } \)

Chapter 26 Geometrical Optics Q.24P
CE An object is placed to the left of a concave mirror, beyond its focal point. In which direction will the image move when the object is moved farther to the left?
Solution:
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Chapter 26 Geometrical Optics Q.25P
CE An object is placed to the left of a convex mirror. In which direction will the image move when the object is moved farther to the left?
Solution:
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We know that the image produced by a convex mirror is always behind mirror.
When the object is moved farther to the left, the image will move to right, i.e., towards the focal point of the lens.

Chapter 26 Geometrical Optics Q.26P
A small object is located 30.0 cm in frontofa concave mirror with a radius of curvature of 40.0 cm, Where will the image be formed?
Solution:
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Chapter 26 Geometrical Optics Q.27P
Use ray diagrams to show whether the image formed by a convex mirror increases or decreases in size as an object is brought closer to the mirror’s surface.
Solution:
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From the ray diagrams we can observe that there is an increase in size of the image if the object is brought closer to the mirror surface.

Chapter 26 Geometrical Optics Q.28P
An object with a height of 46 cm is placed 2.4 m in front of a concave mirror with a focal length of 0.50 m. (a) Determine the approximate location and size of the image using a ray diagram. (b) Is the image upright or inverted?
Solution:
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(b) The image formed just before the concave mirror and the image is inverted.

Chapter 26 Geometrical Optics Q.29P
Find the location and magnification of the image produced by the mirror in Problem using the mirror and magnification equations.
Problem
28. An object with a height of 46 cm is placed 2.4 m in front of a concave mirror with a focal length of 0.50 m. (a) Determine the approximate location and size of the image using a ray diagram. (b) Is the image upright or inverted?
Solution:
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∴ Image is formed at 0.63 m with a magnification -0.26

Chapter 26 Geometrical Optics Q.30P
An object with a height of 46 cm is placed 2.4 m in front of a convex mirror with a focal length of −0.50 m. (a) Determine the approximate location and size of the image using a ray diagram. (b) Is the image upright or inverted?
Solution:
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Chapter 26 Geometrical Optics Q.31P
Find the loca ti on and magnification of Ehe image produced by the mirror in Problem using the mirror and magnification equations.
Problem
30. An object with a height of 46 cm is placed 2.4 m in front of a convex mirror with a focal length of −0.50 m. (a) Determine the approximate location and size of the image using a ray diagram. (b) Is the image upright or inverted?
Solution:
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Chapter 26 Geometrical Optics Q.32P
During a daytime football game you notice that a player’s reflective helmet forms an image of the Sun 4.8 cm behind the surface of the helmet. What is the radius of curvature of the helmet, assuming it to be roughly spherical?
Solution:
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Chapter 26 Geometrical Optics Q.33P
IP A magician wishes to create the illusion of a 2.74-m-tall elephant. He plans to do this by forming a virtual image of a 50.0-cm-tall model elephant with the help of a spherical mirror. (a) Should the mirror be concave or convex? (b) If the model must be placed 3.00 m from the mirror, what radius of curvature is needed? (c) How far from the mirror will the image be formed?
Solution:
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Chapter 26 Geometrical Optics Q.34P
A person 1.7 m tall stands 0.66 m from a reflecting globe in a garden. (a) If the diameter of the globe is 18 cm, where is the image of the person, relative to the surface of the globe? (b) How large is the person’s image?
Solution:
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Chapter 26 Geometrical Optics Q.35P
Shaving /makeup mirrors typically have one flat and one concave (magnifying) surface. You find that you can project a magnified image of a lightbulb onto the wall of your bathroom if you hold the mirror 1.8 m from the bulb and 3.5 m from the wall. (a) What is the magnification of the image? (b) Is the image erect or inverted? (c) What is the focal length of the mirror?
Solution:
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Chapter 26 Geometrical Optics Q.36P
The Hale Telescope The 200-inch-diameter concave mirror of the Hale telescope on Mount Falomar has a focal length of 16.9 m. An astronomer stands 20.0 m in front of this mirror. (a) Where is her image located? is it in front of or behind the mirror? (b) Is her image real or virtual? How do you know? (c) What is the magnification of her image?
Solution:
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Chapter 26 Geometrical Optics Q.37P
A concave mirror produces a virtual image that is three times as tall as the object. (a) If the object is 28 cm in front of the mirror, what is the image distance? (b) What is the focal length of this mirror?
Solution:
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Chapter 26 Geometrical Optics Q.38P
A concave mirror produces a real image that is three times as large as the object. (a) If the object is 22 cm in front of the mirror, what is the image distance? (b) What is the focal length of this mirror?
Solution:
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Chapter 26 Geometrical Optics Q.39P
The virtual image produced by a convex mirror is one-quarter the size of the object. (a) If the object is 36 cm in front of the mirror, what is the image distance? (b) What is the focal length of this mirror?
Solution:
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Chapter 26 Geometrical Optics Q.40P
IP A 5.7-ft tallshopper in a department store is 17 ft from a convex security mirror. The shopper notices that his image in the mirror appears to be only 6.4 in. tali. (a) Ts the shopper’s image upright or inverted? Explain. (b) What is the mirror’s radius of curvature?
Solution:
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Chapter 26 Geometrical Optics Q.41P
You view a nearby tree in a concave mirror. The inverted image of the tree is 3.8 cm high and is located 7.0 cm in front of the mirror. If the tree is 23 m from the mirror, what is its height?
Solution:
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Chapter 26 Geometrical Optics Q.42P
A shaving/makeup mirror produces an erect image that is magnified by a factor of2.2 when your face is 25 cm from the mirror. What is the mirror’s radius of curvature?
Solution:
1403-26-42P SA Code: 6078.SRCode: 5784
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Chapter 26 Geometrical Optics Q.43P
A concave mirror with a focal length of 36 cm produces an image whose distance from the mirror is one-third the object distance. Find the object and image distances.
Solution:
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Chapter 26 Geometrical Optics Q.44P
CE Predict/Explain When a ray of light enters a glass lens surrounded by air, it slows down. (a) As it leaves the glass, does its speed increase, decrease, or stay the same? (b) Choose the best explanation from among the following:
I. Its speed increases because the ray is now propagating in a medium with a smaller index of refraction.
II. The speed decreases because the speed of light decreases whenever light moves from one medium to another.
III. The speed will stay the same because the speed of light is a universal constant.
Solution:
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Chapter 26 Geometrical Optics Q.45P
CE Samurai Fishing A humorous scene in Akira Kurosawa’s classic film The Seven Samumi shows the young samurai Kikuchiyo wading into a small stream and plucking a fish from it for his dinner. (a) As Kikuchiyo looks through the water to the fish, does he sec it in the general direction of point 1or point 2 in Figure? (b) If the fish looks up at Kikuchiyo, does it see Kikuchiyo’s head in the general direction of point 3 or point 4?
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Solution:
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(b)
If the ray from the fish toward the head is extended without any bending, then the fish sees the head along the direction of point 4.
Therefore the right answer is point 4.

Chapter 26 Geometrical Optics Q.46P
CE When color A and color B are sent through a prism, color Ais bent more than color B. Which color travels more rapidly in the prism? Explain.
Solution:
The speed of light in a material is given by \( \frac { c }{ n } \) The color with the smaller reflective index has the greater speed, so from the given data we can say that the color B travels more rapidly because the color B has bent less than the color A.

Chapter 26 Geometrical Optics Q.47P
CE Day Versus Night (a) Imagine for a moment that the Earth has no atmosphere. Over the period of a year, is the number of daylight hours at your home greater than, less than, or equal to the number of nighttime hours? (b) Repeat part (a), only this time take into account the Earth’s atmosphere.
Solution:
a) The atmosphere acts as a spherical mirror by reflecting sunlight. If there is no atmosphere, Sunlight is trapped within the atmosphere due to the reflecting effects of gases. Therefore, day and night is depends upon whether the sun is shining on the planet or not. Since the Earth rotates about its axis with constant speed and resolves around the sun, half of the earth faces the sun at any given time. Thus there should be an equal number of day light times and night time hours.
b) If the atmosphere is present, some of the light is absorbed by the atmosphere of the earth. The light gets refracting during dawn and dusk adding to daylight hours and reducing night time hours, so daylight hours are greater than night time.

Chapter 26 Geometrical Optics Q.48P
CE Predict/Explain A kitchen has twin side-by-side sinks. One sink is filled with water, the other is empty. (a) Does the sink with water appear to be deeper, shallower, or the same depth as the empty sink? (b) Choose the best explanation from among the following:
I. The sink with water appears deeper because you have to look through the water to sec the bottom.
II. Water bends the light, making an object under the water appear to be closer to the surface. Thus the water-filled sink appears shallower.
III. The sinks are identical, and therefore have the same depth. This doesn’t change by putting water in one of them.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics48ps

Chapter 26 Geometrical Optics Q.49P
CE A light beam undergoes total internal reflection at the interface between medium A, in which it propagates, and medium B, on the other side of the interface. Which medium has the greater index of refraction? Explain.
Solution:
Total internal reflection takes place when light goes from denser medium to rarer medium. Here the light ray is going from medium A to medium B and it is undergoing total internal reflection.
Therefore the medium A is denser than medium B. so, the refractive index for the medium A is greater than medium B.

Chapter 26 Geometrical Optics Q.50P
Light travels a distance of 0.960 m in 4.00 ns in a given substance. What is the index of refraction of this substance?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics50ps

Chapter 26 Geometrical Optics Q.51P
Find the ratio of the speed of light in water to the speed of light in diamond.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics51ps

Chapter 26 Geometrical Optics Q.52P
Ptolemy’s Optics One of the many works published by the Greek astronomer Ptolemy (A.D. ca. 100–170) was Optics. In this book Ptolemy reports the results of refraction experiments he conducted by observing light passing from air into water. His results are as follows: angle of incidence = 10.0°, angle of refraction = 8.00°; angle of incidence = 20.0°, angle of refraction = 15.5°. Find the percentage error in the calculated index of refraction of water for each of Ptolemy’s measurements.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics52ps
mastering-physics-solutions-chapter-26-geometrical-optics52ps1

Chapter 26 Geometrical Optics Q.53P
Light enters a container of benzene at an angle of 43° to the normal; the refracted beam makes an angle of 27° with the normal. Calculate the index of refraction of benzene.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics53ps
mastering-physics-solutions-chapter-26-geometrical-optics53ps1
mastering-physics-solutions-chapter-26-geometrical-optics53ps2

Chapter 26 Geometrical Optics Q.54P
The angle of refraction of a ray of light traveling into an ice cube from air is 38°. Find the angle of incidence.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics54ps

Chapter 26 Geometrical Optics Q.55P
IP (a) Referring to Problem, suppose the ice melts, but the angle of refraction remains the same. Is the corresponding angle of incidence greater than, less than, or the same as it was for ice? Explain. (b) Calculate the angle of incidence for part (a).
Problem
54. The angle of refraction of a ray of light traveling into an ice cube from air is 38°. Find the angle of incidence.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics55ps
mastering-physics-solutions-chapter-26-geometrical-optics55ps1

Chapter 26 Geometrical Optics Q.56P
A submerged scuba diver looks up toward the calm surface of a freshwater lake and notes that the Sun. appears to be 35° from the vertical. The diver’s friend is standing on the shore of the lake. At what angle above the horizon does the friend see the sun?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics56ps
mastering-physics-solutions-chapter-26-geometrical-optics56ps1

Chapter 26 Geometrical Optics Q.57P
A pond with a total depth (ice + water) of 3.25 m is covered by a transparent layer of ice, with a thickness of 0.38 m. Find the time required for light to travel vertically from the surface of the ice to the bottom of the pond.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics57ps
mastering-physics-solutions-chapter-26-geometrical-optics57ps1

Chapter 26 Geometrical Optics Q.58P
Light is refracted as it travels from a point A in medium 1 to a point B in medium 2. If the index of refraction is 1.33 in médium 1 and 1.51 in medium 2, how long does it take light to go from A to B, assuming it travels 331 cm in medium 1 and 151 cm in medium 2?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics58ps
mastering-physics-solutions-chapter-26-geometrical-optics58ps1

Chapter 26 Geometrical Optics Q.59P
You have a semicircular disk of glass with an index of refraction of n = 1.52, Find the incident angle θfor which the beam of light in Figure will hit the indicated point on the screen.
mastering-physics-solutions-chapter-26-geometrical-optics59p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics59ps
mastering-physics-solutions-chapter-26-geometrical-optics59ps1

Chapter 26 Geometrical Optics Q.60P
The observer in Figure is positioned so that the far edge of the bottom of the empty glass (not to scale) is just visible. When the glass is filled to the top with water, the center oi the bottom of the glass is just visible to the observer. Find the height, H, of the glass, given that its width is W = 6.2 cm.
mastering-physics-solutions-chapter-26-geometrical-optics60p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics60ps
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mastering-physics-solutions-chapter-26-geometrical-optics60ps2

Chapter 26 Geometrical Optics Q.61P
mastering-physics-solutions-chapter-26-geometrical-optics61p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics61ps
mastering-physics-solutions-chapter-26-geometrical-optics61ps1
mastering-physics-solutions-chapter-26-geometrical-optics61ps2

Chapter 26 Geometrical Optics Q.62P
A ray of light enters the long side of a 45°–90°–45° prism and undergoes two total internal reflections, as indicated in Figure. The result is a reversal in the ray’s direction of propagation. Find the minimum value of the prism’s index of refraction, n, for these internal reflections to be total.
mastering-physics-solutions-chapter-26-geometrical-optics62p
Solution:
From the given figure if the light ray is incident normally on the hypotenuse face, it is incident on the opposite face at an angle of 45 and suffers total internal reflection. The same thing happens at the next face and the ray emerges out of the prism parallel to incident direction from the given figure.
mastering-physics-solutions-chapter-26-geometrical-optics62ps
mastering-physics-solutions-chapter-26-geometrical-optics62ps1

Chapter 26 Geometrical Optics Q.63P
When the prism in Problem is immersed in a fluid with an index of refraction of 1.21, the internal reflections shown in Figure are still total. Thereflections are no longer totaj however, when the prism is immersed in a fluid with n = 1.43. Use this information to set upper and lower limits on the possi-ble values of the prism’s index of refraction.
Problem
62. A ray of light enters the long side of a 45°–90°–45° prism and undergoes two total internal reflections, as indicated in Figure. The result is a reversal in the ray’s direction of propagation. Find the minimum value of the prism’s index of refraction, n, for these internal reflections to be total.
mastering-physics-solutions-chapter-26-geometrical-optics63p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics63ps
mastering-physics-solutions-chapter-26-geometrical-optics63ps1

Chapter 26 Geometrical Optics Q.64P
IP A glass paperweight with an index of refraction n rests un a desk, as shown in Figure. An incident ray of light enters the horizontal top surface of the paperweight at an angle θ = 77° to the vertical. (a) Find the minimum value of n for which the internal reflection on the vertical surface of the paperweight is total. (b) If θ is decreased, is the minimum value of n increased or decreased? Explain.
mastering-physics-solutions-chapter-26-geometrical-optics64p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics64ps
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mastering-physics-solutions-chapter-26-geometrical-optics64ps2
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Chapter 26 Geometrical Optics Q.65P
IP Suppose the glass paperweight in Figure has an index of refraction n = 1.38. (a) Find the value of θ for which the reflection on the vertical surface of the paperweight exactly satisfies the condition for total internal reflection. (b) If θis increased, is the reflection at the vertical surface still total? Explain.
mastering-physics-solutions-chapter-26-geometrical-optics65p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics65ps
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mastering-physics-solutions-chapter-26-geometrical-optics65ps2
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Chapter 26 Geometrical Optics Q.66P
IP Consider the physical system shown in Figure and described in Problem. (a) If the index of refraction of the glass is increased, will the desired value of θ increase or decrease? Explain. (b) Find the value of θfor the case of flint glass (n = 1.66).
Problem
59. You have a semicircular disk of glass with an index of refraction of n = 1.52, Find the incident angle θfor which the beam of light in Figure will hit the indicated point on the screen.
mastering-physics-solutions-chapter-26-geometrical-optics66p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics66ps
mastering-physics-solutions-chapter-26-geometrical-optics66ps1

Chapter 26 Geometrical Optics Q.67P
While studying physics at the library late one night, you notice the image of the desk lamp reflected from the varnished tabletop. When you turn your Polaroid sunglasses sideways, the reflected image disappears. Tf this occurs when the angle between the incident and reflected rays is 110°, what is the index of refraction of the varnish?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics67ps

Chapter 26 Geometrical Optics Q.68P
A horizontal beam of light enters a 45°–90°–45° prism at the center of its long side, as shown in Figure. The emerging ray moves in a direction that is 34° below the horizontal. What is the index of refraction of this prism?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics68p
Therefore, refractive index of the prism is.1.70
mastering-physics-solutions-chapter-26-geometrical-optics68ps
mastering-physics-solutions-chapter-26-geometrical-optics68ps1
mastering-physics-solutions-chapter-26-geometrical-optics68ps2

Chapter 26 Geometrical Optics Q.69P
A laser beam enters one of the sloping faces of the equilateral glass prism (n = 1.42) in Figure and refracts through the prism. Within the prism the light travels horizontally. What is the angle θ between the direction of the incident ray and the direction of the outgoing ray?
mastering-physics-solutions-chapter-26-geometrical-optics69p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics69ps
mastering-physics-solutions-chapter-26-geometrical-optics69ps1
mastering-physics-solutions-chapter-26-geometrical-optics69ps2

Chapter 26 Geometrical Optics Q.70P
mastering-physics-solutions-chapter-26-geometrical-optics70p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics70ps

Chapter 26 Geometrical Optics Q.71P
mastering-physics-solutions-chapter-26-geometrical-optics71p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics71ps

Chapter 26 Geometrical Optics Q.72P
An object is a distance f/2 from a convex lens. (a) Use a ray diagram to find the approximate location of the image. (b) Is the image upright or inverted? (c) Is the image real or virtual? Explain.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics72ps

Chapter 26 Geometrical Optics Q.73P
An object is a distance 2f from a convex lens. (a) Use a ray diagram to find the approximate location of the image. (b) Is the image upright or inverted? (c) Is the image real or virtual? Explain.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics73ps
mastering-physics-solutions-chapter-26-geometrical-optics73ps1

Chapter 26 Geometrical Optics Q.74P
Two lenses that are 35 cm apart are used to form an image, as shown in Figure. Lens 1 is converging and has a focal length f1 = 14 cm; lens 2 is diverging and has a focal length f2 = −7.0 cm. The object is placed 24 cm to the left of lens 1. (a) Use a ray diagram to find the approximate location of the image. (b) Is the image upright or inverted? (c) Is the image real or virtual? Explain.
mastering-physics-solutions-chapter-26-geometrical-optics74p
Solution:
The location of an image by a concave or a convex mirror can be traced with the help of a ray tracing diagram.
The properties of the three principal rays are as follows,
1. The parallel ray travels parallel to the axis and passes through the focal point of the lens if it is convex or traces back to the focal point if it is concave.
2. The focal point ray passes through the focal point of the lens if the lens is convex. For a concave lens, the focal point ray passes through the focal point on the other side of the lens by passing through the lens.
3. The midpoint ray passes through the middle of the ray.
mastering-physics-solutions-chapter-26-geometrical-optics74ps
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Chapter 26 Geometrical Optics Q.75P
Two lenses that are 35 cm apart are used to form an image, as shown in Figure. Lens 1 is diverging and has a focal length f1 = −7.0 cm; lens 2 is converging and has a focal length f2 = 14cm. The object is placed 24 cm to the left of lens 1. (a) Use a ray diagram to find the approximate location of the image. (b) Is the image upright or inverted? (c) Is the image real or virtual? Explain.
mastering-physics-solutions-chapter-26-geometrical-optics75p
Solution:
Concave lens always produce upright and virtual images. Convex lens produce either upright or inverted images depending upon the position of the object, which can be either virtual or real.
mastering-physics-solutions-chapter-26-geometrical-optics75ps
mastering-physics-solutions-chapter-26-geometrical-optics75ps1

Chapter 26 Geometrical Optics Q.76P
A convex lens is held over a piece of paper outdoors on a sunny day. When the paper is held 26 cm below the lens, the sunlight is focused on the paper and the paper ignites. What is the focal length of the lens?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics76ps

Chapter 26 Geometrical Optics Q.77P
A concave lens has a focal length of −32 cm. Find the image distance and magnification that result when an object is placed 29 cm in front of the lens.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics77ps

Chapter 26 Geometrical Optics Q.78P
When an object is located 46 cm to the left of a lens, the image is formed 17 cm to the right of the lens. What is the focal length of the lens?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics78ps

Chapter 26 Geometrical Optics Q.79P
Anobjectwithaheightof2.54cmisplaced36.3mmtotheleft of a lens with a focal length of 35.0 mm. (a) Where is the image located? (b) What is the height of the image?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics79ps
mastering-physics-solutions-chapter-26-geometrical-optics79ps1

Chapter 26 Geometrical Optics Q.80P
A lens for a 35-mm camera has a focal length given by ƒ = 55 mm. (a) How close to the film should the lens be placed to form a sharp image of an object that is 5.0 m away? (b) What is the magnification of the image on the film?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics80ps
mastering-physics-solutions-chapter-26-geometrical-optics80ps1
mastering-physics-solutions-chapter-26-geometrical-optics80ps2

Chapter 26 Geometrical Optics Q.81P
IP An object is located to the left of a convex lens whose focal length is 34 cm. The magnification produced by the lens is m = 3.0. (a) To increase the magnification to 4.0, should the object be moved closer to the lens or farther away? Explain. (b) Calculate the distance through which the object should be moved.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics81ps
mastering-physics-solutions-chapter-26-geometrical-optics81ps1

Chapter 26 Geometrical Optics Q.82P
IP You have two lenses at your disposal, one with a focal length f1 = +40.0 cm, the other with a focal length f2 = −40.0 cm. (a) Which of these two lenses would you use to project an image of a lightbulb onto a wall that is far away? (b) if you want to produce an image of the bulb that is enlarged by a factor of 2.00, how far from the wall should the lens be placed?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics82ps
mastering-physics-solutions-chapter-26-geometrical-optics82ps1

Chapter 26 Geometrical Optics Q.83P
(a) Determine the distance from lens 1 to the final image for the system shown in Figure. (b) What is the magnification of this image?
mastering-physics-solutions-chapter-26-geometrical-optics83p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics83ps
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Chapter 26 Geometrical Optics Q.84P
(a) Determine the distance from lens 1 to the final image for the system shown in Figure. (b) What is the magnification of this image?
mastering-physics-solutions-chapter-26-geometrical-optics84p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics84ps
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Chapter 26 Geometrical Optics Q.85P
mastering-physics-solutions-chapter-26-geometrical-optics85p
Solution:
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mastering-physics-solutions-chapter-26-geometrical-optics85ps1
mastering-physics-solutions-chapter-26-geometrical-optics85ps2

Chapter 26 Geometrical Optics Q.86P
IP BIO Albert is nearsi ghtcd, and wi thou t his eyeglasses he can focus only on objects less than 2.2 m away. (a) Are Albert’s eyeglasses concave or convex? Explain. (b) To correct Albert’s nearsightedness, his eyeglasses must produce a virtual, upright image at a distance of 2.2 m when viewing an infinitely distant object. What is the focal length of Albert’s eyeglasses?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics86ps

Chapter 26 Geometrical Optics Q.87P
A small insect viewed through a convex lens is 1.4 cm from the lens and appears twice its actual size. What is the focal length of the lens?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics87ps

Chapter 26 Geometrical Optics Q.88P
IP A friend tells you that when he takes off his eyeglasses and holds them 23 cm above a printed page the image of the print is erect but reduced to 0:67 of its actual size. (a) Is the image real or virtual? How do you know? (b) What is the focal length of your friend’sglasses? (c) Are the lenses in the glasses concave or convex? Explain.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics88ps

Chapter 26 Geometrical Optics Q.89P
IP A friend tells you that when she takes off her eyeglasses and holds them 23 cm above a printed page the image of the print is erect but enlarged to 1.5 times its actual size. (a) Is the image real or virtual? How do you bow? (b) What is the focal length of your friend’s glasses? (c) Are the lenses in the glasses concave or convex? Explain.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics89ps

Chapter 26 Geometrical Optics Q.90P
CE Predict/Explain You take a picture of a rainbow with an infrared camera, and your friend takes a picture at the same time with visible light. (a) Is the height of the rainbow in the infrared picture greater than, less than, or the same as the height of the rainbow in the visible-light picture? (b) Choose the best explanation from among the following:
I. The height will be greater because the top of a rainbow is red, and so infrared light would be even higher.
II. The height will be less because infrared light is below the visible spectrum.
III. A rainbow is the same whether seen in visible light or infrared; therefore the height is the same.
Solution:
1403-26-90P SA Code: 6078.
SR Code:5784
(a)
Height of the rainbow in the infrared picture is greater than the height of the rainbow in the visible-light picture. Because the top of a rainbow is red and so infrared light would be even higher and also the spectrum of infrared waves region is wider than the spectrum that contains the visible region.
(b)
Therefore the best explanation is I.

Chapter 26 Geometrical Optics Q.91P
The index of refraction for red light in a certain liquid is 1.320; the index of refraction for violet light in the same liquid is 1.332. Find the dispersion (θV − θr)for red and violet light when both arc incident on the flat surface of the liquid at an angle of 45.00° to the normal.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics91ps
mastering-physics-solutions-chapter-26-geometrical-optics91ps1

Chapter 26 Geometrical Optics Q.92P
A horizontal incident beam consisting of white light passes through an equilateral prism, like the one shown in Figure. What is the dispersion (θV − θr)of the outgoing beam if the prism’s index of refraction is nv = 1.505 for violet light and nr = 1.421 for red light?
mastering-physics-solutions-chapter-26-geometrical-optics92p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics92ps
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Chapter 26 Geometrical Optics Q.93P
The focal length of a lens is inversely proportional to the quantity (n − 1), where n is the index of refraction of the lens material. The value of n, however, depends on the wavelength of the light that passes through the lens. For example, one type of flint glass has an index of refraction of nr = 1.572 for red light and nv = 1.605 in violet light. Now, suppose a white object is placed 24.00 an infront of a lens made from this type of glass. If the red light reflected from this object produces a sharp image 55.00 cm from the lens, where will the violet image be found?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics93ps
mastering-physics-solutions-chapter-26-geometrical-optics93ps1

Chapter 26 Geometrical Optics Q.94GP
CE Jurassic Park A T. rex chases the heroes of Steven Spielberg’s Jurassic Park as they desperately try to escape in their Jeep. The T. rex is closing in fast, as they can see in the outside rearview mirror. Near the bottom of the mirror they also see the following helpful message: objects in the mirror are closer than they appear. Is this mirror concave or convex? Explain.
Solution:
Objects in the mirror are closer than they appear because the mirror produces an image that is reduced in size, which makes the object look as if it is farther away. In addition, we know that the rear view mirror always gives an upright image, no matter how close or far away the object. The mirror that produces upright and reduced images is the convex mirror. So the mirror is convex mirror.

Chapter 26 Geometrical Optics Q.95GP
CE The receiver for a dish antenna is placed in front of the concave surface of the dish. If the radius of curvature of the dish is R, how far in fron tof the dish should the receiver be placed? Explain.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics95ps

Chapter 26 Geometrical Optics Q.96GP
CE Predict/Explain if a lens is immersed in water, its focal length changes, as discussed in Conceptual Checkpoint 26−5. (a) if a spherical mirror is immersed in water, does its focal length increase, decrease, or stay the same? (b) Choose the best explanation from among the following:
I. The focal length will increase because the water will cause more bending of light.
II. Water will refract the light. This, combined with the reflection due to the mirror, will result in a decreased focal length.
III. The focal length stays the same because it depends on the fact that the angle of incidence is equal to the angle of reflection for a mirror. Thisis unaffected by the presence of the water.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics96ps

Chapter 26 Geometrical Optics Q.97GP
CE Predict/Explain A glass slab surrounded by air causes a sideways displacement in a beam of light. (a) If die slab is now placed in water, does the displacement it causes increase, decrease, or stay the same? (b) Choose the best explanation from among the following:
I. The displacement of the beam increases because of the in creased refraction due to the water.
II. The displacement of the beam is decreased because with water surrounding the slab there is a smaller difference m index of refraction between the slab and its surroundings
III. The displacement stays the same because it is determined only by the properties of the slab; in particular, the material it is made of and its thickness.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics97ps
(b)Therefore the best explanation is II.

Chapter 26 Geometrical Optics Q.98GP
CE Referring to Conceptual Question, suppose the same type of glass used in an eyedropper is made into a convex lens with a focal length ƒ. If this lens is immersed in the oil of the bottle on the left in the photo, will its focal length be 0, f/2, 2f, or ∞? (Hint: See Conceptual Checkpoint 26–5.)
Question
12. The Disappearing Eyedropper The accompanying photograph shows eyedroppers partially immersed in oil (left) and water (right). Explain why the dropper is invisible in the oil.
mastering-physics-solutions-chapter-26-geometrical-optics98p
Solution:
As the index of refraction of the lens and that of the oil is same the light rays goes straight without bend. If the light rays travels straight, they will be focused at the infinity. So the focus of the lens is infinity

Chapter 26 Geometrical Optics Q.99GP
CE Two identical containers are filled with different bans-parent liquids. The container with liquid A appears to have a greater depth than the container with liquid B. Which liquid has the greater index of refraction? Explain.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics99ps

Chapter 26 Geometrical Optics Q.100GP
CE Is the image you see in a three-dimensional corner refleeter upright or inverted?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics100ps

Chapter 26 Geometrical Optics Q.101GP
CE Inverse Lenses Suppose we mold a hollow piece of plastic into the shape of a double concave lens. The “lens” is watertight, and itsinterior is filled with air. We now place this lens in water and shine a beam of light on it. (a) Does the lens converge or diverge the beam of light? Explain. (b) If our hollow lens is double convex instead, does it converge or diverge a beam of light when immersed in water? Explain.
Solution:
Normally we use a concave lens of greater refractive index. As the light is traveling from less index of refraction to high index of refraction and bends towards the normal and in case of concave it diverges.
a) But here we care using the lens with less index of refraction. So it bends towards the normal and the beam will be converged
c) Similarly, in case of convex lens, the beam will be diverged

Chapter 26 Geometrical Optics Q.102GP
IP Suppose the separation between the two mirrors in Figure is increasedby moving the top mirror upward. (a) Will this affect the number of reflections made by the beam of light? If so, how? (b) What is the total number of reflections made by the beam of light when the separation between the mirrors is 145 cm?
Figure
mastering-physics-solutions-chapter-26-geometrical-optics102p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics102ps
mastering-physics-solutions-chapter-26-geometrical-optics102ps1

Chapter 26 Geometrical Optics Q.103GP
Standing 2.0 m in front of a small vertical mirror you see the reflection of your belt buckle, which is 0.70m below your eyes. If you remain 2.0 m from the mirror but climb onto a stool, how high must the stool be to allow you to see your knees in the mirror? Assume that your knees are 1.2 m below your eyes.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics103ps

Chapter 26 Geometrical Optics Q.104GP
IP Apparent Size of Floats in a Tennoinetro Lentos
The Galileo thermometer, or Termometro Lentos (slow thermometer in Italian), consists of a vertical, cylindrical flask containing a fluid and several glass floats of different color. The floats all have the same dimensions, but they appear to differ in size depending on their location within the cylinder. (a) Does a float near the front surface of the cylinder (the surface closest to you) appear to be larger or smaller than a float near the back surface? (b) Figure shows a ray diagram for a float near the front surface of the cylinder. Draw a ray diagram for a float at the center of the cylinder, and show that the change in apparent size agrees with your answer to part (a).
Figure
Solution:
(a)
When the float is near to the front surface then the angle of incidence more then the angle of divergence is more. Then the point where the light rays are appearing to be diverging is very nearer. So the float near the front surface of the cylinder appears to be smaller than the float near the back surface.
mastering-physics-solutions-chapter-26-geometrical-optics104p
mastering-physics-solutions-chapter-26-geometrical-optics104ps

Chapter 26 Geometrical Optics Q.105GP
(a) Find the two locations where an object can be placed in front of a concave mirror witlva radius of curvature of 39 cm such that its image is twice its size. (b) In each of these cases, state whether the image is real or virtual upright or inverted.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics105ps
mastering-physics-solutions-chapter-26-geometrical-optics105ps1
mastering-physics-solutions-chapter-26-geometrical-optics105ps2

Chapter 26 Geometrical Optics Q.106GP
A convex mirror with a focal length of −85 cm is used to give a truck driver a view behind the vehicle. (a) If a person who is 1.7m tall stands 2.2 m from the mirror, where is the person’s image located? (b) Is the image upright or inverted? (c) What is the size of the image?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics106ps
mastering-physics-solutions-chapter-26-geometrical-optics106ps1
mastering-physics-solutions-chapter-26-geometrical-optics106ps2

Chapter 26 Geometrical Optics Q.107GP
IP The three laser beams shown in Figure meet at a point at the back of a solid, transparent sphere. (a) What is the index of refraction of the sphere? (b) Ts there a finite index of refraction that will make the three beams come to a focus at the center of the sphere? If your answer is yes, give the required index of refraction; if your answer is no, explain why not.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics107p
mastering-physics-solutions-chapter-26-geometrical-optics107ps
mastering-physics-solutions-chapter-26-geometrical-optics107ps1

Chapter 26 Geometrical Optics Q.108GP
The speed of light in substance A is x times greater than the speed of light in substance B. Find the ratio nA/nB interms of x.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics108ps

Chapter 26 Geometrical Optics Q.109GP
IP A film of oil, with an index of refraction of 1.48 and a thickness of 1.50 cm, floats on a pool of water, as shown in Figure. Abeam of light is incident on the oil atan angle of 60.0° to the vertical. (a) Find the angle θ the light beam makes with the vertical as it travels through the water. (b) How does your answer to part (a) depend on the thickness of the oil film? Explain.
mastering-physics-solutions-chapter-26-geometrical-optics109p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics109ps
b)
The answer to the part (a) does not depends upon the thick ness of the film. It depends only upon the original angle of incidence and the refractive index of water and air.

Chapter 26 Geometrical Optics Q.110GP
IP Consider the physical system shown in Figure. For this problem we assume that the angle of incidence at the air-oil interface can be varied from 0° to 90°. (a) What is the maximum possible value for 9, the angle of refraction in the water? (b) If an oil with a larger index of refraction is used, does your answer to part (a) increase or decrease? Explain.
mastering-physics-solutions-chapter-26-geometrical-optics110p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics110ps
mastering-physics-solutions-chapter-26-geometrical-optics110ps1

Chapter 26 Geometrical Optics Q.111GP
IP Consider the physical system shown in Figure, only this time let the direction of the light rays be reversed, (a) Find the angle of incidence θ at the water-oil interface such that the condition for total internal reflection at the oil-air surface is exactly satisfied. (b) If θ is decreased, is the reflection at the oil-air interface still total? Explain.
mastering-physics-solutions-chapter-26-geometrical-optics111p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics111ps
mastering-physics-solutions-chapter-26-geometrical-optics111ps1

Chapter 26 Geometrical Optics Q.112GP
Figure shows a ray of light entering one end of an optical fiber at an angle of incidence θ1 = 50.0°. The index of refraction of the fiber is 1.62. (a) Fine the angle θ the ray makes with the normal when it reaches the curved surface of the fiber. (b) Show that the internal reflection from the curved surface is total.
mastering-physics-solutions-chapter-26-geometrical-optics112p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics112ps
mastering-physics-solutions-chapter-26-geometrical-optics112ps1
mastering-physics-solutions-chapter-26-geometrical-optics112ps2

Chapter 26 Geometrical Optics Q.113GP
Suppose the person’s eyes in Figure are 1.6 m above the ground and that the small plane mirror can be moved up or down. (a) Find the height of the bottom of the mirror such that the lowest point the person can see on the building is 19.6 m above the ground. (b) With the mirror held at the height found inpart (a), what is the highest point on the building the person can see?
mastering-physics-solutions-chapter-26-geometrical-optics113p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics113ps

Chapter 26 Geometrical Optics Q.114GP
An arrow 2.00 cm long is located 75.0 cm from a lens that has a focal length ƒ = 30.0 cm. (a) If the arrow is perpendicular to the principal axis of the lens, as in Figure (a), what is its lateral magnification, defined as hi/ho? (b) Suppose, instead, that the arrow lies along the principal axis, extending from 74.0 cm to 76.0 cm from the lens, as indicated in Figure (b). What is the longitudinalmagnification of the arrow, defined as Li/Lo(Hint: Use the thin-lens equation to locate the image of each end of the arrow.)
mastering-physics-solutions-chapter-26-geometrical-optics114p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics114ps
mastering-physics-solutions-chapter-26-geometrical-optics114ps1
mastering-physics-solutions-chapter-26-geometrical-optics114ps2

Chapter 26 Geometrical Optics Q.115GP
Repeat Problem, this time for a diverging lens with a focal length f = −30.0 cm.
Problem
114. An arrow 2.00 cm long is located 75.0 cm from a lens that has a focal length ƒ = 30.0 cm. (a) If the arrow is perpendicular to the principal axis of the lens, as in Figure (a), what is its lateral magnification, defined as hi/ho? (b) Suppose, instead, that the arrow lies along the principal axis, extending from 74.0 cm to 76.0 cm from the lens, as indicated in Figure (b). What is the longitudinalmagnification of the arrow, defined as Li/Lo(Hint: Use the thin-lens equation to locate the image of each end of the arrow.)
mastering-physics-solutions-chapter-26-geometrical-optics115p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics115ps
mastering-physics-solutions-chapter-26-geometrical-optics115ps1
mastering-physics-solutions-chapter-26-geometrical-optics115ps2

Chapter 26 Geometrical Optics Q.116GP
Aconvex lens with f1 = 20.0cm is mounted 40.0 cm to the left of a concave lens. When an object is placed 30.0 cm to the left of the convex lens, a real image is formed 60.0 cm to the right of the concave lens. What is the focal length f2 of the concave lens?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics116ps
mastering-physics-solutions-chapter-26-geometrical-optics116ps1
mastering-physics-solutions-chapter-26-geometrical-optics116ps2

Chapter 26 Geometrical Optics Q.117GP
Two thin lenses, with focal lengths f1 and ƒ2, are placed in contact. What is the effective focal length of the double lens?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics117ps
mastering-physics-solutions-chapter-26-geometrical-optics117ps1

Chapter 26 Geometrical Optics Q.118GP
When an object is placed a distance do in front of acurved mirror, the resulting image has a magnification m. Find an expression for the focal length of the mirror,f, interms of do and m.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics118ps

Chapter 26 Geometrical Optics Q.119GP
A Slab of Glass Give a symbolic expression for the sideways displacement d of a light ray passing through the slab of glass shown in Figure. The thickness of the glass is t, its index of refraction is n, and the angle of incidence is θ.
mastering-physics-solutions-chapter-26-geometrical-optics119p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics119ps
mastering-physics-solutions-chapter-26-geometrical-optics119ps1
mastering-physics-solutions-chapter-26-geometrical-optics119ps2

Chapter 26 Geometrical Optics Q.120GP
mastering-physics-solutions-chapter-26-geometrical-optics120p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics120ps

Chapter 26 Geometrical Optics Q.121GP
Least Time A beam of light propagates from point A in medium ’1 to point B in medium 2, as shown in Figure. The index of refraction is different in these two media; therefore, the light follows a refracted path that obeys Snell’s law. (a) Calculate the time required for light to travel from. A to B along the refracted path. (b) Compare the time found in part (a) with the time it takes for light to tra vel from A to B along a straight-line path. (Note that the time on the straight-line path is longer than the time on the refracted path. In general, the shortest time between two points in different media is along the path given by Snell’s law.)
mastering-physics-solutions-chapter-26-geometrical-optics121p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics121ps
mastering-physics-solutions-chapter-26-geometrical-optics121ps1
mastering-physics-solutions-chapter-26-geometrical-optics121ps2

Chapter 26 Geometrical Optics Q.122GP
The ray of light shown in Figure passes from medium 1. to medium 2 to medium 3. The index of refraction in medium 1 is nY, in medium 2 it is n2> n1, and in medium 3 it is n3 > n2 Show that medium 2 can be ignored when calculating the angle of refraction in medium 3; that is, show that n1 sin θ1 = n3 sin 03.
mastering-physics-solutions-chapter-26-geometrical-optics122p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics122ps

Chapter 26 Geometrical Optics Q.123GP
IP A beam of light enters the sloping side of a 45°−90°−45° glass prism with an index of refraction n = 1.66. The situation is similar to that shown in Figure, except tha t the angle of incidence of the incoming beam can be varied. (a) Find die angle of incidence for which the reflection on the vertical side of the prism exactly satisfies the condition for total internal reflection. (b) If the angle of incidence is increased, is the reflection at the vertical surface still total? Explain. (c) What Is the minimum value of n such that a horizontal beam like that in Figure undergoes total internalreflection at the vertical side of the prism?
mastering-physics-solutions-chapter-26-geometrical-optics123p
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics123pss
mastering-physics-solutions-chapter-26-geometrical-optics123pss1
mastering-physics-solutions-chapter-26-geometrical-optics123pss2
mastering-physics-solutions-chapter-26-geometrical-optics123pss3
mastering-physics-solutions-chapter-26-geometrical-optics123pss4

Chapter 26 Geometrical Optics Q.124PP
A converging lens with a focal length in air of ƒ = +5.25 cm is made from ice. What is the focal length of this lens if it is immersed in benzene? (Refer to Table.)
A. −20.7 cm
B. −18.1cm
C. −12.8 cm
D. −11.2 cm
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics124ps

Chapter 26 Geometrical Optics Q.125PP
A diverging lens with ƒ = −12.5 cm is made from ice. What is the focal length of this lens if it is immersed in ethyl alcohol? (Refer to Table.)
A. 102 cm
B. 105 cm
C. 118 cm
D. 122 cm
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics125ps

Chapter 26 Geometrical Optics Q.126PP
Calculate the focal length of a lens in water, given that the index of refraction of the lens is nlens = 1.52 and its focal length in air is 25.0 cm. (Refer to Table.)
A. 57.8 cm
B. 66.0 cm
C. 91.0 cm
D. 104 cm
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics126ps

Chapter 26 Geometrical Optics Q.127PP
Suppose a lens is made from fused quartz (glass), and that its focal length in air is −7.75 cm. What is the focal length of this lens if it is immersed in benzene? (Refer to Table.)
A. –130 cm
B. 134 cm
C. 141 cm
D. −145 cm
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics127ps

Chapter 26 Geometrical Optics Q.128IP
Referring to Example Suppose the radius of curvature of the mirror is 5.0 cm. (a) Find the object distance that gives an upright knage with a magnification of 1.5. (b) Find the object distance that gives an inverted image with a magnification of −1.5.
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics128ps
mastering-physics-solutions-chapter-26-geometrical-optics128ps1

Chapter 26 Geometrical Optics Q.129IP
IP Referring to Example An object is 4.5 cm in front of the mirror. (a) What radius of curvature must the mirror have if the image is to be 2.2 cm in front of the mirror? (b) What is the magnification of the image? (c) If the object is moved closer to the mirror, does the magnification of the image increase in magnitude, decrease in magnitude, or stay the same?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics129ps
mastering-physics-solutions-chapter-26-geometrical-optics129ps1

Chapter 26 Geometrical Optics Q.130IP
Referring to Example (a) What object distance is required to give an image with a magnification of +2.0? Assume that the focal length of the lens is +5.0cm. (b) What is the location of the image in this case?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics130ps
mastering-physics-solutions-chapter-26-geometrical-optics130ps1

Chapter 26 Geometrical Optics Q.131IP
IP Referring to Example Suppose the convex lens is replaced with a concave lens with a focal length of −5.0 cm. (a) Where must the objectbe placed to form an image with a mag-nification of 0.50? (b) What is the location of the image in this case? (c) If we now move the object closer to the lens, does the magnification of the image increase, decrease, or stay the same?
Solution:
mastering-physics-solutions-chapter-26-geometrical-optics131ps

Mastering Physics Solutions Chapter 25 Electromagnetic Waves

Mastering Physics Solutions Chapter 25 Electromagnetic Waves

Mastering Physics Solutions

Chapter 25 Electromagnetic Waves Q.1CQ
Explain why the “invisible man” would be unable to see
Solution:
672-25-1CQ RID: 514 I 04/03/2013
RIDI: 2791 31/08/2015
An invisible person allows light to pass through himself/herself uninterruptedly. That makes himself/herself invisible, if incase. there happens any kind of phenomena of light that makes light little bit hindered or unease to pass through. human brain detects that, and tells that there is some medium that hinders light. This way the existence of the invisible person is exposed. and he/she no longer stays invisible. Now for an invisible person to see, the incident light has to be absorbed by the retina, though very small. This absorption can readily be detected if not by naked eye. by sensitive instrument. So the invisibility of the person is exposed. In order for a person to maintain invisibility, the eye of himlher absorbs zero amount of light, which makes himlher blind, completely unable to see.

Chapter 25 Electromagnetic Waves Q.1P
If the electric field in an electromagnetic wave is increasing in magnitude at a particular time, is the magnitude of the magnetic field at the same time increasing or decreasing? Explain.
Solution:
lithe electric field in an electromagnetic wave increases in magnitude at a particular time then the magnitude of the magnetic field at the same time increases because the electric and magnetic fields in an electromagnetic wave vary in phase with one another According to the equation E = cB we can say that if Electric field increases then magnetic field also increases

Chapter 25 Electromagnetic Waves Q.2CQ
The magnitude of the Dopplcr effect tells how rapidly a weather system is moving. What determines whether the system is approaching or receding?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves2cqs

Chapter 25 Electromagnetic Waves Q.2P
The electric field of an electromagnetic wave points in the positive y direction. At the same time, the magnetic field of this wave points in the positive z direction. In what direction is the wave traveling?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves2ps

Chapter 25 Electromagnetic Waves Q.3CQ
Explain why radiation pressure is more significant on a grain of dust in interplanetary space when the grain is very small.
Solution:
As the grain of dust becomes smaller, its volume and mass decreases more rapidly than does its area. The radiation pressure which acts on the surface of the grain, becomes increasingly important as the size of the grain is decreased. So, the gravity which acts on the mass of the grain becomes less important.

Chapter 25 Electromagnetic Waves Q.3P
An electric charge on the x axis oscillates sinusoidally about the origin. A distant observer is located at a point on the +y axis. (a) In what direction will the electric field oscillate at the observer’s location? (b) In what direction will the magnetic field oscillate at the observer’s location? (c) In what direction will the electromagnetic wave propagate at the observer’s location?
Solution:
(a)
The electric field propagates parallel to the oscillations.
In this case, electric charge on the x axis oscillates simultaneously about the origin.
Therefore, electric field oscillates in x direction.
(b)
The direction of electric field and magnetic field perpendicular to each other, they are also perpendicular to the direction of propagation.
In this case, electric field direction is x and direction of propagation is +y. Thus, the only direction perpendicular to x and y directions is the z direction.
Therefore, the magnetic field oscillates in the z direction.
(c)
The observer is located at a point on the +y axis. From the rectilinear propagation of light, light travels in the straight line. Thus, the electromagnetic wave must traveled in the +y direction because of it have traveled from origin to a point on the +y axis.
Therefore, the electromagnetic wave propagates in the +y direction.

Chapter 25 Electromagnetic Waves Q.4CQ
While wearing your Polaroid sunglasses at the beach, you notice that they reduce the glare from the water better when you are sitting upright than when you are lying on your side. Explain.
Solution:
Light reflected from a horizontal surface has a polarization in the horizontal direction. It follows that when you sit upright , with the transmission axis of your glasses in the vertical direction , they will block most of the reflected light when you lie on your side , however , the transmission axis is horizontal. This allows most of the reflected light to enter your eyes.

Chapter 25 Electromagnetic Waves Q.4P
An electric charge on the z axis oscillates sinusoidally about the origin. A distant observer is located at a point on the +y axis. (a) In what direction will the electric field oscillate at the observer’s location? (b) In what direction will the magnetic field oscillate at the observer’s location? (c) In what direction will the electromagnetic wave propagate at the observer’s location?
Solution:
Electric charge is oscillating on z-axis about origin and location of distant observer is
+y-axis. Electric field and magnetic field are mutually perpendicular to each other in electromagnetic wave.
The oscillation of electromagnetic wave is shown in figure.
mastering-physics-solutions-chapter-25-electromagnetic-waves4ps
(a)
Since electric field propagates in the direction of oscillation of charge, hence the direction oscillations of electric field will be z-axis.
(b)
In electromagnetic wave the electric and magnetic field are mutually perpendicular to each other in all condition. And magnetic field is also perpendicular to the direction of propagation. So the magnetic field oscillates in x-axis direction.
(c)
The electromagnetic wave propagates in +y direction, in straight line from origin to observer’s location and also it propagates +y direction after it reaches observer’s location.

Chapter 25 Electromagnetic Waves Q.5CQ
You want to check the time while wearing your Polaroid sunglasses. If you hold your forearm horizontally, you can read the time easily. If you hold your forearm vertically, however, so that you are looking at your watch sideways, you notice that the display is black. Explain.
Solution:
This is because, the light coming from the display is linearly polarized. If the polarization direction of the display and sunglasses align you can read the time. If these directions are at 90 to one another no light will pass through the sunglasses and the display will appear black.
When the direction of the linearly polarized light ray is parallel to the axis of the Polaroid then we can see the light .If it is perpendicular to the axis of the Polaroid we can not see light ray.
Here the watch is a LCD display watch that means the light coming from it will be a linearly polarized light .Now if our four arm is horizontal then the light coming from the watch is parallel to the Polaroid sun glass. Therefore we can see the display of the watch. Now if our four arms are vertical then the light coming from the watch is perpendicular to the Polaroid sun glass. Therefore we can not see the display of the watch .Therefore now the display becomes black.

Chapter 25 Electromagnetic Waves Q.5P
Give the direction (N, S, E, W, up, or down) of the missing quantity for each of the four electromagnetic waves listed in Table 25-1.
mastering-physics-solutions-chapter-25-electromagnetic-waves5p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves5ps

Chapter 25 Electromagnetic Waves Q.6CQ
Polarization and the Ground Spider The ground spider Drassodes cupreus, like many spiders, has several pairs of eyes. It has been discovered that one of these pairs of eyes acts as a set of polarization filters, with one eye’s polarization direction oriented at 90° to the other eye’s polarization direction. In addition,
experiments show that the spider uses these eyes to aid in navigating to and from its burrow. Explain how such eyes might aid navigation.
Solution:
We know that the amount of light received by the polarizer is depend on the orientation of the polarizer with respect to the axis of the source .Here light from the sky is polarized perpendicular to the sun .Thus the amount of light received by the two polarizing eyes depend on the orientation of spider with respect to the sun. By monitoring the amount of light received by each eye, the spider can navigate to and from its burrow.

Chapter 25 Electromagnetic Waves Q.6P
Give the direction (±x, ±y, ±z) of the missing quantity for each of the four electromagnetic waves listed in Table 25-2.
mastering-physics-solutions-chapter-25-electromagnetic-waves6p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves6ps
mastering-physics-solutions-chapter-25-electromagnetic-waves6ps1

Chapter 25 Electromagnetic Waves Q.7CQ
The electromagnetic waves we pick up on our radios are typically polarized. In contrast, the indoor light we see every day is typically unpolarized. Explain.
Solution:
The indoor light we see generally from the sun and from light bulbs is unpolarized light because the atoms which are emitting the light can have any orientation relative to one another. Hence, even if individual atoms emit polarized light, the net result from a group of atoms is light with no preferred direction of polarization.
The radio stations generate their electromagnetic waves with the large vertical antennas. As a result, these waves are polarized in the vertical direction.

Chapter 25 Electromagnetic Waves Q.7P
mastering-physics-solutions-chapter-25-electromagnetic-waves7p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves7ps
mastering-physics-solutions-chapter-25-electromagnetic-waves7ps1

Chapter 25 Electromagnetic Waves Q.8CQ
You are given a sheet of Polaroid material. Describe how to determine the direction of its transmission axis if none is indicated on the sheet.
Solution:
We can determine the direction of its transmission axis by viewing a linearly polarized light whose direction of polarization is known.
For example, the light reflected from the horizontal has a polarization in the horizontal direction. If we view this light through the sheet and rotate it until we get the maximum intensity, at maximum intensity the transmission axis is horizontal.

Chapter 25 Electromagnetic Waves Q.8P
mastering-physics-solutions-chapter-25-electromagnetic-waves8p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves8pss
mastering-physics-solutions-chapter-25-electromagnetic-waves8pss1

Chapter 25 Electromagnetic Waves Q.9CQ
Can sound waves be polarized? Explain.
Solution:
Sound waves are longitudinal waves. So, they cannot be polarized. In case of longitudinal propagation of the wave the particles can moves in only one direction i.e. back and forth along the direction of propagation. For a wave to be polarized there must be two directions of propagations. Hence the sound waves are not polarized.

Chapter 25 Electromagnetic Waves Q.9P
Three electromagnetic waves have electric and magnetic fields pointing in the directions shown in Figure 25-23. For each of the three cases, state whether the wave propagates in the +x, −x, +y, −y, +z, or −z direction.
mastering-physics-solutions-chapter-25-electromagnetic-waves9p
Solution:
From the figure (1) using right hand rule
The propagation of the wave is along x-direction because the electric field is along the y-direction and magnetic field is along z-direction.
From the figure (2) using right hand rule
The propagation of the wave is along the +z-direction because the electric field is along the x-direction and magnetic field is along y-direction.
From the figure (3) using the right hand rule
The propagation of the wave is along the -x – direction because the electric field is along the z-direction and magnetic field is along y-direction.

Chapter 25 Electromagnetic Waves Q.10CQ
At a garage sale you find a pair of “Polaroid” sunglasses priced to sell. You are not sure, however, if the glasses are truly Polaroid, or if they simply have tinted fenses. How can you tell which is the case? Explain.
Solution:
We observe the light from the pair of Polaroid sunglasses and rotate one of the glass with respective to the other. If the intensity of the light is varying then they are Polaroid glasses. If they are tinted lenses then the intensity of the light does not change.

Chapter 25 Electromagnetic Waves Q.10P
The light-year (ly) is a unit of distance commonly used in astronomy. It is defined as the distance traveled by light in a vacuum in one year. (a) Express 1 ly in km. (b) Express the speed of light, c, in units of ly per year. (c) Express the speed of light in feet per nanosecond.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves10ps

Chapter 25 Electromagnetic Waves Q.11CQ
3-D Movies Modern-day 3-D movies are produced by projecting two different images onto the screen, with polarization directions that are at 90° relative to one another. Viewers must wear headsets with polarizing filters to experience the 3-D effect. Explain how this works.
Solution:
The two projected images give a view of a scene from slightly different angles, just as our eyes view a three dimensional object from different angles. Without headsets, the screen is confusing superposition of the two images. With the headsets our right eye sees one view of the scene and our left eye sees the other view. As these views are combined in our brain, then we can experience a realistic three dimensional effect.

Chapter 25 Electromagnetic Waves Q.11P
Alpha Centauri, the closest star to the sun, is 4.3 ly away. How far is this in meters?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves11ps

Chapter 25 Electromagnetic Waves Q.12P
Mars Rover When the Mars rover was deployed on the surface of Mars in July 1997, radio signals took about 12 min to travel from Earth to the rover. How far was Mars from Earth at that time?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves12ps

Chapter 25 Electromagnetic Waves Q.13P
A distant star is traveling directly away from Earth with a speed of 37,500 km/s. By what factor are the wavelengths in this star’s spectrum changed?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves13ps

Chapter 25 Electromagnetic Waves Q.14P
A distant star is traveling directly toward Earth with a speed of 37,500 km/s. (a) When the wavelengths in this star’s spectrum are measured on Earth, are they greater or less than the wavelengths we would find if the star were at rest relative to us? Explain. (b) By what fraction are the wavelengths in this star’s spectrum shifted?
Solution:
Given data:
Velocity of the star u=37,500 km/s
(a) ) When the star is moving towards the Earth, the frequencies of the electromagnetic wave increases. But, since the wave length are inversely proportional to the frequencies. So the measured wavelengths are less than what they would be if the stars are at rest relative to us.
mastering-physics-solutions-chapter-25-electromagnetic-waves14ps

Chapter 25 Electromagnetic Waves Q.15P
The frequency of light reaching Earth from a particular galaxy is 12% lower than the frequency the light had when it was emitted. (a) Is this galaxy moving toward or away from Earth? Explain. (b) What is the speed of this galaxy relative to the Earth? Give your answer as a fraction of the speed of light.
Solution:
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Chapter 25 Electromagnetic Waves Q.16P
Measuring the Speed of Light Galileo attempted to measure the speed of light by measuring the time elapsed between his opening a lantern and his seeing the light return from his assistant’s lantern. The experiment is illustrated in Figure 25–24.
What distance, d, must separate Galileo and his assistant in order for the human reaction time, ∆t = 0.2 s, to introduce no more than a 15% error in the speed of light?
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Solution:
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Chapter 25 Electromagnetic Waves Q.17P
Measuring the Speed of Light: Michelson In 1926, Albert Michelson measured the speed of light with a technique similar to that used by Fizeau. Michelson used an eight-sided mirror rotating at 528 rev/s in place of the toothed wheel, as illustrated in Figure 25-25. The distance from the rotating mirror to a distant reflector was 35.5 km. If the light completed the 71.0-km round trip in the time it took the mirror to complete one-eighth of a revolution, what is the speed of light?
mastering-physics-solutions-chapter-25-electromagnetic-waves17p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves17ps

Chapter 25 Electromagnetic Waves Q.18P
Communicating with the Voyager Spacecraft When the Voyager I and Voyager II spacecraft were exploring the outer planets, NASA flight controllers had to plan the crafts’ moves well in advance. How many seconds elapse between the time a command is sent from Earth and the time the command is received by Voyager at Neptune? Assume the distance from. Earth to Neptune is 4.5 × 1012 m.
Solution:
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Chapter 25 Electromagnetic Waves Q.19P
A father and his daughter are interested in the same baseball game. The father sits next to his radio at home and listens to the game; his daughter attends the game and sits in the outfield bleachers. In the bottom of the ninth inning a home run is hit. If the father’s radio is 132 km from the radio station, and the daughter is 115 m from home plate, who hears the home run first? (Assume that there is no time delay between the baseball being hit and its sound being broadcast by the radio station. In addition, let the speed of sound in the stadium be 343 m/s.)
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves19ps

Chapter 25 Electromagnetic Waves Q.20P
(a) How fast would a motorist have to be traveling for a yellow (λ = 590 nm) traffic light to appear green (λ = 550 nm) because of the Doppler shift? (b) Should the motoristbe traveling toward or away from the traffic light to see this effect? Explain.
Solution:
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Chapter 25 Electromagnetic Waves Q.21P
Most of the galaxies in the universe are observed to be moving away from Earth. Suppose a particular galaxy emits orange light with a frequency of 5.000 × 1014 Hz. If the galaxy is receding from Earth with a speed of 3325 km/s, what is the frequency of the light when it reaches Earth?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves21ps

Chapter 25 Electromagnetic Waves Q.22P
Two starships, the Enterprise and the Constitution, are approaching each other head-on from a great distance. The separation between them is decreasing at a rate of 782.5 km/s. The Enterprise sends a laser signal toward the Constitution. If the Constitution observes a wavelength λ = 670.3 nm, what wavelength was emitted by the Enterprise?
Solution:
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Chapter 25 Electromagnetic Waves Q.23P
Baseball scouts often use a radar gun to measure the speed of a pitch. One particular model of radar gun emits a microwave signal at a frequency of 10.525 GHz. What will be the increase in frequency if these waves are reflected from a 90.0-mi/h fastball headed straight toward the gun? (Note: 1 mi/h = 0.447 m/s)
Solution:
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Chapter 25 Electromagnetic Waves Q.24P
A state highway patrol car radar unit uses a frequency of 8.00 × 109 Hz. What frequency difference will the unit detect from a car receding at a speed of 44.5 m/s from a stationary patrol car?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves24ps

Chapter 25 Electromagnetic Waves Q.25P
Consider a spiral galaxy that is moving directly away from Earth with a speed V = 3.600 × 105 m/s at its center, as shown in Figure 25-26. The galaxy is also rotating about its center, so that points in its spiral arms are moving with a speed v = 6.400 × 105 m/s relative to the center. If light with a frequency of 8.230 × 1014 Hz is emitted in both arms of the galaxy, what frequency is detected by astronomers observing the arm that is moving (a) toward and (b) away from Earth? (Measurements of this type are used to map out the speed of various regions in distant, rotating galaxies.)
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Solution:
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Chapter 25 Electromagnetic Waves Q.26P
A highway patrolman sends a 24.150-GHz radar beam toward a speeding car. The reflected wave is lower in frequency by 4.04 kHz. (a) Is the car moving toward or away from the radar gun? Explain. (b) What is the speed of the car? [Hint:For small values of x, the following approximation may be used: (1 + x)2≈ 1 + 2x.]
Solution:
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Chapter 25 Electromagnetic Waves Q.27P
Dental X-rays X-rays produced in the dentist’s office typically have a wavelength of 0.30 nm. What is the frequency of these rays?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves27ps

Chapter 25 Electromagnetic Waves Q.28P
Find the frequency of blue light with a wavelength of 460 nm.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves28ps

Chapter 25 Electromagnetic Waves Q.29P
Yellow light has a wavelength λ = 590 nm. How many of these waves would span the 1.0-mm thickness of a dime?
Solution:
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Chapter 25 Electromagnetic Waves Q.30P
How many red wavelengths (λ = 705 nm) tall are you?
Solution:
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Chapter 25 Electromagnetic Waves Q.31P
A cell phone transmite at a frequency of 1.75 × 108 Hz. What is the wavelength of the electromagnetic wave used by this phone?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves31ps

Chapter 25 Electromagnetic Waves Q.32P
Human Radiation Under normal conditions, humans radiate electromagnetic waves with a wavelength of about 9.0 microns. (a) What is the frequency of these waves? (b) To what portion of the electromagnetic spectrum do these waves belong?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves32ps

Chapter 25 Electromagnetic Waves Q.33P
UV Radiation. Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm. and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 × 1014 Hz belong?
Solution:
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Chapter 25 Electromagnetic Waves Q.34P
Communicating with a Submarine Normal radiofrequency waves cannot penetrate more than a few meters below the surface of the ocean. One method of communicating with submerged submarines uses very low frequency (VLF) radio waves. What is the wavelength (in air) of a 10.0-kHz VLF radio wave?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves34ps

Chapter 25 Electromagnetic Waves Q.35P
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Solution:
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Chapter 25 Electromagnetic Waves Q.36P
(a) Whichcolor of light has the higher frequency, red or violet? (b) Calculate the frequency of blue light with a wavelength of 470 nm, and red light with a wavelength of 680 nm.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves36ps
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Chapter 25 Electromagnetic Waves Q.37P
ULF (ultra low frequency) electromagnetic waves, produced in the depths of outer space, have been observed with wavelengths inexcess of 29 million kilometers. What is the period of such a wave?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves37ps

Chapter 25 Electromagnetic Waves Q.38P
A television is tuned to a station broadcasting at a frequency of 6.60 × 107 Hz. For best reception, the rabbit-ear antenna used by the TV should be adjusted to have a tip-to-tip length equal to half a wavelength of the broadcast signal. Find the optimum length of the antenna.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves38ps

Chapter 25 Electromagnetic Waves Q.39P
An AM radio station’s antenna is constructed to be λ/4 tall, where λ is the wavelength of the radio waves. How tall should the antenna be for a station broadcasting at a frequency of 810 kHz?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves39ps

Chapter 25 Electromagnetic Waves Q.40P
As you drive by an AM radio station, you notice a sign saying that its antenna is 112 m high. If this height represents one quarter-wavelength of its signal, what is the frequency of the station?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves40ps

Chapter 25 Electromagnetic Waves Q.41P
Find the difference in wavelength (λ1 − λ2) for each of the following pairs of radio waves: (a) f1 = 50 kHz and f2 = 52 kHz, (b) f1 = 500 kHz and f2 = 502 kHz.
Solution:
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Chapter 25 Electromagnetic Waves Q.42P
Find the difference in frequency (f1 − f2) for each of the following pairs of radio waves: (a) λ1 = 300.0 m and λ2 = 300.5 m, (b) λ1 = 30.0 m and λ2 = 30.5 m.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves42ps
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Chapter 25 Electromagnetic Waves Q.43P
If the rms value of the electric field in an electromagnetic wave is doubled, (a) by what factor does the rms value of the magnetic field change? (b) By what factor does the average intensity of the wave change?
Solution:
a) If the rms value of the electric field doubled then the rms value of the magnetic field also get doubled since the electric and magnetic fields in an electromagnetic field are proportional to one another according to the relation.E = cB
b) The intensity of an electromagnetic wave depends on the square of the electric and magnetic fields. As a result, doubling the fields quadruples the intensity of the wave.

Chapter 25 Electromagnetic Waves Q.44P
The radiation pressure exerted by beam of light 1 is half the radiation pressure of beam of light 2. If the rms electric field of beam 1 has the value E0, what is the rms electric field in beam 2?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves44ps

Chapter 25 Electromagnetic Waves Q.45P
The maximum magnitude of the electric field in an electromagnetic wave is 0.0400 V/m. What is the maximum magnitude of the magnetic field in this wave?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves45ps

Chapter 25 Electromagnetic Waves Q.46P
What is the rms value of the electric field in a sinusoidal electromagnetic wave that has a maximum electric field of 88 V/m?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves46ps

Chapter 25 Electromagnetic Waves Q.47P
The magnetic field in an electromagnetic wave has a peak value given by B = 3.7 μ T. For this wave, find (a) the peak electric field strength, (b) the peak intensity, and (c) the average intensity.
Solution:
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Chapter 25 Electromagnetic Waves Q.48P
What is the maximum value of the electric field in an electromagnetic wave whose maximum intensity is 5.00 W/m2?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves48ps

Chapter 25 Electromagnetic Waves Q.49P
What is the maximum value of the electric field in an ele tromagnetic wave whose average intensityis 5.00 W/m2?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves49ps

Chapter 25 Electromagnetic Waves Q.50P
Electromagnetic wave 1 has a maximum electric field of E0 = 52 V/m, and electromagnetic wave 2 has a maximum magnetic field of B0 = 1.5 μ T.(a) Which wave has the greater intensity? (b) Calculate the intensity of each wave.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves50ps
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Chapter 25 Electromagnetic Waves Q.51P
A 65-kW radio station broadcasts its signal uniformly in all directions. (a) What is the average intensity of its signal at a distance of 250 m from the antenna? (b) What is the average intensity of its signal at a distance of 2500 m from the antenna?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves51ps

Chapter 25 Electromagnetic Waves Q.52P
At what distance will a 45-W lightbulb have the same apparent brightness as a 120-W bulb viewed from a distance of 25 m? (Assume that both bulbs convert electrical power to light with the same efficiency, and radiate light uniformly in all directions.)
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves52ps

Chapter 25 Electromagnetic Waves Q.53P
What is the ratio of the sunlight intensity reaching Pluto compared with the sunlight intensity reaching Earth? (On average, Pluto is 39 times as far from the Sun as is Earth.)
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves53ps

Chapter 25 Electromagnetic Waves Q.54P
In the following, assume that lightbulbs radiate uniformly in all directions and that 5.0% of their power is converted to light. (a) Find the average intensity of light at a point 2.0 m from a 120-W red lightbulb (λ = 710 nm). (b) Is the average intensity 2.0 m from a 120-W blue lightbulb (λ = 480 nm) greater than, less than, or the same as the intensity found in part (a)? Explain. (c) Calculate the average intensity for part (b).
Solution:
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Chapter 25 Electromagnetic Waves Q.55P
A 5.0-mW laser produces a narrow beam of light. How much energy is contained in a 1.0-m length of its beam?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves55ps

Chapter 25 Electromagnetic Waves Q.56P
What length of a 5.0-mW laser’s beam will contain 9.5 mJ of energy?
Solution:
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Chapter 25 Electromagnetic Waves Q.57P
Sunlight Intensity After filtering through the atmosphere, the Sun’s radiation illuminates Earth’s surface with an average intensity of 1.0 kW/m2. Assuming this radiation strikes the 15-m × 45-m black, flat roof of a building at normal incidence, calculate the average force the radiation exerts on the roof.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves57ps

Chapter 25 Electromagnetic Waves Q.58P
(a) Find the electric and magnetic field amplitudes inan electromagnetic wave that has an average energy density of 1.0 J/m3. (b) By what factor must the field amplitudes be increased if the average energy density is to be doubled to 2.0 J/m3?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves58ps
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Chapter 25 Electromagnetic Waves Q.59P
Lasers for Fusion Two of the most powerful lasers in the world are used in nuclear fusion experiments. The NOVA laser produces 40.0 kJ of energy in a pulse that lasts 2.50 ns, and the N1F laser (under construction) will produce a 10.0-ns pulse with 3.00 MJ of energy. (a) Which laser produces more energy in each pulse? (b) Which laser produces the greater average power during each pulse? (c) If the beam diameters are the same, which laser produces the greater average intensity?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves59ps
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Chapter 25 Electromagnetic Waves Q.60P
You are standing 2.5 m from a 150-W lightbulb. (a) If the pupil of your eye is a circle 5.0 mm in diameter, how much energy enters your eye per second? (Assume that 5.0% of the lightbulb’s power is converted to light.) (b) Repeat part (a) for the case of a 1.0-mm-diameter laser beam with a power of 0.50 mW.
Solution:
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Chapter 25 Electromagnetic Waves Q.61P
Laser Safety A 0.75-mW laser emits a narrow beam of light that enters the eye, as shown in Figure 25−27. (a) How much energy is absorbed by the eye in 0.2 s? (b) The eye focuses this beam to a tiny spot on the retina, perhaps 5.0 μ m in diameter. What is the average intensity of light (in W/cm2) at this spot? (c) Damage to the retina can occur if the average intensity of light exceeds 1.0 × 10−2 W/cm2. By what factor has the intensity of this laser beam exceeded the safe value?
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Solution:
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Chapter 25 Electromagnetic Waves Q.62P
Find the rms electric and magnetic fields at a point 2.50 m from a lightbulb that radiates 75.0 W of light uniformly in all directions.
Solution:
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Chapter 25 Electromagnetic Waves Q.63P
A 0.50-mW laser produces abeam of light with a diameter of 1.5 mm. (a) What is the average intensity of this beam? (b) At what distance does a 150-W lightbulb have the same average intensity as that found for the laser beam in part (a)? (Assume that 5.0% of the bulb’s power is converted to light.)
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves63ps

Chapter 25 Electromagnetic Waves Q.64P
A laser emits a cylindrical beam of light 2.4 mm in diameter. If the average power of the laser is 2.8 mW, what is the rms value of the electric field in the laser beam?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves64ps

Chapter 25 Electromagnetic Waves Q.65P
(a) If the laser in Problem 64 shines its light on a perfectly absorbing surface, how much energy does the surface receive in 12 s? (b) What is the radiation pressure exerted by the beam?
Solution:
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Chapter 25 Electromagnetic Waves Q.66P
Laser Surgery Each pulse produced by an argon-fluoride excimer laser used in PRK and LASIK ophthalmic surgery lasts only 10.0 ns but delivers an energy of 2.50 mJ. (a) What is the power produced during each pulse? (b) If the beam has a diameter of 0.850 mm, what is the average intensity of the beam during each pulse? (c) If the laser emits 55 pulses per second, what is the average power it generates?
Solution:
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Chapter 25 Electromagnetic Waves Q.67P
A pulsed laser produces brief bursts of light. One such laser emits pulses that carry 0.350 J of energy but last only 225 fs. (a) What is the average power during one of these pulses? (b) Assuming the energy is emitted in a cylindrical beam of light 2.00 mm in diameter, calculate the average intensity of this laser beam. (c) What is the rms electric field in this wave?
Solution:
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Chapter 25 Electromagnetic Waves Q.68P
Consider the two polarization experiments shown in Figure 25–28. (a) If the incident light is unpolar-ized, is the transmitted intensity in case A greater than, less than, or the same as the transmittedintensity in case B? (b) Choose the best explanation from among the following:
I. The transmitted intensify is the same in either case; the first polarizer lets through one-half the incident intensity, and the second polarizer is at an angle θ relative to the first.
II. Case A has a smaller transmitted intensity than case B because the first polarizer is at an angle θ relative to the incident beam.
III. Case B has a smaller transmitted intensity than case A because the direction of polarization is rotated by an angle θ in the clockwise direction in case B.
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Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves68ps
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Chapter 25 Electromagnetic Waves Q.69P
Consider the two polarization experiments shown in Figure 25–28. (a) If the incident light is polarized in the horizontal direction, is the transmitted intensity in case A greater than, less than, or the same as the transmitted intensity in case B? (b) Choose the best explanation from among the following:
I. The two cases havethe same transmitted intensity because the angle between the polarizers is θ in each case.
II. The transmitted intensity is greater in case B because all of the initial beam gets through the first polarizer.
III. The transmitted in tensi ty in case B is smaller than in case A; in fact, the transmitted intensity in case B is zero because the first polarizer is oriented vertically.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves69ps

Chapter 25 Electromagnetic Waves Q.70P
Suppose linearly polarized light is incident on the polarization experiments shown in Figure25–28. In what direction, relative to the vertical, must the incident light be polarized if the transmitted intensity is to be the same in both experiments? Explain.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves70ps

Chapter 25 Electromagnetic Waves Q.71P
An incident beam of light with an intensityI0 passes through a polarizing filter whose transmissionaxis is at an angle θ to the vertical. As the angle is changed from θ = 0to θ = 90°, the intensity as a function of angle is given by one of the curves in Figure 25–29. Give the color of the curve corresponding to an incident beam that is (a) unpolarized, (b) vertically polarized, and (c) horizontally polarized.
mastering-physics-solutions-chapter-25-electromagnetic-waves71p
Solution:
a) The color of the curve corresponding to an incident beam that is unpolarized is Green. Unpolarized light will be reduced in intensity by a factor of 2 regardless of the angle of the transmission axis.
b) The color of the curve corresponding to an incident beam that is vertically polarized is Red. At θ = 0 we have complete transmission, meaning that the light is vertically polarized.
c) The color of the curve corresponding to an incident beam that is horizontally polarized is Blue. At θ = 0 there is no transmission, which means the incident light is horizontally polarized.

Chapter 25 Electromagnetic Waves Q.72P
Vertically polarized light with an intensity of 0.55 W/m2 passes through a polarizer whose transmission axis is at an angle of 65.0° with the vertical. Whatis the intensity of the transmitted light?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves72ps

Chapter 25 Electromagnetic Waves Q.73P
A person riding in a boat observes that the sunlight reflected by the water is polarized parallel to the surface of the water. The person is wearing polarized sunglasses with the polarization axis vertical. If the wearer leans at an angle of 21.5° to the vertical, what fraction of the reflected light intensity will pass through the sunglasses?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves73ps

Chapter 25 Electromagnetic Waves Q.74P
Unpolarized light passes through two polarizers whose transmission axes are at an angle of 30.0° with respect to each other. What fraction of the incident intensity is transmitted through the polarizers?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves74ps

Chapter 25 Electromagnetic Waves Q.75P
In Problem, what should be the angle between the transmission axes of the polarizers if it is desired that one-tenth of the incident intensity be transmitted?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves75ps

Chapter 25 Electromagnetic Waves Q.76P
Unpolarized iight with intensity I0 falls on a polarizing filter whose transmission axis is vertical. The axis of a second polarizing filter makes an angle of θ with the vertical. Plot a graph that shows the intensity of light transmitted by the second filter (expressed as a fraction of I0) as a function of θ. Your graph should cover the range θ = 0° to θ = 360°.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves76ps

Chapter 25 Electromagnetic Waves Q.77P
A beam of vertically polarized light encounters two polarizing filters, as shown in Figure 25–30. (a) Rank the three cases, A, B, and C, in order of increasing transmitted intensity. Indicate ties where appropriate. (b) Calculate the transmitted intensity for each of the cases in Figure 25–30, assuming that the incident intensity is 37.0 W/m2. Verify that your numerical results agree with the rankings in part (a).
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Solution:
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mastering-physics-solutions-chapter-25-electromagnetic-waves77ps1

Chapter 25 Electromagnetic Waves Q.78P
This time assuming that the polarizers to the left in Figure 25–30 are at an angle of 22.5° to the vertical rather than 45°. The incident intensity is again 37.0 W/m2.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves78ps
mastering-physics-solutions-chapter-25-electromagnetic-waves78ps1

Chapter 25 Electromagnetic Waves Q.79P
BIO Optical Activity Optically active molecules have the property of rotating the direction of polarization of linearly polarized light. Many biologically important molecules have this property, some causing a counterclockwise rotation (negative rotation angle), others causing a clockwise rotation (positive rotation angle). For example, a 5.00 gram per 100 mL solution of l-leucine causes a rotation of −0.550°; the same concentration of d-glutamic acid causes a rotation of 0.620”. (a) If placed between crossed polarizers, which of these solutions transmits the greater intensity? Explain. (b) Find the transmitted intensity for each of these solutions when placed between crossed polarizers. The incident beam is unpolarized and has an intensity of 12.5 W/m2.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves79ps

Chapter 25 Electromagnetic Waves Q.80P
A helium-neon laser emits a beam of unpolarized light that passes through three Polaroid filters, as shown in Figure 25–31. The intensity of the laser beam is I0. (a) What is the intensity of the beam at point A? (b) What is the intensity of the beam at point B? (c) What is the intensity of the beam at point C? (d) If filter 2 is removed, what is the intensity of the beam at point C?
mastering-physics-solutions-chapter-25-electromagnetic-waves80p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves80ps
mastering-physics-solutions-chapter-25-electromagnetic-waves80ps1

Chapter 25 Electromagnetic Waves Q.81P
Referring to Figure 25–31, suppose that filter 3 is at a general angle θ with the vertical, rather than the angle 90°. (a) Find an expression for the transmitted intensity as a function of θ. (b) Plot your result from part (a), and determine the maximum transmitted intensity. (c) At what angle θ does maximum transmission occur?
mastering-physics-solutions-chapter-25-electromagnetic-waves81p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves81ps
mastering-physics-solutions-chapter-25-electromagnetic-waves81ps1
mastering-physics-solutions-chapter-25-electromagnetic-waves81ps2

Chapter 25 Electromagnetic Waves Q.82GP
Suppose the magnitude of the electric field in an electro-magnetic wave is doubled. (a) By what factor does the magnitude of the magnetic field change? (b) By what factor does the maximum intensity of the wave change?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves82ps

Chapter 25 Electromagnetic Waves Q.83GP
If “sailors” of the future use radiation pressure to propel their ships, should the surfaces of their sails be absorbing or re fleeting? Explain.
Solution:
We know that there will be a greater transfer of momentum when the beam is reflected than when it is merely absorbed. Thus if the sailors of the future use the radiation pressure to propel their ships then surface of their sails should be reflecting.

Chapter 25 Electromagnetic Waves Q.84GP
Sunlight at the surface of Earth has an average intensity of about 1.00 × 103 W/m2. Find the rms values of the electric and magnetic fields in the sunlight.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves84ps
>mastering-physics-solutions-chapter-25-electromagnetic-waves84ps1

Chapter 25 Electromagnetic Waves Q.85GP
A typical medical X-ray has a frequency of 1.50 × 1019 Hz. What is the wavelength of such an X-ray?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves85ps

Chapter 25 Electromagnetic Waves Q.86GP
How many hydrogen atoms, 0.10 nm indiameter, must be placed end to end to fit into one wavelength of 410-nm violet light?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves86ps

Chapter 25 Electromagnetic Waves Q.87GP
Radiofrequency Ablation In radiofrequency (RF) ablation, a small needle is inserted into a cancerous tumor. When radiofrequency oscillating currents are sent into the needle, ions in the neighboring tissue respond by vibrating rapidly, causing local heating to temperatures as high as 100°C. This kills the cancerous cells and, because of the small size of the needle, relatively few of the surrounding healthy cells. A typical RF ablation treatment uses a frequency of 750 kHz. What is the wavelength that such radio waves would have in a vacuum?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves87ps

Chapter 25 Electromagnetic Waves Q.88GP
Figure 25–32 shows four polarization experiments in which unpolarized incident light passes throughtwo polarizing filters with different orientations. Rank the four cases in order of increasing amount of transmitted light. Indicate ties where appropriate.
mastering-physics-solutions-chapter-25-electromagnetic-waves88p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves88ps
mastering-physics-solutions-chapter-25-electromagnetic-waves88ps1

Chapter 25 Electromagnetic Waves Q.89GP
(a) What minimum intensity must a laser beam have if it is to levitate a tiny black (100% absorbing) sphere of radius r = 0.5 μ m and mass = 1.6 × 10−15 kg? Comment on the feasibility of such levitation. (b) If the radius of the sphere is doubled but its mass remains the same, will the minimum intensity be greater than, less than, or equal to the value found in part (a)? Explain. (c) Find the minimum intensity for the situation described in part (b).
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves89ps
mastering-physics-solutions-chapter-25-electromagnetic-waves89ps1

Chapter 25 Electromagnetic Waves Q.90GP
The Apollo 11 Reflector One of the experiments placed on the Moon’s surface by Apollo 11 astronauts was a reflector that is used to measure the Earth–Moon distance with high accuracy. A laser beam on Earth is bounced off the reflector, and its round-trip travel time is recorded. If the travel time can be measured to within an accuracy of 0.030 ns, what is the uncertainty in the Earth–Moon distance?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves90ps

Chapter 25 Electromagnetic Waves Q.91GP
The Hβ line of the hydrogen atom’s spectrum has a normal wavelength λβ = 486 nm. This same line is observed in the spectrum of a distant quasar, but lengthened by 20.0 nm. What is the speed of the quasar relative to Earth, assuming it is moving along our line of sight?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves91ps
mastering-physics-solutions-chapter-25-electromagnetic-waves91ps1

Chapter 25 Electromagnetic Waves Q.92GP
Suppose the distance to the fixed mirror in Figure 25–25 is decreased to 20.5 km. (a) Should the angular speed of the rotating mirror be increased or decreased to ensure that the experiment works as described in Problem 17? (b) Find the required angular speed, assuming the speed of light is 3.00 × 108 m/s.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves92ps
mastering-physics-solutions-chapter-25-electromagnetic-waves92ps1

Chapter 25 Electromagnetic Waves Q.93GP
Suppose the speed of the galaxy in Problem is increased by a factor of 10; that is, V = 3.600 × 106 m/s. The speed of the arms, v, and the frequency of the light remain the same. (a) Does the arm near the top of Figure 25–26 show a red shift (toward lower frequency) or a blue shift (toward higher frequency)? Does the lower arm show a red or a blue shift? Explain. What frequency is detected by astronomers observing (b) the upper arm and (c) the lower arm?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves93ps
mastering-physics-solutions-chapter-25-electromagnetic-waves93ps1

Chapter 25 Electromagnetic Waves Q.94GP
Consider the physical situation illustrated in Figure 25–27. (a) Is Erms in the incident laser beam greater than, less than, or the same as Erms where the beam hits the retina? Explain. (b) If the intensity of the beam at the retina is equal to the damage threshold, 1.0 × 10−2 W/cm2, what is the value of Erms at that location? (c) If the diameter of the spot on the retina is reduced by a factor of 2, by what factor does the intensity increase? By what factor does Erms increase?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves94ps
mastering-physics-solutions-chapter-25-electromagnetic-waves94ps1
mastering-physics-solutions-chapter-25-electromagnetic-waves94ps2
mastering-physics-solutions-chapter-25-electromagnetic-waves94ps3

Chapter 25 Electromagnetic Waves Q.95GP
Polaroid Vision in a Spider Experiments show that the ground spider Drassodes cupreus uses one of its several pairs of eyes as a polarization detector. In fact, the two eyes in this pair have polarization directions that are at right angles to one another. Suppose linearly polarized light with an intensity of 825 W/m2 shines from the sky onto the spider, and that the intensity transmitted by one of the polarizing eyes is 232 W/m2. (a) For this eye, what is the angle between the polarization direction of the eye and the polarization direction of the incident light? (b) What is the intensity transmitted by the other polarizing eye?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves95ps
mastering-physics-solutions-chapter-25-electromagnetic-waves95ps1

Chapter 25 Electromagnetic Waves Q.96GP
A state highway patrol car radar unit uses a frequency of 9.00 × 109 Hz. What frequency difference will the unit detect from a car approaching a parked patrol car with a speed of 35.0 m/s?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves96ps

Chapter 25 Electromagnetic Waves Q.97GP
What is the ratio of the sunlight intensity reaching Mercury compared with the sunlight intensity reaching Earth? (On average, Mercury’s distance from the Sun is 0.39 that of Earth’s.)
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves97ps

Chapter 25 Electromagnetic Waves Q.98GP
What area is needed for a solar collector to absorb 45.0 kW of power from the Sun’sradiation if the collector is 75.0% efficient? (At the surface of Earth, sunlight has an average intensity of 1.00 × 103 W/m2.)
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves98ps

Chapter 25 Electromagnetic Waves Q.99GP
Near-Infrared Brain Scans Light in the near-infrared (close to visible red) can penetrate surprisingly far through human tissue, a fact that is being used to “illuminate” the interior of the brain in a noninvasive technique known as near-infrared spectroscopy (NIRS). In this procedure, illustrated in Figure 25–33, an optical fiber carrying a beam of infrared laser light with a power of 1.5 mW and a cross-sectional diameter of 1.2 mm is placed against the skull. Some of the light enters the brain, where it scatters from hemoglobin in the blood. The scattered light is picked up by a detector and analyzed by a computer. (a) According to the Beer-Lambert law, the intensity of light, I, decreases with penetration distance, d, as I = I0e−μd where I0 is the initial intensify of the beam and μ = 4.7 cm−1 for a typical case. Find the intensity ofthe laser beam after it penetrates through 3.0 cm of tissue. (b) Find the electric field of the initial light beam.
mastering-physics-solutions-chapter-25-electromagnetic-waves99p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves99pss
mastering-physics-solutions-chapter-25-electromagnetic-waves99pss1

Chapter 25 Electromagnetic Waves Q.100GP
Three polarizers are arranged as shown in Figure 25–31. If the incident beam of light is unpolarized and has an intensity of 1.60 W/m2, find the transmitted intensify (a) when θ2 = 25.0° and θ3 = 50.0°, and (b) when θ2 = 50.0° and θ3 = 25.0°.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves100ps
mastering-physics-solutions-chapter-25-electromagnetic-waves100ps1
mastering-physics-solutions-chapter-25-electromagnetic-waves100ps2
mastering-physics-solutions-chapter-25-electromagnetic-waves100ps3

Chapter 25 Electromagnetic Waves Q.101GP
This time assuming an incident beam that is vertically polarized. The intensity of the incident beam is 1.60 W/m2.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves101ps
mastering-physics-solutions-chapter-25-electromagnetic-waves101ps1
mastering-physics-solutions-chapter-25-electromagnetic-waves101ps2
mastering-physics-solutions-chapter-25-electromagnetic-waves101ps3

Chapter 25 Electromagnetic Waves Q.102GP
A lightbulb emits light uniformly in all directions. If the rms electric field of this light is 16.0 N/C at a distance of 1.35 m from the bulb, what is the average total power radiated by the bulb?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves102ps

Chapter 25 Electromagnetic Waves Q.103GP
A beam of light is a mixture of unpolarized light with intensity Iu and linearly polarized light with intensity Ip. The polarization direction for the polarized light is vertical. When this mixed beam of light is passed through a polarizer that is vertical, the transmitted intensity is 1.6.8 W/m2; when the polarizer is at an angle of 55.0° with the vertical, the transmitted intensity is 8.68 W/m2. (a) Is Iu greater than, less than, or equal to Ip? Explain. (b) Calculate Iu and Ip.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves103ps
mastering-physics-solutions-chapter-25-electromagnetic-waves103ps1

Chapter 25 Electromagnetic Waves Q.104GP
As mentioned in Problem 95, one pair of eyes in a particular species of ground spider has polarization directions that are at right angles to one another. Suppose that linearly polarized light is incident on such a spider. (a) Prove that the transmitted intensity of one eye plus the transmitted intensity from the other eye is equal to the incident intensity. (b) If the transmitted intensities for the two eyes are 163 W/m2 and 662 W/m2, through what angle must the spider rotate to make the transmitted intensities equal to one another?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves104ps
mastering-physics-solutions-chapter-25-electromagnetic-waves104ps1

Chapter 25 Electromagnetic Waves Q.105GP
A typical home may require a total of 2.00 × 103 kWh of energy per month. Suppose you would like to obtain this energy from sunlight, which has an average daylight intensity of 1.00 × 103 W/m2. Assuming that sunlight is available 8.0 h per day, 25 d per month (accounting for cloudy days), and that you have a way to store energy from your collector when the Sun isn’t shining, determine the smallest collector size that will provide the needed energy, given a conversion efficiency of 25%.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves105ps

Chapter 25 Electromagnetic Waves Q.106GP
At the top of Earth’s atmosphere, sunlight has an average intensify of 1360 W/m2. If the average distance from Earth to the Sun is 1.50 × 1011 m, at what rate does the Sun radiate energy?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves106ps

Chapter 25 Electromagnetic Waves Q.107GP
A typical laser used in introductory physics laboratories produces a continuous beam of light about 1.0 mm in diameter. The average power of such a laser is 0.75 mW. What are (a) the average intensity, (b) the peak intensity, and (c) the average energy density of this beam? (d) If the beam is reflected from a mirror, what is the maximum force the laser beam can exert on it? (e) Describe the orientation of the laser beam relative to the mirror for the case of maximum force.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves107ps
mastering-physics-solutions-chapter-25-electromagnetic-waves107ps1

Chapter 25 Electromagnetic Waves Q.108GP
Four polarizers are set up so that the transmission axis of each successive polarizer is rotated clockwise by an angle θ relative to the previous polarizer. Find the angle θ for whichunpolarized light is transmitted through these four polarizers with its intensity reduced by a factor of 25.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves108ps

Chapter 25 Electromagnetic Waves Q.109GP
Optical Activity of Sugar The sugar concentration in a solution (e.g., in a urine specimen) can be measured conveniently by vising the optical activity of sugar and other asymmetric molecules. In general, anoptically active molecule, like sugar, will rotate the plane of polarization through an angle that is proportional to the thickness of the sample and to the concentration of the molecule. To measure the concentration of a given solution, a sample of known thickness is placed between two polarizing filters that are at right angles to each other, as shown in Figure 25–34. The intensity of light transmitted through the two filters can be compared with a calibration chart to determine the concentration. (a) What percentage of the incident (unpolarized) light will pass through the first filter? (b) If no sample is present, what percentage of the initial light will pass through the second filter? (c) When a particular sample is placed between the twofilters, the intensity of light emerging from the second filter is 40.0% of the incident intensity. Through what angle did the sample rotate the plane ofpolarization? (d) A second sample has half the sugar concentration of the first sample. Find the intensity of light emerging from the second filter in this
mastering-physics-solutions-chapter-25-electromagnetic-waves109p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves109ps

Chapter 25 Electromagnetic Waves Q.110PP
An essential part of modern dentistry is visible-light curing (VLC), a procedure that hardens the restorative materials used in fillings, veneers, and other applications. These “curing lights” work by activating molecules known as photoinitiators within the restorative materials. The photoinitiators, in turn, start a process of polymerization that causes monomers to link together to form a tough, solid polymer network. Thus, with VLC a dentist can apply and shape soft restorative materials as desired, shine a bright light on the result as shown in Figure 25–35, and in 20 seconds have a completely hardened—or cured—final product.
The most common photoinitiator is camphoroquinone (CPQ). To cure CPQ in the least time, one should illuminate it with light having a wavelength of approximately 465 nm. Many VLC units use a halogen light, but there are some draw-backs to this approach. First, the filament in a halogen light is heated to a temperature of about 3000 K, which can cause heat degradation of components in the curing unit itself. Second less than 1 % of the energy given off by a halogen bulb is visible light, so a halogen bulb must have a high power rating to produce the desired light intensity.
More recently, VLC units have begun to use LEDs as their light source. These lights stay cool, emit their energy output as visible light at the desired wavelength, and provide light with an intensity as high as 1000 mW/cm2, which is about 10 times the intensity of sunlight on the surface of the Earth.
mastering-physics-solutions-chapter-25-electromagnetic-waves110ps
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves110ps1

Chapter 25 Electromagnetic Waves Q.111PP
An essential part of modern dentistry is visible-light curing (VLC), a procedure that hardens the restorative materials used in fillings, veneers, and other applications. These “curing lights” work by activating molecules known as photoinitiators within the restorative materials. The photoinitiators, in turn, start a process of polymerization that causes monomers to link together to form a tough, solid polymer network. Thus, with VLC a dentist can apply and shape soft restorative materials as desired, shine a bright light on the result as shown in Figure 25–35, and in 20 seconds have a completely hardened—or cured—final product.
The most common photoinitiator is camphoroquinone (CPQ). To cure CPQ in the least time, one should illuminate it with light having a wavelength of approximately 465 nm. Many VLC units use a halogen light, but there are some draw-backs to this approach. First, the filament in a halogen light is heated to a temperature of about 3000 K, which can cause heat degradation of components in the curing unit itself. Second less than 1 % of the energy given off by a halogen bulb is visible light, so a halogen bulb must have a high power rating to produce the desired light intensity.
More recently, VLC units have begun to use LEDs as their light source. These lights stay cool, emit their energy output as visible light at the desired wavelength, and provide light with an intensity as high as 1000 mW/cm2, which is about 10 times the intensity of sunlight on the surface of the Earth.
mastering-physics-solutions-chapter-25-electromagnetic-waves111p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves111ps

Chapter 25 Electromagnetic Waves Q.112PP
An essential part of modern dentistry is visible-light curing (VLC), a procedure that hardens the restorative materials used in fillings, veneers, and other applications. These “curing lights” work by activating molecules known as photoinitiators within the restorative materials. The photoinitiators, in turn, start a process of polymerization that causes monomers to link together to form a tough, solid polymer network. Thus, with VLC a dentist can apply and shape soft restorative materials as desired, shine a bright light on the result as shown in Figure 25–35, and in 20 seconds have a completely hardened—or cured—final product.
The most common photoinitiator is camphoroquinone (CPQ). To cure CPQ in the least time, one should illuminate it with light having a wavelength of approximately 465 nm. Many VLC units use a halogen light, but there are some draw-backs to this approach. First, the filament in a halogen light is heated to a temperature of about 3000 K, which can cause heat degradation of components in the curing unit itself. Second less than 1 % of the energy given off by a halogen bulb is visible light, so a halogen bulb must have a high power rating to produce the desired light intensity.
More recently, VLC units have begun to use LEDs as their light source. These lights stay cool, emit their energy output as visible light at the desired wavelength, and provide light with an intensity as high as 1000 mW/cm2, which is about 10 times the intensity of sunlight on the surface of the Earth.
mastering-physics-solutions-chapter-25-electromagnetic-waves112p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves112ps

Chapter 25 Electromagnetic Waves Q.113PP
An essential part of modern dentistry is visible-light curing (VLC), a procedure that hardens the restorative materials used in fillings, veneers, and other applications. These “curing lights” work by activating molecules known as photoinitiators within the restorative materials. The photoinitiators, in turn, start a process of polymerization that causes monomers to link together to form a tough, solid polymer network. Thus, with VLC a dentist can apply and shape soft restorative materials as desired, shine a bright light on the result as shown in Figure 25–35, and in 20 seconds have a completely hardened—or cured—final product.
The most common photoinitiator is camphoroquinone (CPQ). To cure CPQ in the least time, one should illuminate it with light having a wavelength of approximately 465 nm. Many VLC units use a halogen light, but there are some draw-backs to this approach. First, the filament in a halogen light is heated to a temperature of about 3000 K, which can cause heat degradation of components in the curing unit itself. Second less than 1 % of the energy given off by a halogen bulb is visible light, so a halogen bulb must have a high power rating to produce the desired light intensity.
More recently, VLC units have begun to use LEDs as their light source. These lights stay cool, emit their energy output as visible light at the desired wavelength, and provide light with an intensity as high as 1000 mW/cm2, which is about 10 times the intensity of sunlight on the surface of the Earth.
mastering-physics-solutions-chapter-25-electromagnetic-waves113p
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves113ps

Chapter 25 Electromagnetic Waves Q.114IP
Suppose the incident beam of light is linearly polarized in the same direction θ as the transmission axis of the analyzer. The transmission axis of the polarizer remains vertical. (a) What value must θ have if the transmitted intensity is to be 0.200 I0? (b) If θ is increased from the value found in part (a), docs the transmittedintensity increase, decrease, or stay the same? Explain.
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves114ps

Chapter 25 Electromagnetic Waves Q.115IP
Suppose the incident beam of light is linearly polarized in the vertical direction. In addition, the transmission axis of the analyzer is at an angle of 80.0° to the vertical. What angle should the transmission axis of the polarizer make with the vertical if the transmitted intensity is to be a maximum?
Solution:
mastering-physics-solutions-chapter-25-electromagnetic-waves115ps

Mastering Physics Solutions Chapter 24 Alternating Current Circuits

Mastering Physics Solutions Chapter 24 Alternating Current Circuits

Mastering Physics Solutions

Chapter 24 Alternating Current Circuits Q.1CQ
How can the rms voltage of an ac circuit be nonzero when its average value is zero? Explain.
Solution:
For a complete cycle the voltage oscillates between positive and negative symmetrically. Therefore the sum of symmetric positive and negative values will become zero and therefore the average value of ac – voltage is zero for a complete cycle.
But when we calculate the rms value of ac – voltage we first calculate the square of the voltage Therefore whether the voltage is positive or negative its square is a positive Therefore the rms voltage of ac is always non zero.
Chapter 24 Alternating Current Circuits Q.1P
An ac generator produces a peak voltage of 55 V. What is the rms voltage of this generator?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits1ps

Chapter 24 Alternating Current Circuits Q.2CQ
Why is the current in an ac circuit not always in phase with its voltage?
Solution:
If we take a pure resistive circuit (circuit with out inductors and capacitors). Then the circuit will respond to the current it self. Therefore in a pure resistive circuit current and voltage will be in
same phase.
Where as if we consider the capacitors they do not respond to the current. They respond to the charge The charge will not build up on the capacitor instantaneously. It will take some time to
build up. Therefore for a capacitor when alternative current is applied, the voltage across it lags behind the current.
Also if we consider the inductors. They respond to the rate of change of current. When the current reaches from its largest value to least value the rate of change is more. Therefore the
voltage across the inductor leads the current Therefore if a circuit is having all these elements definitely the circuit cannot be always in phase with its voltage.

Chapter 24 Alternating Current Circuits Q.2P
European Electricity In many European homes the rms voltage available from a wall socket is 240 V. What is the maximum voltage in this case?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits2ps

Chapter 24 Alternating Current Circuits Q.3CQ
Does an LC circuit consume any power? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits3cqs

Chapter 24 Alternating Current Circuits Q.3P
An rms voltage of 120 V produces a maximum current of 2.1 A in a certain resistor. Find the resistance of this resistor.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits3ps

Chapter 24 Alternating Current Circuits Q.4CQ
An LC circuit is driven at a frequency higher than its resonance frequency. What can be said about the phase angle, ϕ, for this circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits4cqs

Chapter 24 Alternating Current Circuits Q.4P
The rms current in an ac circuit with a resistance of 150 Ω is 0.85 A. What are (a) the average and (b) the maximum power consumed by this circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits4ps

Chapter 24 Alternating Current Circuits Q.5CQ
An LC circuit is driven at a frequency lower than its resonance frequency. What can be said about the phase angle, ϕ, for this circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits5cqs

Chapter 24 Alternating Current Circuits Q.5P
A 3.33-kΩ resistor is connected to a generator with a maximum voltage of 141 V. Find (a) the average and (b) the maximum power delivered to this circuit.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits5ps

Chapter 24 Alternating Current Circuits Q.6CQ
In Conceptual Checkpoint 24-3 we considered an ac circuit consisting of a lightbulb in series with an inductor. The effect of the inductor was to cause the bulb to shine less brightly.
Would the same be true in a direct-current (dc) circuit? Explain.
Solution:
No, the same is not true in a direct current.
In the ac circuit we used the inductor to increase the resistance of the circuit. So, that the bulb shines less brightly. But for dc, frequency is zero. Therefore the inductive reactance of the inductor for dc is zero. So the impedance of the circuit will not decrease. Therefore the bulb will shine with same brightness.

Chapter 24 Alternating Current Circuits Q.6P
A“75-watt” lightbulb uses an average power of 75 W when connected to an rms voltage of 120 V. (a) What is the resistance of the lightbulb? (b) What is the maximum current in the bulb? (c) What is the maximum power used by the bulb at any given instant of time?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits6ps
mastering-physics-solutions-chapter-24-alternating-current-circuits6ps1

Chapter 24 Alternating Current Circuits Q.7CQ
How do the resistance, capacitive reactance, and inductive reactance change when the frequency in a circuit is increased?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits7cqs

Chapter 24 Alternating Current Circuits Q.7P
Square-Wave Voltage I The relationship is valid only for voltages that vary sinusoidally. Find the relationship between Vrms and Vmax for the “square-wave” voltage shown in Figure.
mastering-physics-solutions-chapter-24-alternating-current-circuits7p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits7ps
mastering-physics-solutions-chapter-24-alternating-current-circuits7ps1

Chapter 24 Alternating Current Circuits Q.8CQ
In the analogy between an RLC circuit and a mass on a spring, what is the analog of the current in the circuit? Explain.
Solution:
The current in an RLC circuit is the analog of velocity of the particle in mass spring system.
We have in an RLC circuit charge is analogous to position in spring mass system.
∴ The current = Rate of change of charge is analogous to velocity = Rate of change of position.

Chapter 24 Alternating Current Circuits Q.8P
The reactance of a capacitor is 65 Ω at a frequency of 57Hz. What is its capacitance?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits8ps

Chapter 24 Alternating Current Circuits Q.9CQ
In the analogy between an RLC circuit and a mass on a spring, the mass is analogous to the inductance, and the spring constant is analogous to the inverse of the capacitance. Explain.
Solution:
As from the concept of inertia in mechanics the mass of a body resists changes in its motion.
Also from the concept of electromagnetic induction and inductor resists the change in the current flowing through it. As these two quantities mass and inductance are having similar function.
Therefore mass is analogous to inductive of the inductor.
Also we know that a spring having large spring constant requires more force to stretch it. Therefore for a spring having large spring constant gets little stretch for a given force. i.e., a more is the spring constant less is the displacement of the spring for a given force. At the same time the capacitor with small capacitance will store a small charge for a given voltage.
∴ More is the spring constant less is the displacement for a given force. Less is the capacitance less is the charge stored for a given voltage. But we know that charge and displacement are analogous to each other therefore the spring constant and inverse of capacitance are analogous to each other.

Chapter 24 Alternating Current Circuits Q.9P
The capacitive reactance of a capacitor at 60.0 Hz is 105 Ω. At what frequency is its capacitive reactance 72.5 Ω?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits9ps

Chapter 24 Alternating Current Circuits Q.10CQ
Two RLC circuits have different values of L and C. Is it possible for these two circuits to have the same resonance frequency? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits10cqs

Chapter 24 Alternating Current Circuits Q.10P
A 105-μ Fcapacitor is connected to an ac generator with an rms voltage of 20.0 V and a frequency of 100.0 Hz. What is the rms current in this circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits10ps

Chapter 24 Alternating Current Circuits Q.11CQ
Can an RLC circuit have the same impedance at two different frequencies? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits11cqs

Chapter 24 Alternating Current Circuits Q.11P
The rms voltage across a 0.010-μ F capacitor is 1.8 V at a frequency of 52 Hz. What are (a) the rms and (b) the maximum current through the capacitor?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits11ps

Chapter 24 Alternating Current Circuits Q.12P
An ac generator with a frequency of 30.0 Hz and an rms voltage of 12.0 V is connected to a 45.5-μ F capacitor, (a) What is the maximum current in this circuit? (b) What is the current in the circuit when the voltage across the capacitor is 5.25 V and increasing? (c) What is the current in the circuit when the voltage across the capacitor is 5.25 V and decreasing?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits12ps
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Chapter 24 Alternating Current Circuits Q.13P
The maximum current in a 22-μ F capacitor connected to an ac generator with a frequency of 120 Hz is 0.15 A. (a) What is the maximum voltage of the generator? (b) What is the voltage across the capacitor when the current in the circuit is 0.10 A and increasing? (c) What is the voltage across the capacitor when the current in the circuit is 0.10 A and decreasing?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits13ps
mastering-physics-solutions-chapter-24-alternating-current-circuits13ps1

Chapter 24 Alternating Current Circuits Q.14P
An rms voltage of 20.5 V with a frequency of 1.00 kHz is applied to a 0.395-μ F capacitor. (a) What is the rms current in this circuit? (b) By what factor does the current change if the frequency of the voltage is doubled? (c) Calculate the current for a frequency of 2.00 kHz.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits14ps
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Chapter 24 Alternating Current Circuits Q.15P
A circuit consists of a 1.00-kHz generator and a capacitor. When the rms voltage of the generator is 0.500 V, the rms current in the circuit is 0.430 mA. (a) What is the reactance of the capacitor at 1.00 kHz? (b) What is the capacitance of the capacitor? (c) If the rms voltage is maintained at 0.500 V, what is the rms current at 2.00 kHz? At 10.0 kHz?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits15ps
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Chapter 24 Alternating Current Circuits Q.16P
A capacitor has an rms current of 21 mA at a frequency of 60.0 Hz when the rms voltage across it is 14 V. (a) What is the capacitance of this capacitor? (b) If the frequency is increased, will the current in the capacitor increase, decrease, or stay the same? Explain. (c) Find the rms cm-rent in this capacitor at a frequency of 410 Hz.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits16ps
mastering-physics-solutions-chapter-24-alternating-current-circuits16ps1

Chapter 24 Alternating Current Circuits Q.17P
A 0.22-μ F capacitor is connected to an ac generator with an rms voltage of 12 V. For what range of frequencies will the rms current in the circuit be less than 1.0 mA?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits17ps

Chapter 24 Alternating Current Circuits Q.18P
At what frequency will a generator with an rms voltage of 504 V produce an rms current of 7.50 mA in a 0.0150-μF capacitor?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits18ps

Chapter 24 Alternating Current Circuits Q.19P
Find the impedance of a60.0-Hz circuit with a 45.5-Ω resistor connected in series with a 95.0-μ F capacitor.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits19ps

Chapter 24 Alternating Current Circuits Q.20P
An ac generator with a frequency of 105 Hz and an rms voltage of 22.5 V is connected in series with a 10.0-kΩ resistor and a 0.250-μ F capacitor. What is the rms current in this circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits20ps
mastering-physics-solutions-chapter-24-alternating-current-circuits20ps1

Chapter 24 Alternating Current Circuits Q.21P
The rms current in an RC circuit is 0.72 A. The capacitor in this circuit has a capacitance of 13 μ Fand the ac generator has a frequency of 150 Hz and an rms voltage of 95 V. What is the resistance in this circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits21ps
mastering-physics-solutions-chapter-24-alternating-current-circuits21ps1

Chapter 24 Alternating Current Circuits Q.22P
A 65.0-Hz generator with an rms voltage of 135 V is connected in series to a3.35-kΩresistor and a 1.50-μ F capacitor. Find (a) the rms current in the circuit and (b) the phase angle, ϕ, between the currentand the voltage.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits22ps
mastering-physics-solutions-chapter-24-alternating-current-circuits22ps1

Chapter 24 Alternating Current Circuits Q.23P
(a) At what frequency must the circuit in Problem be operated for the current to lead the voltage by 23.0°? (b) Using the frequency found in part (a), find the average power consumed by this circuit.
A 65.0-Hz generator with an rms voltage of 135 V is connected in series to a3.35-kΩresistor and a 1.50-μ F capacitor. Find (a) the rms current in the circuit and (b) the phase angle, , between the currentand the voltage.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits23ps
mastering-physics-solutions-chapter-24-alternating-current-circuits23ps1
mastering-physics-solutions-chapter-24-alternating-current-circuits23ps2

Chapter 24 Alternating Current Circuits Q.24P
(a) Sketch the phasor diagramfor an ac circuit with a 105-Ω resistor in series with a 32.2-μ F capacitor. The frequency of the generator is 60.0 Hz. (b) If the rms voltage of the generator is 120 V, what is the average power consumed by the circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits24ps
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Chapter 24 Alternating Current Circuits Q.25P
Find the power factor for an RC circuit connected to a 70.0-Hz generator with an rms voltage of 155 V. The values of R and C in this circuit are 105 Ω and 82.4 μ F, respectively.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits25ps

Chapter 24 Alternating Current Circuits Q.26P
(a) Determine the power factor for an RC circuit with R = 4.0 kΩ and C = 0.35 μ Fthatis connected to an ac generator with an rms voltage of 24 V and a frequency of 150 Hz. (b) Will the power factor for this circuit increase, decrease, or stay the same if the frequency of the generator is increased? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits26ps
mastering-physics-solutions-chapter-24-alternating-current-circuits26ps1

Chapter 24 Alternating Current Circuits Q.27P
mastering-physics-solutions-chapter-24-alternating-current-circuits27p
Solution:
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Chapter 24 Alternating Current Circuits Q.28P
CE Predict/Explain When a long copper wire of finite resistance is connected to an ac generator, as shown in Figure,a certain amount of current flows through the wire. The wire is now wound into a coil of many loops and reconnected to the generator, as indicated in Figure. (a) Is the current supplied to the coil greater than, less than, or the same as the current supplied to the uncoiled wire? (b) Choose the best explanation from among the following:
I. More currentflows in the circuit because the coiled wire is an inductor, and inductors tend to keep the current flowing in an ac circuit.
II. The current supplied to the circuit is the same because the wire is the same. Simply wrapping the wire in a coil changes nothing.
III. Less current is supplied to the circuit because the coiled wire acts as an inductor, which increases the impedance of the circuit.
mastering-physics-solutions-chapter-24-alternating-current-circuits28p
Solution:
a)
In the circuit (a) the resistance of the wire is R.
In circuit (b) the resistance of the inductor is called as inductive reactance and it is due to the wounding of copper wire.
The inductor impedance is more than the resistance of the wire. So the current supplied to the coil is less than the current supplied to the single copper wire.
b)
As the circuit (b) consists of inductor the impedance of the circuit is increased. So less current is supplied to the circuit.
Hence best explanation is (III)

Chapter 24 Alternating Current Circuits Q.29P
An inductor has a reactance of 56.5 Ω at 75.0 Hz. What is its reactance at 60.0 Hz?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits28ps
mastering-physics-solutions-chapter-24-alternating-current-circuits28ps1

Chapter 24 Alternating Current Circuits Q.30P
What is the rms current in a 77.5-mH inductor when it is connected to a 60.0-Hz generator with an rms voltage of 115 V?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits30ps

Chapter 24 Alternating Current Circuits Q.31P
What rms voltage is required to produce an rms current of 2.1 A in a 66-mH inductor at a frequency of 25 Hz?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits31ps

Chapter 24 Alternating Current Circuits Q.32P
A 525-Ω resistor and a 295-mH inductor are connected in series with an ac generator with an rms voltage of 20.0 V and a frequency of 60.0 Hz. What is the rms current in this circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits32ps

Chapter 24 Alternating Current Circuits Q.33P
mastering-physics-solutions-chapter-24-alternating-current-circuits33ps
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits33ps1
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Chapter 24 Alternating Current Circuits Q.34P
An ac generator with a frequency of 1.34 kHz and an rms voltage of 24.2 V is connected in series with a 2.00-kΩ resistor and a 315-mH inductor. (a) What is the power factor for this circuit? (b) What is the average power consumed by this circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits34ps
mastering-physics-solutions-chapter-24-alternating-current-circuits34ps1

Chapter 24 Alternating Current Circuits Q.35P
An rms voltage of 22.2 V with a frequency of 1.00 kHz is applied to a0.290-mH inductor. (a) What is the rms current in this circuit? (b) By what factor does the current change if the frequency of the voltage is doubled? (c) Calculate the current for a frequency of 2.00 kHz.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits35ps
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Chapter 24 Alternating Current Circuits Q.36P
A 0.22-μ H inductor is connected to an ac generator with an rms voltage of 12 V. For what range of frequencies will the rms current in the circuit be less than 1.0 mA?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits36ps

Chapter 24 Alternating Current Circuits Q.37P
The phase angle in a certain RL circuit is 76° at a frequency of 60.0 Hz. If R = 2.7 Ω for this circuit, what is the value of the inductance, L?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits37ps
mastering-physics-solutions-chapter-24-alternating-current-circuits37ps1

Chapter 24 Alternating Current Circuits Q.38P
mastering-physics-solutions-chapter-24-alternating-current-circuits38p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits38ps
mastering-physics-solutions-chapter-24-alternating-current-circuits38ps1

Chapter 24 Alternating Current Circuits Q.39P
(a) Sketch the phasor diagram for an ac circuit with a 105-Ω resistor in scries with a 22.5-mH inductor. Thefrequency of the generator is 60.0 Hz. (b) If the rms voltage of the generator is 120 V, what is the average power consumed by the circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits39ps
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mastering-physics-solutions-chapter-24-alternating-current-circuits39ps2
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Chapter 24 Alternating Current Circuits Q.40P
In Problem, does the phase angle increase, decrease, or stay the same when the frequency is increased? Verify your answer by calculating the phase angle at 70.0 Hz.
(a) Sketch the phasor diagram for an ac circuit with a 105-Ω resistor in scries with a 22.5-mH inductor. Thefrequency of the generator is 60.0 Hz. (b) If the rms voltage of the generator is 120 V, what is the average power consumed by the circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits40ps

Chapter 24 Alternating Current Circuits Q.41P
A large air conditioner has a resistance of 7.0 Ω and an inductive reactance of 15 Ω. If the air conditioner is powered by a 60.0-Hz generator with an rms voltage of 240 V, find (a) the impedance of the air conditioner, (b) its rms current, and (c) the average power consumed by the air conditioner.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits41ps
mastering-physics-solutions-chapter-24-alternating-current-circuits41ps1

Chapter 24 Alternating Current Circuits Q.42P
Square-Wave Voltage III The “square-wave” voltage shown in Figure is applied to an RL circuit. Sketch the shape of the instantaneous voltage across the inductor, assuming the time constant of the circuit is much less than the period of the applied voltage.
mastering-physics-solutions-chapter-24-alternating-current-circuits42p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits42ps
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Chapter 24 Alternating Current Circuits Q.43P
CE An inductor and a capacitor are to be connected to a generator. Will the generator supply more current at high frequency if the inductor and capacitor arc connected in series or in parallel? Explain.
Solution:
When the frequency of the ac generator is high, then the reactance of the inductor is high and the reactance of the capacitor is low.
For very high frequencies the inductor behaves as an open circuit as its reactance is very high and the capacitor behaves as a very low reactance path.
But when a high resistance is connected in series, the resultant resistance of the circuit will becomes high and there by current in the circuit is low.
And when these two are connected in parallel then the current in the circuit will take the low reactance path and therefore the current in the circuit is very high.
Therefore at high frequency the inductor and capacitor should be connected in parallel to get more current from the generator supply.

Chapter 24 Alternating Current Circuits Q.44P
An inductor and a capacitor arc to be connected to a generator. Will the generator supply more current at low frequency if the inductor and capacitor are connected in series or in parallel? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits44ps

Chapter 24 Alternating Current Circuits Q.45P
Predict/Explain (a) When the ac generator in Figure operates at high frequency, is the rms current in the circuit greater than, less than, or the same as when the generator operates at low frequency? (b) Choose the best explanation from among the following:
I. The current is the same because at high frequency the inductor is like an open circuit, and at low frequency the capacitor is like an open circuit. In either case the resistance of the circuit is R.
II. Less current flows at high frequency because in that limit the inductor acts like an open circuit, allowing no current to flow.
III. More current flows at high frequency because in that limit the capacitor acts like an ideal wire of zero resistance.
mastering-physics-solutions-chapter-24-alternating-current-circuits45p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits45ps
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mastering-physics-solutions-chapter-24-alternating-current-circuits45ps2

Chapter 24 Alternating Current Circuits Q.46P
Predict/Explain (a) When the ac generator in Figure operates at high frequency, is the rms current in the circuit greater than, less than, or the same as when the generator operates at low frequency? (b) Choose the best explanation from among the following:
I. The current at high frequency is greater because the higher the frequency the more chargethat flows through a circuit.
II. Less current flows at high frequency because in that limit the inductoris like an open circuit and current has only one path to flow through.
III. The inductor has zero resistance, and therefore the resistance of the circuit is the same at all frequencies. As a result the current is the same at all frequencies.
mastering-physics-solutions-chapter-24-alternating-current-circuits46p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits46ps
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Chapter 24 Alternating Current Circuits Q.47P
Predict/Explain (a) When the ac generator in Figure operates at high frequency, is the rms current in the circuit greater than, less than, or the same as when the generator operates at low frequency? (b) Choose the best explanation from among the following:
I. The capacitor has no resistance, and therefore the resistance of the circuit is the same at all frequencies. As a result the current is the same at all frequencies.
II. Less current flows at high frequency because in that limit the capacitor is like an open circuit and current has only one path to flow through.
III. More current flows at high frequency because in that limit the capacitor is like a short circuit and current has two parallel paths to flow through.
mastering-physics-solutions-chapter-24-alternating-current-circuits47p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits47ps
mastering-physics-solutions-chapter-24-alternating-current-circuits47ps1

Chapter 24 Alternating Current Circuits Q.48P
Find the rms voltage across the element in an RLC circuit with R = 9.9 kΩ, C = 0.15 μ F, and L = 25 mH. The generator supplies an rms voltage of 115 V at a frequency of 60.0 Hz.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits48ps
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Chapter 24 Alternating Current Circuits Q.49P
What is the impedance of a1.50-k Ω resistor, a 105-mH inductor, and a 12.8-μ F capacitor connected in series with a 60.0-Hz ac generator?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits49ps

Chapter 24 Alternating Current Circuits Q.50P
Consider the circuit shown in Figure. The ac generator in tills circuit has an rms voltage of 65 V. Given that R = 15 Ω and L = 0.22 mH, find the rms current in this circuit in the limit of (a) high frequency and (b) low frequency.
mastering-physics-solutions-chapter-24-alternating-current-circuits50p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits50ps
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Chapter 24 Alternating Current Circuits Q.51P
Consider the circuit shown in Figure. The ac generator in this circuit has an rms voltage of 75 V. Given that R = 15 Ω and C = 41 μ F, find the rms current in this circuit in the limit of (a) high frequency and (b) low frequency.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits51ps

Chapter 24 Alternating Current Circuits Q.52P
What is the phase angle in an RLC circuit with R = 9.9 kΩ, C = 1.5 μ F, and L = 250 mH? The generator supplies an rms voltage of 115 V at a frequency of 60.0 Hz.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits52ps
mastering-physics-solutions-chapter-24-alternating-current-circuits52ps1

Chapter 24 Alternating Current Circuits Q.53P
IP An RLC circuit has a resistance of 105 Ω, an inductance of 85.0 mH, and a capacitance of 13.2 μ F.(a) What is the power factor for this circuit when it is connected to a 125-Hz ac generator? (b) Will the power factor increase, decrease, or stay the same if the resistance is increased? Explain. (c) Calculate the power factor for a resistance of 525 Ω.
Solution:
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Chapter 24 Alternating Current Circuits Q.54P
An ac voltmeter, which displays the rms voltage between the two points touched by its leads, is used to measure voltages in the circuit shown in Figure. In this circuit, the ac generator has an rms voltage of 6.00 V and a frequency of 30.0 kHz. The inductance in the circuit is 0.300 mH,the capacitance is 0.100 μ F, and the resistance is 2.50 Ω. What is the reading on a voltmeter when it is connected to points (a) A and B, (b) B and C, (c) A and C, and (d) A and D?
mastering-physics-solutions-chapter-24-alternating-current-circuits54p
Solution:
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Chapter 24 Alternating Current Circuits Q.55P
IP Consider the ac circuit shown in Figure, where we assume that the values of R, L, and C are the same as in the previous problem, and that the rms voltage of the generator is still 6.00 V. The frequency of the generator, however, is doubled to 60.0 kHz. Calculate the rms voltage across (a) the resistor, R, (b) the inductor, L, and (c) the capacitor, C. (d) Do you expect the sum of the rms voltages in parts (a), (b), and (c) to be greater than, less than, or equal to 6.00 V? Explain.
mastering-physics-solutions-chapter-24-alternating-current-circuits55p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits55ps
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Chapter 24 Alternating Current Circuits Q.56P
(a) Sketch the phasor diagram for an ac circuit with a 105-Ω resistor in series with a 22.5-mH inductor and a 32.2-μ F capacitor. The frequency of the generator is 60.0 Hz. (b) If the rms voltage of the generator is 120 V, what is the average power consumed by the circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits56ps
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Chapter 24 Alternating Current Circuits Q.57P
A generator connected to an RLC circuit has an rms voltage of 120 V and an rms current of 34 mA. If the resistance inthe circuit is 3.3 kΩ and the capacitive reactance is 6.6 kΩ, what is the inductive reactance of the circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits57ps
mastering-physics-solutions-chapter-24-alternating-current-circuits57ps1

Chapter 24 Alternating Current Circuits Q.58P
Manufacturing Plant Power A manufacturing plant uses 2.22 kW of electric power provided by a 60.0-Hz ac generator with an rms voltage of 485 V. The plant uses this power to run a number of high-inductance electric motors. The plant’s total resistance is R = 25.0 Ω and its inductive reactance is XL = 45.0 Ω. (a) What is the total impedance of the plant? (b) What is the plant’s power factor? (c) What is the rms current used by the plant? (d) Whatcapacitance, connected in series with the power line, will increase the plant’s power factor to unity? (e) If the power factor is unity, how much current is needed to provide the 2.22 kW of power needed by the plant? Compare your answer with the current found in part (c). (Because power-line losses arc proportional to the square of the current, a utility company will charge an industrial user with a low power factor a higher rate per kWh than a company with a power factor close to unity.)
Solution:
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Chapter 24 Alternating Current Circuits Q.59P
CE A capacitor and an inductor connected in series have a period of oscillation given by T. At the time t = 0 the capacitor has its maximum charge. In terms of T, what is the first time after t = 0 that (a) the current in the circuit has its maximum value and (b) the energy stored in the electric field is a maximum?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits59ps
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mastering-physics-solutions-chapter-24-alternating-current-circuits59ps2
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Chapter 24 Alternating Current Circuits Q.60P
Predict/Explain In an RLC circuit a second capacitor is added in series to the capacitor already present. (a) Does the resonance frequency increase, decrease, or stay the same? (b) Choose the best explanation from among the following:
I. The resonance frequency stays the same because it depends only on the resistance in the circuit.
II. Adding a capacitor in series increases the equivalent capacitance, and this decreases the resonance frequency.
III. Adding a capacitor in series decreases the equivalent capacitance, and this increases the resonance frequency.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits60ps

Chapter 24 Alternating Current Circuits Q.61P
In an RLC circuit a second capacitor is added in parallel to the capacitor already present. Does the resonance frequency increase, decrease, or stay the same? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits61ps

Chapter 24 Alternating Current Circuits Q.62P
An RLC circuit has a resonance frequency of 2.4 kHz. If the capacitance is 47 μ F, what is the inductance?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits62ps

Chapter 24 Alternating Current Circuits Q.63P
At resonance, the rms current in an RLC circuit is 2.8 A. If the rms voltage of the generator is 120 V, what is the resistance, R?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits63ps

Chapter 24 Alternating Current Circuits Q.64P
The resistance in an RLC circuit is doubled. (a) Does the resonance frequency increase, decrease, or stay the same? Explain. (b) Does the maximum current in the circuit increase, decrease, or stay the same? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits64ps

Chapter 24 Alternating Current Circuits Q.65P
The voltage in a sinusoidally driven RLC circuit leads the current. (a) If we want to bring this circuit into resonance by changing the frequency of the generator, should the frequency be increased or decreased? Explain. (b) If we want to bring this circuit into resonance by changing the inductance instead, should the inductance be increased or decreased? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits65ps

Chapter 24 Alternating Current Circuits Q.66P
A 115-Ω resistor, a 67.6-mH inductor, and a 189-μ F capacitor are connected in series to an ac generator. (a) At what frequency will the current in the circuit be a maximum? (b) At what frequency will the impedance of the circuit be a minimum?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits66ps
mastering-physics-solutions-chapter-24-alternating-current-circuits66ps1
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Chapter 24 Alternating Current Circuits Q.67P
An ac generator of variable frequency is connected to an RLC circuit with R = 12 Ω, L = 0.15 mH, and C = 0.20 mF. At a frequency of 1.0 kHz, the rms current in the circuit is larger than desired. Should the frequency of the generator be increased or decreased to reduce the current? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits67ps

Chapter 24 Alternating Current Circuits Q.68P
(a) Find the frequency at which a 33-μ F capacitor has the same reactance as a 33-mH inductor. (b) What is the resonance frequency of an LC circuit made with this inductorand capacitor?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits68ps
mastering-physics-solutions-chapter-24-alternating-current-circuits68ps1

Chapter 24 Alternating Current Circuits Q.69P
Consider an RLC circuit with R = 105 Ω, L = 518 mH, and C = 0.200 μ F. (a) At what frequency is this circuit in resonance? (b) Find the impedance of this circuit if the frequency has the value found in part (a), but the capacitance is increased to 0.220 μ F. (c) What is the power factor for the situation described in part (b)?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits69ps
mastering-physics-solutions-chapter-24-alternating-current-circuits69ps1

Chapter 24 Alternating Current Circuits Q.70P
An RLC circuit has a resonance frequency of 155 Hz. (a) If both L and C are doubled, does the resonance frequency increase, decrease, or stay the same? Explain. (b) Find the resonance frequency when L and C are doubled.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits70ps

Chapter 24 Alternating Current Circuits Q.71P
An RLC circuit has a capacitance of 0.29 μ F. (a) What inductance will produce a resonance frequency of 95 MHz? (b) It is desired that the impedance at resonance be one-fifth the impedance at 11 kHz. What value of R should be used to obtain this result?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits71ps1
mastering-physics-solutions-chapter-24-alternating-current-circuits71ps

Chapter 24 Alternating Current Circuits Q.72GP
CE BIO Persistence of Vision Although an incandescent lightbulb appears to shine with constant intensity, this is an artifact of the eye’s persistence of vision. In fact, the intensity of a bulb’s light rises and falls with time due to the alternating current used in household circuits. If you could perceive these oscillations, would you see the light attain maximum brightness 60 or 120 times per second? Explain.
Solution:
In alternating current : the current attains positive maximum and negative maximum in one complete cycle. But the bulb will be brightest when the current is at maximum in either direction.
But the alternating currents used in house hold circuits have the frequency.60Hz
Therefore in one second the current reaches 60 times positive maximum and 60 times negative maximum.
The light will attain maximum brightness for 120 times in a second.

Chapter 24 Alternating Current Circuits Q.73GP
An inductor in an LC circuit has a maximum current of 2.4 A and a maximum energy of 36 mJ. When the current in the inductor is 1.2 A, what is the energy stored in the capacitor.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits73ps

Chapter 24 Alternating Current Circuits Q.74GP
An RLC circuit is driven at its resonance frequency. Is its impedance greater than, less than, or equal to R? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits74ps

Chapter 24 Alternating Current Circuits Q.75GP
Predict/Explain Suppose the circuits shown in Figures and are connected to identical batteries, rather than to ac generators. (a) Assuming the value of R is the same in the two circuits, is the current in Figure greater than, less than, or the same as the current in Figure? (b) Choose the best explanation from among the following:
I. The circuits have the same current because the capacitor acts like an open circuit and the inductor acts like a short circuit.
II. The current in Figure is larger because it has more circuit elements, each of which can carry current.
III. The current in Figure is larger because it has fewer circuit elements, meaning less resistance to current flow.
mastering-physics-solutions-chapter-24-alternating-current-circuits75p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits75ps
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mastering-physics-solutions-chapter-24-alternating-current-circuits75ps2

Chapter 24 Alternating Current Circuits Q.76GP
Suppose the circuits shown in Figures and are connected to identical batteries, rather than to ac generators. Assuming the value of R is the same in the two circuits, is the current in Figure greater than, less than, or the same as the current in Figure? Explain.
mastering-physics-solutions-chapter-24-alternating-current-circuits76p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits76ps

Chapter 24 Alternating Current Circuits Q.77GP
Predict/Explain Consider a circuit consisting of a lightbulb and a capacitor, as shown in circuit 2 of Conceptual Checkpoint 24-2. (a) If the frequency of the generator is increased, does the intensity of the lightbulb increase, decrease, or stay the same? (b) Choose the best explanation from among the following:
I. As the frequency increases it becomes harder to force current through the capacitor, and therefore the intensity of the lightbulb decreases.
II. The intensity of the lightbulb increases because as the frequency becomes higher the capacitor acts more like a short circuit, allowing more current to flow.
III. The intensity of the lightbulb is independent of frequency because the circuit contains a capacitor but not an inductor.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits77ps

Chapter 24 Alternating Current Circuits Q.78GP
Consider a circuit consisting of a lightbulb and an inductor, as shown in Conceptual Checkpoint 24-3. If the frequency of the generator is increased, does the intensity of the lightbulb increase, decrease, or stay the same? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits78ps

Chapter 24 Alternating Current Circuits Q.79GP
A 4.40-μ F and an 8.80-μ F capacitor are connected in parallel to a 60.0-Hz generator operating with an rms voltage of 115 V. What is the rms currentsupplied by the generator?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits79ps

Chapter 24 Alternating Current Circuits Q.80GP
A 40.40-μ F and an 8.80-μ F capacitor are connected in series to a 60.0-Hz generator operating with an rms voltage of 115 V. What is the rms current supplied by the generator?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits80ps

Chapter 24 Alternating Current Circuits Q.81GP
A 10.0-μ F capacitor and a 30.0-μ F capacitor are connected in parallel to an ac generator with a frequency of 60.0 Hz. What is the capacitivc reactance of this pair of capacitors?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits81ps
mastering-physics-solutions-chapter-24-alternating-current-circuits81ps1

Chapter 24 Alternating Current Circuits Q.82GP
A generator drives an RLC circuit with the voltage V shown in Figure. The correspond ing current I is also shown in the figure. (a) Is the inductive reactance of this circuit greater than, less than, or equal to its capacitive reactance? Explain. (b) Is the frequency of this generator greater than, less than, or equal to the resonance frequency of the circuit? Explain.
mastering-physics-solutions-chapter-24-alternating-current-circuits82p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits82ps

Chapter 24 Alternating Current Circuits Q.83GP
Consider the RLC circuit shown in Example, and the corresponding phasor diagram given inthe Insight. (a) On the basis of the phasor diagram, can you conclude that the resonance frequency of this circuit is greater than, less than, or equal to 60.0 Hz? Explain. (b) Calculate the resonance frequency for
this circuit. (c) The impedance of this circuit at 60.0 Hz is 226 Ω. What is the impedance at resonance?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits83ps
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mastering-physics-solutions-chapter-24-alternating-current-circuits83ps2
mastering-physics-solutions-chapter-24-alternating-current-circuits83ps3

Chapter 24 Alternating Current Circuits Q.84GP
When a certain resistor is connected to an ac generator with a maximum voltage of 15 V, the average power dissipated in the resistor is 22 W. (a) What is the resistance of the resistor? (b) What is the rms current in the circuit? (c) We know that Pav = l2rmsR, and hence it seems that reducing the resistance should reduce the average power. On the other hand, we also know that Pav = V2rms/R,which suggests that reducing R increases Pav. Which conclusion is correct? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits84ps
mastering-physics-solutions-chapter-24-alternating-current-circuits84ps1

Chapter 24 Alternating Current Circuits Q.85GP
A 9.5-Hz generator is connected to a capacitor. If the current in the generator has its maximum value at t = 0, what is the earliest possible time that the voltage across the capacitor is a maximum?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits85ps

Chapter 24 Alternating Current Circuits Q.86GP
The voltage across an inductor reaches its maximum value 25 ms before the current supplied by the generator reaches its maximum value. What is the lowest possible frequency at which the generator operates?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits86ps

Chapter 24 Alternating Current Circuits Q.87GP
Find the average power consumed by an RC circuit connected to a 60.0-Hz generator with an rms voltage of 122 V. The values of R and C in this circuit are 3.30 kΩ and 2.75 μ F,respectively.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits87ps

Chapter 24 Alternating Current Circuits Q.88GP
A 1.15-k Ω resistor and a 505-mH inductor are connected in series to a 1250-Hz generator with an rms voltage of 14.2 V. (a) What is the rms current in the circuit? (b) What capacitance must be inserted in series with the resistor and inductorto reduce the rms current to half the value found in part (a)?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits88ps
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Chapter 24 Alternating Current Circuits Q.89GP
Phasor The phasor diagram for an RLC circuit is shown in Figure. (a) If the resistance in this circuit is 525 Ω, what is the impedance? (b) If the frequency inthis circuit is increased, will the impedance increase, decrease, or stay the same? Explain.
mastering-physics-solutions-chapter-24-alternating-current-circuits89p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits89ps
mastering-physics-solutions-chapter-24-alternating-current-circuits89ps1

Chapter 24 Alternating Current Circuits Q.90GP
Figure shows the phasor diagram for an RLC circuit in whichthe impedance is 337 Ω. (a) What is the resistance, R, in this circuit? (b) Is this circuit driven at a frequency that is greater than, less than, or equal to the resonance frequency of the circuit? Explain.
mastering-physics-solutions-chapter-24-alternating-current-circuits90p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits90ps
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Chapter 24 Alternating Current Circuits Q.91GP
An RLC circuit has a resistance R = 25 Ω and an inductance L = 160 mH, and is connected to an ac generator with a frequency of 55 Hz. The phasor diagram for this circuit is shown in Figure. Find (a) the impedance, Z, and (b) the capacitance, C, for this circuit. (c) If the value of C is decreased, will the impedance of the circuit increase, decrease, or stay the same? Explain.
mastering-physics-solutions-chapter-24-alternating-current-circuits91p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits91pss
mastering-physics-solutions-chapter-24-alternating-current-circuits91ps1
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Chapter 24 Alternating Current Circuits Q.92GP
IP Black-Box Experiment You are given a sealed box with two electrical terminals. The box contains a 5.00-Ω resistor in series with either an inductor or a capacitor. When you attach an ac generator with an rms voltage of 0.750 V to the terminals of the box, you find that the current increases with increasing frequency. (a) Does the box contain an inductor or a capacitor?
Explain. (b) When the frequency of the generator is 25.0 kHz, the rms current is 87.2 mA. What is the capacitance or inductance of the unknown component in the box?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits92ps
mastering-physics-solutions-chapter-24-alternating-current-circuits92ps1
mastering-physics-solutions-chapter-24-alternating-current-circuits92ps2

Chapter 24 Alternating Current Circuits Q.93GP
A circuit is constructed by connecting a 1.00-kΩ resistor, a 252-μ F capacitor, and a 515-mH inductor in series. (a) What is the highest frequency at which the impedance of this circuit is equal to 2.00 kΩ? (b) To reduce the impedance of this circuit, should the frequency be increased or decreased from itsvalue in part (a)? Explain.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits93ps
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mastering-physics-solutions-chapter-24-alternating-current-circuits93ps3
mastering-physics-solutions-chapter-24-alternating-current-circuits93ps4

Chapter 24 Alternating Current Circuits Q.94GP
An RLC circuit with R = 25.0 Ω, L = 325 mH, and C = 45.2 μ Fis connected to an ac generator with an rms voltage of 24 V. Determine the average power delivered to this circuit when the frequency of the generator is (a) equal to the resonance frequency, (b) twice the resonance frequency, and (c) half the resonance frequency.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits94ps
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mastering-physics-solutions-chapter-24-alternating-current-circuits94ps2
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Chapter 24 Alternating Current Circuits Q.95GP
A Light-Dimmer Circuit The intensity of a lightbulb with aresistance of 120 Ω is controlled by connecting it in series with an inductor whose inductance can be varied from L = 0 to L = Lmax. This “light dimmer”circuit is connected to an ac generator with a frequency of 60.0 Hz and an rms voltage of 110 V. (a) What is the average power dissipated in the lightbulb when L = 0? (b) The inductor is now adjusted so that L = Lmax. In this case, the average power dissipated in the lightbulb is one-fourth the value found in part(a). What is the value of Lmax?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits95ps
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Chapter 24 Alternating Current Circuits Q.96GP
An electric motor with a resistance of 15 Ω and an inductance of 53 mH is connected to a 60.0-Hz ac generator. (a) What is the power factor for this circuit? (b) In order to increase the power factor of this circuit to 0.80, a capacitor is connected in series with the motor and inductor. Find the required value of the capacitance.
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits96ps
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Chapter 24 Alternating Current Circuits Q.97GP
Tuning a Radio A radio tuning circuit contains an RLC circuit with R = 5.0 Ω and L = 2.8 μ H.(a) What capacitance is needed to produce a resonance frequency of 85 MHz?
(b) If the capacitance is increased above the value found in part (a), will the impedance increase, decrease, or stay the same? Explain. (c) Find the impedance of the circuit at resonance. (d) Find the impedance of the circuit when the capacitance is 1% higherthan the value found in part (a).
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits97ps
mastering-physics-solutions-chapter-24-alternating-current-circuits97ps1
mastering-physics-solutions-chapter-24-alternating-current-circuits97ps2

Chapter 24 Alternating Current Circuits Q.98GP
If the maximum voltage in the square wave shown in Figure is Vmax, what are (a) the average voltage, Vav, and (b) the rms voltage, Vrms?
mastering-physics-solutions-chapter-24-alternating-current-circuits98p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits98ps
mastering-physics-solutions-chapter-24-alternating-current-circuits98ps1

Chapter 24 Alternating Current Circuits Q.99GP
An ac generator supplies anrms voltage of 5.00 V to an RC circuit. At a frequency of 20.0 kHz the rms current in the circuit is 45.0 mA; at a frequency of 25.0 kHz the rms current is 50.0 mA. What are the vahtes of R and C in this circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits99ps
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Chapter 24 Alternating Current Circuits Q.100GP
An ac generator supplies an rms voltage of 5.00 V to an RL circuit. At a frequency of 20.0 kHz the rms current in the circuit is 45.0 mA; at afrequency of 25.0 kHz the rms current is 40.0 mA. What are the values of R and L in this circuit?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits100ps
mastering-physics-solutions-chapter-24-alternating-current-circuits100ps1

Chapter 24 Alternating Current Circuits Q.101GP
mastering-physics-solutions-chapter-24-alternating-current-circuits101p
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits101ps
mastering-physics-solutions-chapter-24-alternating-current-circuits101ps1

Chapter 24 Alternating Current Circuits Q.102PP
Suppose a theremin uses an oscillator with a fixed frequency of 90.1 MHz and an RLC circuit with R = 1.5 Ω, L = 2.08 μH, and C = 1.50pF. What is the beat frequency of these two oscillators? (Audio frequencies range from about 20 Hz to 20,000 Hz.)
A. 3740 Hz
B. 5100 Hz
C. 4760 Hz
D. 9000 Hz
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits101ps2
mastering-physics-solutions-chapter-24-alternating-current-circuits102ps

Chapter 24 Alternating Current Circuits Q.103PP
If the thereminist moves one of her fingers and increases the capacitance of the system slightly, does the beat frequency increase, decrease, or stay the same?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits103ps

Chapter 24 Alternating Current Circuits Q.104PP
Find the new beat frequency if the thereminist increases the capacitance by 0.100% over its value inProblem 102. All other quantities stay the same.
A. 761 Hz
B. 41,300 Hz
C. 41,900 Hz
D. 86,300 Hz
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits104ps

Chapter 24 Alternating Current Circuits Q.105PP
What is the rms current in the therernm’s RLC circuit (Problem) if it is attached to an ac generator with an rms voltage of 25.0 V and a frequency of 90.0 MHz?
A. 2.14 mA
B. 3.46 mA
C. 8.06 A
D. 16.7 A
Suppose a theremin uses an oscillator with a fixed frequency of 90.1 MHz and an RLC circuit with R = 1.5 Ω, L = 2.08 μH, and C = 1.50pF. What is the beat frequency of these two oscillators? (Audio frequencies range from about 20 Hz to 20,000 Hz.)
A. 3740 Hz
B. 5100 Hz
C. 4760 Hz
D. 9000 Hz
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits105ps

Chapter 24 Alternating Current Circuits Q.106IP
IP Referring to Example Suppose we would like to change the phase angle for this circuit to ϕ = −25.0°, and that we would like to accomplish this by changing the resistor to a value other than 175 Ω. The inductor is still 90.0 mH, the capacitor is 15.0 μ F, the rms voltage is 120.0 V, and the ac frequency is 60.0 Hz. (a) Should the resistance be increased or decreased? Explain. (b) Find the resistance that gives the desired phase angle. (c) What is the rms current in the circuit with the resistance found in part (b)?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits106ps
mastering-physics-solutions-chapter-24-alternating-current-circuits106ps1
mastering-physics-solutions-chapter-24-alternating-current-circuits106ps2

Chapter 24 Alternating Current Circuits Q.107IP
Referring to Example You plan to change the frequency of the generator in this circuit to produce a phase angle of smaller magnitude. The resistor is still 175 Ω, the inductor is 90.0 mH, the capacitor is 15.0 μ F,and the rms voltage is 120.0 V.
(a) Should you increase or decrease the frequency? Explain.
(b) Find the frequency that gives aphase angle of −22.5°.
(c) What is the rms current in the circuit at the frequency found in part (b)?
Solution:
mastering-physics-solutions-chapter-24-alternating-current-circuits107ps
mastering-physics-solutions-chapter-24-alternating-current-circuits107ps1
mastering-physics-solutions-chapter-24-alternating-current-circuits107ps2

Mastering Physics Solutions Chapter 23 Magnetic Flux and Faraday’s Law of Induction

Mastering Physics Solutions Chapter 23 Magnetic Flux and Faraday’s Law of Induction

Mastering Physics Solutions

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.1CQ
Explain the difference between a magnetic field and a magnetic flux.
Solution:
Magnetic field:
It is the amount of magnetic force experience by a charged particle moving with a velocity at a
given point in space
Magnetic flux:
It is the measure of the amount of magnetic field passing through a given area of any coil

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.1P
A 0.055-T magnetic field passes through a circular ring of radius 3.1 cm at an angle of 16° with the normal. Find the magnitude of the magnetic flux through the ring.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction1ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.2CQ
(Answers to odd-numbered Conceptual Questions can be found in the back of the book)
A metal ring with a break in its perimeter is dropped from a field-free region of space into a region with a magnetic fieId What effect does the magnetic field have on the ring?
Solution:
An induced emf will be developed in a conductor if it is moving in a magnetic field and hence some current flows through the conductor In the case of a broken ring, the magnetic field does
induce an emf between the ends of the broken ring, but the flow of the current will be prevented along the circumference because of the break in the ring.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.2P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction2p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction2ps
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Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.3CQ
In a common classroom demonstration, a magnet is dropped down a long. vertical copper tube. The magnet moves very slowly as it moves through the tube, taking several seconds to reach the bottom Explain this behavior
Solution:
The eddy current in the copper tube produces a magnetic field that opposes the direction of fall. Due to this repulsion. the magnet falls slowly taking much time to reach the bottom.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.3P
A magnetic field is oriented at an angle of 47° to the normal of a rectangular area 5.1 cm by 6.8 cm. If the magnetic flux through this surface has a magnitude of 4.8 × 10−5 T· m2, what is the strength of the magnetic field?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction3ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.4CQ
Many equal-arm balances have a small metal plate attached to one of the two arms. The plate passes between the poles of a magnet mounted in the base of the balance. Explain the purpose of this arrangement.
Solution:
An electrical current is induced in a piece of metal due to the relative motion of a nearby magnet is known as eddy current
The metal plate moving between the poles of a magnet experiences eddy currents that retard its motion. This helps to damp out oscillations in the balance, resulting in the more accurate readings.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.4P
Find the magnitude of the magnetic flux through the floor of a house that measures 22 m by 18 m. Assume that the Earth’s magnetic field at the location of the house has a horizontal component of 2.6 × 10−5 T pointing north, and a downward vertical component of 4.2 × 10−5 T.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction4ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.5CQ
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction5cqq
Solution:
When the switched in closed, the current in the wire coil produces a magnetic field in the iron rod. This increases the magnetic flux through the metal ring and a corresponding induced e.m.f.
The current produced by the induced e.m.f. generates a magnetic field opposite in direction to the field in the rod causing the ring to fly into the air.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.5P
The magnetic field produced by an MRI solenoid 2.5 m long and 1.2 m in diameter is 1.7 T. Find the magnitude of the magnetic flux through the core of this solenoid.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction5ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.6CQ
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction6cq
Solution:
The break prevents a current from circulating around the ring. This in turn will prevent the ring from experiencing a force that would throw it into the air.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.6P
At a certain location, the Earth’s magnetic field has a magnitude of 5.9 × 10−5 T and points in a direction that is 72° below the horizontal. Find the magnitude of the magnetic flux through the top of a desk at this location that measures 130 cm by 82 cm.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction6ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.7CQ
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction7cq
Solution:
The rod initially moves to left due to the downward current. As it moves, the motional emf which it generates will begin to oppose the emf of the battery. When both the emfs are balanced, the current stops flowing in the rod, from this point it moves with constant speed.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.7P
A solenoid with 385 turns per meter and a diameter of 17.0 cm has a magnetic flux through its core of magnitude 1.28 × 10−4 T · m2. (a) Find the current in this solenoid, (b) How would your answer to part (a) change if the diameter of the solenoid were doubled? Explain.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction7ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction7ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.8CQ
A penny is placed on edge in the powerful magnetic field of an MR1 solenoid. If the penny is tipped over, it takes several seconds for it to land on one of its faces. Explain.
Solution:
As the penny begins to tip over; there is a large change in magnetic flux due to magnetic field of the solenoid.
This change in flux produces induced current in the penny which opposes it from falling down. So it takes more seconds to land.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.8P
A single-turn square loop of side L is centered on the axis of a long solenoid. In addition, the plane of the square loop is perpendicular to the axis of the solenoid. The solenoid has 1250 turns per meter and a diameter of 6.00 cm, and carries a current of 2.50 A. Find the magnetic flux through the loop when (a) L = 3.00 cm, (b) L = 6.00 cm, and (c) L = 12.0 cm.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction8ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction8ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.9CQ
Recently, NASA tested a power generation system that involves connecting a small satellite to the space shuttle with a conducting wire several miles long. Explain how such a system can generate electrical power.
Solution:
Since the e.m.f. is given as the product of length of the wire, speed of the shuttle and perpendicular component of magnetic field.
The long conducting wire connected to the shuttle moves through a field can generate an induced e.m.f.
With large value of speed and length the induced emf is great enough to provide electrical power.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.9P
A 0.45-T magnetic field is perpendicular to a circular loop of wire with 53 turns and a radius of 15 cm. If the magnetic field is reduced to zero in 0.12 s, what is the magnitude of the induced emf?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction9ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.10CQ
Explain what happens when the angular speed of the coil in an electric generator is increased.
Solution:
When the angular speed of the coil in an electric generator is increased; the magnitude of the induced emf increases because induced emf is directly proportional to angular speed.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.10P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction10p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction10ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction10ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.11CQ
The inductor in an RL circuit determines how long it takes for the current to reach a given value, but it has no effect on the final value of the current. Explain.
Solution:
When current reaches a given value in an RL circuit, it stops changing; the back emf in an inductor vanishes.
So the final current in the circuit is determined by the resistor and the e.m.f. of the battery.
The inductor behaves like an ideal wire with zero resistance when the current reaches a given value.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.11P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction11p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction11ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction11ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.12CQ
When the switch in a circuit containing an inductor is opened, it is common for a spark to jump across the contacts of the switch. Why?
Solution:
An inductor acts to resist any change in the current whether increasing or decreasing.
When the switch in a circuit containing an inductor is opened, the inductor tries to maintain the original current, therefore the continuing current causes a spark to jump the gap.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.12P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction12p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction12ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.13P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction13p
Solution:
Given that, the figure in the given question shows four different situations in which a metal ring moves to the right with constant speed through a region with a varying magnetic field. The intensity of the color indicates the intensity of the field and in each case the field either increases or decreases at a uniform rate from the left edge of the colored region to the right edge.
For figure (1):–
The magnetic field is coming out of the page. As the ring moves and leaves the magnetic field which is coming out of the page, it will create an EMF that tries to account for this change. Hence the induced emf will be a anti clock wise or counter clockwise in order to create a magnetic field out of the page through the ring.
For figure (2):–
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction13ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.14P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction14p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction14ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction14ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.15P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction15p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction15ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction15ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction15ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.16P
A single conducting loop of wire has an area of 7.2 × 10−2 m2 and a resistance of 110 Ω. Perpendicular to the plane of the loop is a magnetic field of strength 0.48 T. At what rate (in T/s) must this field change if the induced current in the loop is to be 0.32 A?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction16ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction16ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.17P
The area of a 120-turn coil oriented with its plane perpendicular to a 0.20-T magnetic field is 0.050 m2. Find the average induced emf in this coil if the magnetic field reverses its direction in 0.34 s.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction17ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.18P
An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 4.25 s and the local 0.125-T magnetic field is perpendicular to the plane of the loop.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction18ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction18ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction18ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.19P
A magnetic field increases from 0 to 0.25 T in 1.8 s. How many turns of wire are needed in a circular coil 12 cm in diameter to produce an induced emf of 6.0 V?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction19ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction19ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.20P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction20p
Solution:
(a)
According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
At location (1) the ring is entering in the field and the induced current should oppose the entering the field by creating induced magnetic field. So the induced current is in
clock wise direction.
At location (2) the ring is in the uniform magnetic field. So the induced current is zero.
At location (3) the ring is leaving the field and the induced current should oppose the leaving the field by creating induced magnetic field. So the induced current is in
counter clock wise direction.
(b) So the at location (1) the induced current is clock wise direction and, at location (2) it is zero and at location (3) it is in counter clock wise direction.
Hence best explanation is (I).

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.21P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction21p
Solution:
(a) According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
At location (1) the ring is entering in the field and the induced current should oppose the entering the field by creating induced magnetic field. The induced magnetic field opposes the ring so it is in upward direction.
At location (2) the ring is in the uniform magnetic field. So the force is zero.
At location (3) the ring is leaving the field and the induced current should oppose the leaving the field by creating induced magnetic field. So the induced magnetic field opposes the ring to leave from field .So it is in upward direction.
(b) So the at location (1) the force is in upward direction and, at location (2) it is zero and at location (3) it is in upward direction.
Hence best explanation is (III)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.22P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction22p
(a) Is the retarding effect of eddy currents on the solid disk greater than, less than, or equal to the retarding effect on the slotted disk?
(b) Choose the best explanation from among the following:
I. The solid disk experiences a greater retarding force because eddy currents in it flow freely and are not interrupted by the slots.
II. The slotted disk experiences the greater retarding force because the slots allow more magnetic field to penetrate the disk.
III. The disks are the same size and made of the same material· therefore, they experience the same retarding force.
Solution:
(a) The retarding effect is greatest on the solid disk. Because the holes in the slotted disk interrupt the flow of eddy currents, and it is the eddy currents that produce the opposing magnetic field responsible for the retarding effect.
(b) The retarding effect is greatest on the solid disk. So the solid disk experiences a greater retarding force.
So the best explanation is (I)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.23P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction23p
Solution:
When the disk swing to the right as far as it can go, it is still in the constant magnetic field. Therefore the change in the magnetic field during its swing is zero. So the induced current in the disk is minimum (zero).

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.24P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction24p
Solution:
(a) According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
As the solid disk is swings from right to left where the magnetic the field is in to the page. So the induced current should produce the induced magnetic field which is opposite to the filed, that is filed must directed out of the page. Hence the induced current is in counter clock wise direction.
(b)The induced current should be in counter clock wise direction in order oppose the field by pointing out of the page.
Best explanation is(II)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.25P
A bar magnet with its north pole pointing downward is falling toward the center of a horizontal conducting ring. As viewed from above, is the direction of the induced current in the ring clockwise or counterclockwise? Explain.
Solution:
According to the Lenz’s law, the polarity of the induced EMF is such that it opposes the change in the magnetic flux which is the cause of its production.
When the north pole of a bar magnet is moving downward then the amount of magnetic flux linked with the ring increases. So, the current induced in such a direction that it opposes the increases in flux. This will happen when current is in the counterclockwise direction. Hence, the direction of the induced current is in the Counterclockwise direction.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.26P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction26p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction26pss
a) When the loop is above the magnet , the magnetic field increases and is directed
out of the page. According to Lenz law the current in the loop will oppose the
Increasing field by flowing clockwise
b) When the loop is below the magnet, the magnetic field decreases and is directed out of the page. According to Lenz law the current opposes the decreasing magnetic field by flowing in counter clock wise

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.27P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction27p
Solution:
a) The poles of the loop field line up with the magnet causing repulsive force, resisting to the loop which is moving down towards the magnet. So the tension in the string is less than the loops weight.
b) The poles of the loop field line up with the magnet causing attraction force , resisting to the loop to move down away from magnet So the tension in the string is again less than the loops weight.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.28P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction28p
Solution:
a) The poles of the loop field line up with the magnet causing repulsion and resisting to the magnet. So the tension in the string is greater than the loops weight.
b) The poles of the loop field line up with the magnet causing attraction and resisting the loop to move upward away from the magnet. So the tension again greater than the loops weight.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.29P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction29p
Solution:
a) Since the current in the wire is constant, the magnetic field does not vary with Time, so the induced current is zero
b) Since the current in the wire is increasing, the magnetic field through the circuit is increasing. Since the magnetic field is directed out of the page, the induced, is the circuit will induce magnetic field into the page. So the current flows is clock wise

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.30P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction30p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction30ps
If the current in the wire changes its direction, the direction of magnetic field will be into the page. According to Lenz law, the current induced in the circuit will oppose this change by flowing counter clock wise generating a field which is directed out of the page.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.1CQ
Explain the difference between a magnetic field and a magnetic flux.
Solution:
Magnetic field:
It is the amount of magnetic force experience by a charged particle moving with a velocity at a
given point in space
Magnetic flux:
It is the measure of the amount of magnetic field passing through a given area of any coil

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.1P
A 0.055-T magnetic field passes through a circular ring of radius 3.1 cm at an angle of 16° with the normal. Find the magnitude of the magnetic flux through the ring.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction1ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.2CQ
(Answers to odd-numbered Conceptual Questions can be found in the back of the book)
A metal ring with a break in its perimeter is dropped from a field-free region of space into a region with a magnetic fieId What effect does the magnetic field have on the ring?
Solution:
An induced emf will be developed in a conductor if it is moving in a magnetic field and hence some current flows through the conductor In the case of a broken ring, the magnetic field does
induce an emf between the ends of the broken ring, but the flow of the current will be prevented along the circumference because of the break in the ring.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.2P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction2p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction2ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction2ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction2ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.3CQ
In a common classroom demonstration, a magnet is dropped down a long. vertical copper tube. The magnet moves very slowly as it moves through the tube, taking several seconds to reach the bottom Explain this behavior
Solution:
The eddy current in the copper tube produces a magnetic field that opposes the direction of fall. Due to this repulsion. the magnet falls slowly taking much time to reach the bottom.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.3P
A magnetic field is oriented at an angle of 47° to the normal of a rectangular area 5.1 cm by 6.8 cm. If the magnetic flux through this surface has a magnitude of 4.8 × 10−5 T· m2, what is the strength of the magnetic field?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction3ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.4CQ
Many equal-arm balances have a small metal plate attached to one of the two arms. The plate passes between the poles of a magnet mounted in the base of the balance. Explain the purpose of this arrangement.
Solution:
An electrical current is induced in a piece of metal due to the relative motion of a nearby magnet is known as eddy current
The metal plate moving between the poles of a magnet experiences eddy currents that retard its motion. This helps to damp out oscillations in the balance, resulting in the more accurate readings.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.4P
Find the magnitude of the magnetic flux through the floor of a house that measures 22 m by 18 m. Assume that the Earth’s magnetic field at the location of the house has a horizontal component of 2.6 × 10−5 T pointing north, and a downward vertical component of 4.2 × 10−5 T.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction4ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.5CQ
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction5cqq
Solution:
When the switched in closed, the current in the wire coil produces a magnetic field in the iron rod. This increases the magnetic flux through the metal ring and a corresponding induced e.m.f.
The current produced by the induced e.m.f. generates a magnetic field opposite in direction to the field in the rod causing the ring to fly into the air.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.5P
The magnetic field produced by an MRI solenoid 2.5 m long and 1.2 m in diameter is 1.7 T. Find the magnitude of the magnetic flux through the core of this solenoid.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction5ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.6CQ
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction6cq
Solution:
The break prevents a current from circulating around the ring. This in turn will prevent the ring from experiencing a force that would throw it into the air.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.6P
At a certain location, the Earth’s magnetic field has a magnitude of 5.9 × 10−5 T and points in a direction that is 72° below the horizontal. Find the magnitude of the magnetic flux through the top of a desk at this location that measures 130 cm by 82 cm.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction6ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.7CQ
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction7cq
Solution:
The rod initially moves to left due to the downward current. As it moves, the motional emf which it generates will begin to oppose the emf of the battery. When both the emfs are balanced, the current stops flowing in the rod, from this point it moves with constant speed.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.7P
A solenoid with 385 turns per meter and a diameter of 17.0 cm has a magnetic flux through its core of magnitude 1.28 × 10−4 T · m2. (a) Find the current in this solenoid, (b) How would your answer to part (a) change if the diameter of the solenoid were doubled? Explain.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction7ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction7ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.8CQ
A penny is placed on edge in the powerful magnetic field of an MR1 solenoid. If the penny is tipped over, it takes several seconds for it to land on one of its faces. Explain.
Solution:
As the penny begins to tip over; there is a large change in magnetic flux due to magnetic field of the solenoid.
This change in flux produces induced current in the penny which opposes it from falling down. So it takes more seconds to land.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.8P
A single-turn square loop of side L is centered on the axis of a long solenoid. In addition, the plane of the square loop is perpendicular to the axis of the solenoid. The solenoid has 1250 turns per meter and a diameter of 6.00 cm, and carries a current of 2.50 A. Find the magnetic flux through the loop when (a) L = 3.00 cm, (b) L = 6.00 cm, and (c) L = 12.0 cm.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction8ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction8ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.9CQ
Recently, NASA tested a power generation system that involves connecting a small satellite to the space shuttle with a conducting wire several miles long. Explain how such a system can generate electrical power.
Solution:
Since the e.m.f. is given as the product of length of the wire, speed of the shuttle and perpendicular component of magnetic field.
The long conducting wire connected to the shuttle moves through a field can generate an induced e.m.f.
With large value of speed and length the induced emf is great enough to provide electrical power.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.9P
A 0.45-T magnetic field is perpendicular to a circular loop of wire with 53 turns and a radius of 15 cm. If the magnetic field is reduced to zero in 0.12 s, what is the magnitude of the induced emf?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction9ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.10CQ
Explain what happens when the angular speed of the coil in an electric generator is increased.
Solution:
When the angular speed of the coil in an electric generator is increased; the magnitude of the induced emf increases because induced emf is directly proportional to angular speed.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.10P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction10p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction10ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction10ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.11CQ
The inductor in an RL circuit determines how long it takes for the current to reach a given value, but it has no effect on the final value of the current. Explain.
Solution:
When current reaches a given value in an RL circuit, it stops changing; the back emf in an inductor vanishes.
So the final current in the circuit is determined by the resistor and the e.m.f. of the battery.
The inductor behaves like an ideal wire with zero resistance when the current reaches a given value.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.11P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction11p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction11ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction11ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.12CQ
When the switch in a circuit containing an inductor is opened, it is common for a spark to jump across the contacts of the switch. Why?
Solution:
An inductor acts to resist any change in the current whether increasing or decreasing.
When the switch in a circuit containing an inductor is opened, the inductor tries to maintain the original current, therefore the continuing current causes a spark to jump the gap.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.12P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction12p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction12ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.13P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction13p
Solution:
Given that, the figure in the given question shows four different situations in which a metal ring moves to the right with constant speed through a region with a varying magnetic field. The intensity of the color indicates the intensity of the field and in each case the field either increases or decreases at a uniform rate from the left edge of the colored region to the right edge.
For figure (1):–
The magnetic field is coming out of the page. As the ring moves and leaves the magnetic field which is coming out of the page, it will create an EMF that tries to account for this change. Hence the induced emf will be a anti clock wise or counter clockwise in order to create a magnetic field out of the page through the ring.
For figure (2):–
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction13ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.14P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction14p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction14ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction14ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.15P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction15p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction15ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction15ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction15ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.16P
A single conducting loop of wire has an area of 7.2 × 10−2 m2 and a resistance of 110 Ω. Perpendicular to the plane of the loop is a magnetic field of strength 0.48 T. At what rate (in T/s) must this field change if the induced current in the loop is to be 0.32 A?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction16ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction16ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.17P
The area of a 120-turn coil oriented with its plane perpendicular to a 0.20-T magnetic field is 0.050 m2. Find the average induced emf in this coil if the magnetic field reverses its direction in 0.34 s.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction17ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.18P
An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 4.25 s and the local 0.125-T magnetic field is perpendicular to the plane of the loop.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction18ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction18ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction18ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.19P
A magnetic field increases from 0 to 0.25 T in 1.8 s. How many turns of wire are needed in a circular coil 12 cm in diameter to produce an induced emf of 6.0 V?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction19ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction19ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.20P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction20p
Solution:
(a)
According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
At location (1) the ring is entering in the field and the induced current should oppose the entering the field by creating induced magnetic field. So the induced current is in
clock wise direction.
At location (2) the ring is in the uniform magnetic field. So the induced current is zero.
At location (3) the ring is leaving the field and the induced current should oppose the leaving the field by creating induced magnetic field. So the induced current is in
counter clock wise direction.
(b) So the at location (1) the induced current is clock wise direction and, at location (2) it is zero and at location (3) it is in counter clock wise direction.
Hence best explanation is (I).

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.21P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction21p
Solution:
(a) According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
At location (1) the ring is entering in the field and the induced current should oppose the entering the field by creating induced magnetic field. The induced magnetic field opposes the ring so it is in upward direction.
At location (2) the ring is in the uniform magnetic field. So the force is zero.
At location (3) the ring is leaving the field and the induced current should oppose the leaving the field by creating induced magnetic field. So the induced magnetic field opposes the ring to leave from field .So it is in upward direction.
(b) So the at location (1) the force is in upward direction and, at location (2) it is zero and at location (3) it is in upward direction.
Hence best explanation is (III)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.22P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction22p
(a) Is the retarding effect of eddy currents on the solid disk greater than, less than, or equal to the retarding effect on the slotted disk?
(b) Choose the best explanation from among the following:
I. The solid disk experiences a greater retarding force because eddy currents in it flow freely and are not interrupted by the slots.
II. The slotted disk experiences the greater retarding force because the slots allow more magnetic field to penetrate the disk.
III. The disks are the same size and made of the same material· therefore, they experience the same retarding force.
Solution:
(a) The retarding effect is greatest on the solid disk. Because the holes in the slotted disk interrupt the flow of eddy currents, and it is the eddy currents that produce the opposing magnetic field responsible for the retarding effect.
(b) The retarding effect is greatest on the solid disk. So the solid disk experiences a greater retarding force.
So the best explanation is (I)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.23P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction23p
Solution:
When the disk swing to the right as far as it can go, it is still in the constant magnetic field. Therefore the change in the magnetic field during its swing is zero. So the induced current in the disk is minimum (zero).

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.24P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction24p
Solution:
(a) According to Lenz law that an induced current flows in the direction that opposes the change that caused current.
As the solid disk is swings from right to left where the magnetic the field is in to the page. So the induced current should produce the induced magnetic field which is opposite to the filed, that is filed must directed out of the page. Hence the induced current is in counter clock wise direction.
(b)The induced current should be in counter clock wise direction in order oppose the field by pointing out of the page.
Best explanation is(II)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.25P
A bar magnet with its north pole pointing downward is falling toward the center of a horizontal conducting ring. As viewed from above, is the direction of the induced current in the ring clockwise or counterclockwise? Explain.
Solution:
According to the Lenz’s law, the polarity of the induced EMF is such that it opposes the change in the magnetic flux which is the cause of its production.
When the north pole of a bar magnet is moving downward then the amount of magnetic flux linked with the ring increases. So, the current induced in such a direction that it opposes the increases in flux. This will happen when current is in the counterclockwise direction. Hence, the direction of the induced current is in the Counterclockwise direction.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.26P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction26p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction26pss
a) When the loop is above the magnet , the magnetic field increases and is directed
out of the page. According to Lenz law the current in the loop will oppose the
Increasing field by flowing clockwise
b) When the loop is below the magnet, the magnetic field decreases and is directed out of the page. According to Lenz law the current opposes the decreasing magnetic field by flowing in counter clock wise

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.27P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction27p
Solution:
a) The poles of the loop field line up with the magnet causing repulsive force, resisting to the loop which is moving down towards the magnet. So the tension in the string is less than the loops weight.
b) The poles of the loop field line up with the magnet causing attraction force , resisting to the loop to move down away from magnet So the tension in the string is again less than the loops weight.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.28P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction28p
Solution:
a) The poles of the loop field line up with the magnet causing repulsion and resisting to the magnet. So the tension in the string is greater than the loops weight.
b) The poles of the loop field line up with the magnet causing attraction and resisting the loop to move upward away from the magnet. So the tension again greater than the loops weight.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.29P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction29p
Solution:
a) Since the current in the wire is constant, the magnetic field does not vary with Time, so the induced current is zero
b) Since the current in the wire is increasing, the magnetic field through the circuit is increasing. Since the magnetic field is directed out of the page, the induced, is the circuit will induce magnetic field into the page. So the current flows is clock wise

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.30P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction30p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction30ps
If the current in the wire changes its direction, the direction of magnetic field will be into the page. According to Lenz law, the current induced in the circuit will oppose this change by flowing counter clock wise generating a field which is directed out of the page.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.31P
A long, straight, current-carrying wire passes through the center of a circular coil. The wire is perpendicular to the plane of the coil, (a) If the current in the wire is constant, is the induced emf in the coil zero or nonzero? Explain, (b) If the current in the wire increases, is the induced emf in the coil zero or nonzero? Explain, (c) Does your answer to part (b) change if the wire no longer passes through the center of the coil but is still perpendicular to its plane? Explain.
Solution:
a) The magnetic field is parallel to the plane of the loop so the induced emf is zero
b) Though the current increases , the magnetic field is still parallel to the plane of the loop so the induced emf is zero
c)
Since the magnetic field is still parallel to the plane of the loop. So the answer part (b) does not change

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.32P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction32p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction32ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.33P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction33p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction33ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.34P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction34p
Solution:
Current carrying conductor produces magnetic field around the conductor The direction of magnetic force is given by magnetic-field right hand rule According to right hand rule, point your right hand thumb in the direction of current and curl your ringers around the wire The direction of ringers gives the direction of magnetic fieId
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction34ps
As shown in the above figure, the magnetic field due to current carrying conductor at ring A is out of the page and increasing. According to Faraday’s law, this increasing magnetic field creates the induced emf in the ring A.
According to Lenz’s law, the direction of magnetic field created by induced emf must oppose the original magnetic field. So this magnetic field due to induced emf must be into the page.
From right hand rule, to create the magnetic field that directed into the page, the induced current must directed clock wise.
Magnetic field in upper half of the ring B directed out of the page and lower half of the ring directed into the page. Hence the net magnetic field is equal to zero. So the induced emf in this loop is equal to zero.
As shown in the above figure, the magnetic field due to current carrying conductor at ring B is into the page and increasing. According to Faraday’s law, this increasing magnetic field creates the induced emf in the ring B.According to Lenz’s law, the direction of magnetic field created by induced emf must oppose the original magnetic field. So this magnetic field due to induced emf must be out of the page.
From right hand rule, to create the magnetic field that directed out the page, the induced current must directed counter clock wise.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.35P
A conducting rod slides on two wires in a region with a magnetic field. The two wires arc not connected. Is a force required to keep the rod moving with constant speed? Explain.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction35ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.36P
A metal rod 0.76 m long moves with a speed Of 2.0 m/s perpendicular to a magnetic field. If the induced emf between the ends of the rod is 0.45 V, what is the strength of the magnetic field?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction36ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.37P
A Boeing KC-135A airplane has a wingspan of 39.9 m and flies at constant altitude in a northerly direction with a speed of 850 km/h. If the vertical component of the Earth’s magnetic field is 5.0 × 10−6 T, and its horizontal component is 1.4 × 10−6 T, what is the induced emf between the wing tips?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction37ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction37ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.38P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction38p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction38ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.39P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction39p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction39ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction39ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.40P
(a) Find the current that flows in the circuit shown in Example. (b) What speed must the rod have if the current in the circuit is to be 1.0 A?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction40ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction40ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.41P
Suppose the mechanical power delivered to the rod in Example is 8.9 W. Find (a) the current in the circuit and (b) the speed of the rod.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction41ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction41ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.42P
The maximum induced em f in a generator rotating at 210 rpm is 45 V. Flow fast must the rotor of the generator rotate if it is to generate a maximum induced emf of 55 V?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction42ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.43P
A rectangular coil 25 cm by 35 cm has 120 turns. This coil produces a maximum emf of 65 V when it rotates with an angular speed of 190 rad/s in a magnetic field of strength B. Find the value of B.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction43ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.44P
A 1.6-m wire is wound into a coil with a radius of 3.2 cm. If this coil is rotated at 85 rpm in a 0.075-T magnetic field, what is its maximum emf?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction44ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.45P
A circular coil with a diameter of 22.0 cm and 155 turns rotates about a vertical axis with an angular speed of 1250 rpm. The only magnetic field in this system is that of the Earth. At the location of the coil, the horizontal component of the magnetic field is 3.80 × 10-5 T, and the vertical component is 2.85 × 10−5 T. (a) Which component of the magnetic field is important when calculating the induced emf in this coil? Explain, (b) Find the maximum emf induced in the coil.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction45ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.46P
A generator is designed to produce a maximum emf of 170 V while rotating with an angular speed of 3600 rpm. Each coil of the generator has an area of 0.016 m2. If the magnetic field used in the generator has a magnitude of 0.050 T, how many turns of wire are needed?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction46ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.47P
Find the induced em f when the current ill a 45.0-mH inductor increases from 0 to 515 mA in 16.5 ms.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction47ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.48P
How many turns should a solenoid of cross-sectional area 0.035 m2 and length 0.22 m have if its inductance is to be 45 mH?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction48ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.49P
The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoid? (b) What is the induced emf in the solenoid if its current drops from 3.2 A to 0 in 55 ms?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction49ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.50P
Determine the inductance of a solenoid with 640 turns in a length of 25 cm. The circular cross section of the solenoid has a radius of 4.3 cm.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction50ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.51P
A solenoid with a cross-sectional area of 1.81 × 10−3 m2 is 0.750 m long and has 455 turns per meter. Find the induced emf in this solenoid if the current in it is increased from 0 to 2.00 A in 45.5 ms.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction51ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction51ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction51ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.52P
A solenoid has N turns of area A distributed uniformly along its length, ℓ. When the current in this solenoid increases at the rate of 2.0 A/s, an induced emf of 75 mV is observed, (a) What is the inductance of this solenoid? (b) Suppose the spacing between coils is doubled. The result is a solenoid that is twice as long but with the same area and number of turns. Will the induced emf in this new solenoid be greater than, less than, or equal to 75 mV when the current changes at the rate of 2.0 A/s? Explain, (c) Calculate the induccd emf for part (b).
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction52ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction52ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.53P
How long does it take for the current in an RL circuit with R = 130 Ω and L = 68 mH to reach half its final value?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction53ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.54P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction54p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction54ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction54ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.55P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction55p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction55ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction55ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction55ps2
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction55ps3
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction55ps4

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.56P
The current in an RL circuit increases to 95% of its filial value 2.24 s after the switch is closed, (a) What is the time constant for this circuit? (b) If the inductance in the circuit is 0.275 H, what is the resistance?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction56ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.57P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction57p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction57ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction57ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction57ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.58P
The number of turns per meter in a solenoid of fixed length is doubled. At the same time, the current in the solenoid is halved. Does the energy stored in the inductor increase, decrease, or stay the same? Explain.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction58ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.59P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction59p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction59pss

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.60P
A solenoid is 1.5 m long and has 470 turns per meter. What is the cross-sectional area of this solenoid if it stores 0.31 J of energy when it carries a current of 12 A?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction60ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.61P
In the Alcator fusion experiment at MIT, a magnetic field of 50.0 T is produced, (a) What is the magnetic energy density in this field? (b) Find the magnitude of the electric field that would have the same energy density found in part (a).
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction61ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction61ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.62P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction62p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction62ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction62ps1
b) From the above equation, the energy stored in an inductor is inversely proportional to the square of the equivalent resistance. So the value of R should be less to store more energy in the inductor.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.63P
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction63p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction63pss
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction63pss1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction63pss2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.64P
Consider the circuit shown in Figure, which contains a 6.0-V battery, a 37-mH inductor, and four 55-Ω resistors, (a) Is more energy stored in the inductor just elfter the switch is closed or long after the switch is closed? Explain, (b) Calculate the energy stored in the inductor one characteristic time interval after the switch is closed, (c) Calculate the energy stored in the inductor long after the switch is closed.
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction64p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction64ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction64ps2
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction64ps3
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction64ps4
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction64ps5
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction64ps6

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.65P
You would like to store 9.9 J of energy in the magnetic field of a solenoid. The solenoid has 580 circular turns of diameter 7.2 cm distributed uniformly along its 28-cm length, (a) How much current is needed? (b) What is the magnitude of the magnetic field inside the solenoid? (c) What is the energy density (energy/volume) inside the solenoid?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction65ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction65ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction65ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.66P
Transformer 1 has a primary voltage Vp, and a secondary voltage Vs. Transformer 2 has twice the number of turns on both its primary and secondary coils compared with transformer 1. If the primary voltage on transformer 2 is 2Vp, what is its secondary voltage? Explain.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction66ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.67P
Transformer 1 has a primary current Ip and a secondary current Is. Transformer 2 has twice as many turns on its primary coil as transformer 1, and both transformers have the same number of turns on the secondary coil. If the primary current on transformer 2 is 3Ip, what is its secondary current? Explain.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction67ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction67ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.68P
The electric motor in a toy train requires a voltage of 3.0 V. Find the ratio of turns on the primary coil to turns on the secondary coil in a transformer that will step the 110-V household voltage down to 3.0 V.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction68ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.69P
A disk drive plugged into a 120-V outlet operates on a voltage of 9.0 V. The transformer that powers the disk drive has 125 turns on its primary coil, (a) Should the number of turns on the secondary coil be greater than or less than 125? Explain, (b) Find the number of turns on the secondary coil.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction69ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.70P
A transformer with a turns ratio (secondary/primary) of 1:18 is used to step down the voltage from a 120-V wall socket to be used in a battery recharging unit. What is the voltage supplied to the recharger?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction70ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.71P
A neon sign that requires a voltage of 11,000 V is plugged into a 120-V wall outlet. What turns ratio (secondary/primary) must a transformer have to power the sign?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction71ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.72P
A step-down transformer produces a voltage of 6.0 V across the secondary coil when the voltage across the primary coil is 120 V. What voltage appears across the primary coil of this transformer if 120 V is applied to the secondary coil?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction72ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction72ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.73P
A step-up transformer has 25 turns on the primary coil and 750 turns on the secondary coil. If this transformer is to produce an output of 4800 V with a 12-mA current, what input current and voltage are needed?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction73ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.74GP
An airplane flies level to the ground toward the north pole, (a) Is the induced emf from wing tip to wing tip when the plane is a t the equator greater them, less than, or equal to the wing-tip-to-wing-tip emf when it is at the latitude of New York? (b) Choose the best explanation from among the following:
I. The induced emf is the same because the strength of the Earth’s magnetic field is the same at the equator and at New York.
II. The induced emf is greater at New York because the vertical component of the Earth’s magnetic field is greater there than at the equator.
III. The induced emf is less at New York because at the equator the plane is flying parallel to the magnetic field lines.
Solution:
(a) At equator the plane is moving in the magnetic field direction so the induced emf is low.
When plane flying at the it moving perpendicular to the field as it is in latitude so the induced emf is high.
(Since the rate of change of magnetic flux for the air plane’s wings is the product of downward component of the magnetic field, wingspan and speed of plane.)
(b) At latitude of the vertical component of earth magnetic field is high and it is low at equator, so the induced emf is high at latitude of .
Hence best explanation is .(II)

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.75GP
You hold a circular loop of wire at the north magnetic pole of the Earth. Consider the magnetic flux through this loop due to the Earth’s magnetic field. Is the flux when the normal to the loop points horizontally greater than, less than, or equal to the flux when the normal points vertically downward? Explain.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction75ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.76GP
You hold a circular loop of wire at the equator. Consider the magnetic flux through this loop due to the Earth’s magnetic field. Is the flux when the normal to the loop points north greater than, less than, or equal to the flux when the normal points vertically upward? Explain.
Solution:
The flux through the loop is the greatest when the normal to the loop points the direction of the field.
We know that at the equator the field points to the north. When the normal to the loop points north, the direction of magnetic field and the normal are parallel. So the flux is maximum. But when the normal to the loop points vertically up, then the direction of magnetic field and the normal are perpendicular. Then the magnetic flux becomes zero. So flux when the normal point north is greater than when the normal points vertical upwards.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.77GP
The inductor shown in Figure is connected to an electric circuit with a changing current. At the moment in question, the inductor has an induced emf with the indicated direction. Is the current in the circuit at this time increasing and to the right, increasing and to the left, decreasing and to the right, or decreasing and to the left?
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction77p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction77ps
When the inductor has an inducei emf with the indicated direction, then the current in the inductor increases in the direction opposite to the increase in current of the circuit. Therefore. the current in the circuit can be either increasing to the right or decreasing to the left because the current in the inductor is directed towards left, that is. it is leaving the positive terminal

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.78GP
The Voyager I spacecraft moves through interstellar space with a speed of 8.0 × 103 m/s. The magnetic field in this region of space has a magnitude of 2.0 × 10-10 T. Assuming that the 5.0-m-long antenna on the spacecraft is at right angles to the magnetic field, find the induced emf between its ends.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction78ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.79GP
The coils used to measure the movements of a blowfly, as described in Section 23-5, have a diameter of 2.0 mm. In addition, the fly is immersed in a magnetic field of magnitude 0.15 mT. Find the maximum magnetic flux experienced by one of these coils.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction79ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.80GP
Computerized jaw tracking, or electrognathography (EGN), is air important tool for diagnosing and treating temporomandibular disorders (TMDs) that affect a person’s ability to bite effectively. The first step in applying EGN is to attach a small permanent magnet to the patient’s gum below the lower incisors. Then, as the jaw undergoes a biting motion, the resulting change in magnetic flux is picked up by wire coils placed on either side of the mouth, as shown in Figure. Suppose this person’s jaw moves to her right and that the north pole of the permanent magnet also points to her right. From her point of view, is the induced current in the coil to (a) her right and (b) her left clockwise or counterclockwise? Explain.
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction80p
Solution:
Given that, computerized jaw tracking or electrognathograph (EGN) is an important tool for diagnosing and treating temporomandibular disorders (TMDs) that affect a person’s ability to bite effectively.
The first step in applying EGN is to attach a small permanent magnet to the patients gum below the lower incisors. Then, as the jaw undergoes a biting motion, the resulting change in magnetic flux is picked up by wire coils place don either side of the mouth, as shown in figure given in the question.
Suppose this person’s jaw moves to her right and that the north pole of the permanent magnet also points to her right.
(a)
Here, the magnetic field lines leave the north pole and enter the south pole, there is a change in the magnetic flux, specifically and increase to the right. This change by decreasing the magnetic flux on her left and increasing on her right will be accounted by the coils. Thus, from her point of view, the current in the coil on her right will turn counter clock wise, forming a magnetic field that opposes the field created by the magnet in her teeth
(b)
To her left, the current in the coil will turn anti-clock wise forming a magnetic field that accounts for the increase in the magnetic flux on her left side.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.81GP
A rectangular loop of wire 24 cm by 72 cm is bent into em L shape, as shown in Figure. The magnetic field in the vicinity of the loop has a magnitude of 0.035 Tand points in a direction 25° below the y axis. The magnetic field has no x component. Find the magnitude of the magnetic flux through the loop.
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction81p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction81pss
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction81pss1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.82GP
A circular loop with a radius of 3.7 cm lies in the x-y plane. The magnetic field in this region of space is uniform and given by (a) What is the magnitude of the magnetic flux through this loop? (b) Suppose we now increase the x component of , leaving the other components unchanged. Does the magnitude of the magnetic flux increase, decrease, or stay the same? Explain, (c) Suppose, instead, that we increase the z component of , leaving the other components unchanged. Does the magnitude of the magnetic flux increase, decrease, or stay the same? Explain.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction82ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.83GP
Consider a rectangular loop of wire 5.8 cm by 8.2 cm in a uniform magnetic field of magnitude 1.3 T. The loop is rotated from a position of zero magnetic flux to a position of maximum flux in 21 ms. What is the average induced emf in the loop?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction83ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction83ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction83ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.84GP
A car with a vertical radio antenna 85 cm long drives due east at 25 m/s. The Earth’s magnetic field at this location has a magnitude of 5.9 × 10−5 T and points northward, 72° below the horizontal, (a) Ts the top or the bottom of the antenna at the higher potential? Explain, (b) Find the induced emf between the ends of the antenna.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction84ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.85GP
The rectangular coils in a 325-turn generator are 11 cm by 17 cm. What is the maximum emf produced by this generator when it rotates with an angular speed of 525 rpm in a magnetic field of 0.45 T?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction85ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.86GP
A cubical box 22 cm 011 a side is placed in a uniform 0.35-T magnetic field. Find the net magnetic flux through the box.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction86ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.87GP
Transcranial magnetic stimulation (IMS) is a non invasive method for studying brain function, and possibly for treatment as well. In this technique, a conducting loop is held near a person’s head, as shown in Figure. When the current in the loop is changed rapidly, the magnetic field it croates can change at the rate of 3.00 X 104 T/s. This rapidly changing magnetic field induces an electric current in a restricted region of the brain that can cause a finger to twitch, bright spots to appear in the visual field (magnetophosphenes), or a feeling of complete happiness to overwhelm a person. If the magnetic field changes at the previously mentioned rate over an area of 1.13 × 10 2 m2, what is the induced emf?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction87p
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction87ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.88GP
A magnetic field with the time dependence shown in Figure is at right angles to a 155-tum circular coil with a diameter of 3.75 cm. What is the induced emf in the coil at (a) t = 2.50 ms, (b) t = 7.50 ms, (c) t = 15.0 ms, and (d) t = 25.0 ms?
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction88p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction88ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction88ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction88ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.89GP
You would like to construct a 50.0-mH inductor by wrapping insulated copper wire (diameter = 0.0332 cm) onto a tube with a circular cross section of radius 2.67 cm. What length of wire is required if it is wrapped onto the tube in a single, close- packed layer?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction89ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction89ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.90GP
The time constant of an RL circuit with L = 25 mH is twice the time constant of an RC circuit with C = 45 μ F. Both circuits have the same resistance R. Find (a) the value of R and (b) the time constant of the RL circuit.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction90ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction90ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.91GP
A 6.0-V battery is connected in series with a 29-mH inductor a 110-Ω resistor, and an open switch, (a) How long after the switch is closed will the current in the circuit be equal to 12 mA? (b) How much energy is stored in the inductor when the current reaches its maximum value?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction91ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction91ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.92GP
A 9.0-V battery is connected in series with a 31-mH inductor, a 180-Ω. resistor, and an open switch, (a) What is the current in the circuit 0.120 ms after the switch is closed? (b) How much energy is stored in the inductor at this time?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction92ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.93GP
Suppose the fly described in Problem 79 turns through an angle of 90° in 37 ms. If the magnetic flux through one of the coils on the insect goes from a maximum to zero during this maneuver, and the coil has 85 turns of wire, find the magnitude of the induced emf.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction93ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.94GP
A conducting rod of mass m is in contact with two vertical conducting rails separated by a distance L, as shown in Figure. The entire system is immersed in a magnetic field of magnitude B pointing out of the page. Assuming the rod slides without friction, (a) describe the motion of the rod after it is released from rest, (b) What is the direction of the induced current (clockwise or countcrclockwise) in the circuit? (c) Find the speed of the rod after it has fallen for a long time.
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction94p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction94ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction94ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.95GP
A single-turn rectangular loop of width W and length L moves parallel to its length with a speed v. The loop moves from a region with a magnetic field perpendicular to the plane of the loop to a region where the magnetic field is zero, as shown in Figure. Find the rate of change in the magnetic flux through the loop (a) before it enters the region of zero field, (b) just after it enters the region of zero field, and (c) once it is fully within the region of zero field, (d) For each of the cases considered in parts (a), (b), and (c), state whether the induced current in the loop is clockwise, counterclockwise, or zero. Explain in each case.
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction95p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction95pss
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction95ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction95ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.96GP
The switch in the circuit shown in Figure is open initially, (a) Find the current in the circuit a long time after the switch is closed, (b) Describe the behavior of the lightbulb from the time the switch is closed until the current reaches the value found in part (a), (c) Now, suppose the switch is opened after having been closed for a long time, if the inductor is large, it is observed that the light flashes brightly and then burns out. Explain this behavior, (d) Find the voltage across the lightbulb just before and just after the switch is opened.
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction96p
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction96ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction96ps1
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction96ps2

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.97GP
An electric field E and a magnetic field B have the same energy density, (a) Express the ratio E/B in terms of the fundamental constants ε0 and μ0. (b) Evaluate E/B numerically, and compare your result with the speed of light.
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction97ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.98PP
“Smart” traffic lights are controlled by loops of wire embedded in the road (Figure). These “loop detectors” sense the change in magnetic field as a large metal object—such as a car or a truck—moves over the loop. Once the object is detected, electric circuits in the controller check for cross traffic, and then turn the light from red to green.
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction98p
A typical loop detector consists of three or four loops of 14- gauge wire buried 3 in. below the pavement. You can see the marks on the road where the pavement has been cut to allow for installation of the wires. There may be more than one loop detector at a given intersection; this allows the system to recognize that an object is moving as it activates first one detector and then another over a short period of time. If the system determines that a car has entered the intersection while the light is red, it can activate one camcra to take a picture of the car from the front—to see the driver’s face—and then a second camera to take a picture of the car and its license plate from behind. This red-light camera system was used to good effect during an exciting chase scene through the streets of London in the movie National Treasure: Book of Secrets.
Motorcycles are small enough that they often fail to activate the detectors, leaving the cyclist waiting and waiting for a green light. Some companies have begun selling powerful neodymium magnets to mount on the bottom of a motorcycle to ensure that they are “seen” by the detectors.
Suppose the downward vertical component of the magnetic field increases as a car drives over a loop detector. As viewed from above, is the induced current in the loop clockwise, counterclockwise, or zero?
Solution:
As the downward vertical component of magnetic field is increased due to the car, according to Lenz law the induced current opposes the increase in the magnetic field.
So from right hand rule (RHR) the current direction must be in counter clock wise

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.99PP
“Smart” traffic lights are controlled by loops of wire embedded in the road (Figure). These “loop detectors” sense the change in magnetic field as a large metal object—such as a car or a truck—moves over the loop. Once the object is detected, electric circuits in the controller check for cross traffic, and then turn the light from red to green.
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction99p
FIGURE
A typical loop detector consists of three or four loops of 14- gauge wire buried 3 in. below the pavement. You can see the marks on the road where the pavement has been cut to allow for installation of the wires. There may be more than one loop detector at a given intersection; this allows the system to recognize that an object is moving as it activates first one detector and then another over a short period of time. If the system determines that a car has entered the intersection while the light is red, it can activate one camcra to take a picture of the car from the front—to see the driver’s face—and then a second camera to take a picture of the car and its license plate from behind. This red-light camera system was used to good effect during an exciting chase scene through the streets of London in the movie National Treasure: Book of Secrets.
Motorcycles are small enough that they often fail to activate the detectors, leaving the cyclist waiting and waiting for a green light. Some companies have begun selling powerful neodymium magnets to mount on the bottom of a motorcycle to ensure that they are “seen” by the detectors.
A car drives onto a loop detector and increases the downward component of the magnetic field within the loop from 1.2 × 10−5, T to 2.6 × 10−5 T in 0.38 s. What is the induced emf in the detector if it is circular, has a radius of 0.67 m, and consists of four loops of wire?
A. 0.66 × 10−4 V
B. 1.5 × 10−4 V
C. 2.1 × 10−4 V
D. 6.2 × 10−4 V
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction99ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.100PP
“Smart” traffic lights are controlled by loops of wire embedded in the road (Figure). These “loop detectors” sense the change in magnetic field as a large metal object—such as a car or a truck—moves over the loop. Once the object is detected, electric circuits in the controller check for cross traffic, and then turn the light from red to green.
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction100p
FIGURE
A typical loop detector consists of three or four loops of 14- gauge wire buried 3 in. below the pavement. You can see the marks on the road where the pavement has been cut to allow for installation of the wires. There may be more than one loop detector at a given intersection; this allows the system to recognize that an object is moving as it activates first one detector and then another over a short period of time. If the system determines that a car has entered the intersection while the light is red, it can activate one camcra to take a picture of the car from the front—to see the driver’s face—and then a second camera to take a picture of the car and its license plate from behind. This red-light camera system was used to good effect during an exciting chase scene through the streets of London in the movie National Treasure: Book of Secrets.
Motorcycles are small enough that they often fail to activate the detectors, leaving the cyclist waiting and waiting for a green light. Some companies have begun selling powerful neodymium magnets to mount on the bottom of a motorcycle to ensure that they are “seen” by the detectors.
A truck drives onto a loop detector and increases the downward component of the magnetic field within the loop from 1.2 × 10−5 T to the larger value B in 0.38 s. The detector is circular, has a radius of 0.67 m, and consists of three loops of wire. What is B, given that the induced emf is 8.1 × 10−4 V?
A. 3.6 × 10−5 T
B. 7.3 × 10−5 T
C. 8.5×10−5T
D. 24 × 10−5T
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction100ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction100ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.101PP
“Smart” traffic lights are controlled by loops of wire embedded in the road (Figure). These “loop detectors” sense the change in magnetic field as a large metal object—such as a car or a truck—moves over the loop. Once the object is detected, electric circuits in the controller check for cross traffic, and then turn the light from red to green.
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction101p
FIGURE
A typical loop detector consists of three or four loops of 14- gauge wire buried 3 in. below the pavement. You can see the marks on the road where the pavement has been cut to allow for installation of the wires. There may be more than one loop detector at a given intersection; this allows the system to recognize that an object is moving as it activates first one detector and then another over a short period of time. If the system determines that a car has entered the intersection while the light is red, it can activate one camcra to take a picture of the car from the front—to see the driver’s face—and then a second camera to take a picture of the car and its license plate from behind. This red-light camera system was used to good effect during an exciting chase scene through the streets of London in the movie National Treasure: Book of Secrets.
Motorcycles are small enough that they often fail to activate the detectors, leaving the cyclist waiting and waiting for a green light. Some companies have begun selling powerful neodymium magnets to mount on the bottom of a motorcycle to ensure that they are “seen” by the detectors.
Suppose a motorcycle increases the downward component of the magnetic field within a loop only from 1.2 × 10-5T to 1.9 × 10-5 T. The detector is square, is 0.75 m on a side, and has four loops of wire. Over what period of time must the magnetic field increase if it is to induce an emf of 1.4 × 10-4 V?
A. 0.028 s
B. 0.11 s
C. 0.35 s
D. 0.60 s
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction101ps1

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.102IP
Suppose the ring is initially to the left of the field region, where there is no field, and is moving to the right. When the ring is partway into the field region, (a) is the induced current in the ring clockwise, counterclockwise, or zero, and (b) is the magnetic force exerted on the ring to the right, to the left, or zero? Explain.
Solution:
a) According to right handed rule ; the motion of the ring is towards the right
So, the current will be clock wise
b)
The magnetic forced exerted on the ring is to left the because v, B & F are mutually perpendicular to each other.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.103IP
Suppose the ring is completely inside the field region initially and is moving to the right, (a) Is the induced current in the ring clockwise, counterclockwise, or zero, and (b) is the magnetic force on the ring to the right, to the left, or zero? Explain. The ring now begins to emerge from the field region, still moving to the right, (c) Is the induced current in the ring clockwise, counterclockwise, or zero, and (d) is the magnetic force on the ring to the right, to the left, or zero? Explain.
Solution:
Given that, suppose the ring is completely inside the field region initially and is moving to the right
a) The induced current in the ring is zero actually since if the ring is completely inside the field region and moves to the right, then there is no change in the magnetic field and hence there is no change in the magnetic flux.
b) There is no force on the ring since there is no induced emf., there is no current is produced
c) If the ring is emerging from the field moving to the right, then the ring experiences a decrease in the magnetic field and hence produces or change in the magnetic flux. The ring will compensate for this change by generating a counter clockwise current to form a magnetic field out of the page.
d) As the ring leaves the magnetic field, a force will be generated to the left on the left side of the ring, while the right side of the ring will have zero force since it has dropped out of the field. Hence the net force generated by the ring will be to the left, counter acting the force moving the ring to the right.

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.104IP
(a) What external force is required to give the rod a speed of 3.49 m/s, every tiling else remaining the same? (b) What is the current in the circuit in this case?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction104ps

Chapter 23 Magnetic Flux and Faraday’s Law of Induction Q.105IP
Suppose the direction of the magnetic field is reversed. Everything else in the system remains the same, (a) Is the magnetic force exerted on the rod to the right, to the left, or zero? Explain, (b) Is the direction of the induced current clockwise, counterclockwise, or zero? Explain. (c) Suppose we now adjust the strength of the magnetic field until the speed of the rod is 2.49 m/s, keeping the force equal to 1.60 N. What is the new magnitude of the magnetic field?
Solution:
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction105ps
mastering-physics-solutions-chapter-23-magnetic-flux-and-faradays-law-of-induction105ps1

Mastering Physics Solutions Chapter 22 Magnetism

Mastering Physics Solutions Chapter 22 Magnetism

Mastering Physics Solutions

Chapter 22 Magnetism Q.1CQ
Two charged particles move at light angles to a magnetic field and deflect in opposite directions Can one conclude that the particles have opposite charges?
Solution:
No
The particles may have charge of the same sign but move in opposite directions along the same lina In this way, they would both move perpendicular to the field, but would deflect in opposite direction.

Chapter 22 Magnetism Q.1P
CE Predict/Explain Proton 1 moves with a speed v from the east coast to the west coast in the continental United States; proton 2 moves with the same speed from the southern United States toward Canada. (a) Is the magnitude of the magnetic force experienced by proton 2 greater than, less than, or equal to the force experienced by proton 1? (b) Choose the best explanation from among the following:
I. The protons experience the same force because the magnetic field is the same and their speeds are the same.
II. Proton 1 experiences the greater force because it moves at right angles to the magnetic field.
III. Proton 2 experiences the greater force because it moves in the same direction as the magnetic field.
Solution:
mastering-physics-solutions-chapter-22-magnetism1ps

Chapter 22 Magnetism Q.2CQ
An electron moves with constant velocity through a region of space that is free of magnetic fields. Can one conclude that the electric field is zero in this region? Explain.
Solution:
Ans:
Yes,
If an electric field exist in this region of space, and no magnetic field is present, the electric field will exert a force on the electron and causes it to accelerate.

Chapter 22 Magnetism Q.2P
CE An electron moves west to east in the continental United States Does the magnetic force experienced by the electron point ma direction that is generally north. south, east, west. upward. or downward? Explain
Solution:
Recall the Earth has its own magnetic fiela The field generated is similar to the one generated by a bar magnet with its pole near to the geographical pole of the EarthS Currently. the magnetic
north is located in northern Canada So. the magnetic field in the continental United States points primarily towards the north Consider an electron moving from west to east in the continental United States The velocity vector will point towards east and magnetic field point towards north
Use the right hand rule to find the direction of magnetic force. According to right hand rule, the direction of magnetic force will be the direction of your thumb when you curl your right hand
fingers from velocity vector to magnetic field vector So. here by right hand rule, the magnetic force will be towards upwards. However here the particle is an electron which is a negatively
charged particle. So. reverse the direction found by right hand rule to get the true direction of magnetic force. Therefore, here the magnetic force is on the electron is downwards

Chapter 22 Magnetism Q.3CQ
An electron moves with constant velocity through a region of space that is Free of electric fields. Can one conclude that the magnetic field is zero in this region? Explain.
Solution:
mastering-physics-solutions-chapter-22-magnetism3cqs

Chapter 22 Magnetism Q.3P
CE An electron moving in the positive x direction, at right angles to a magnetic field. experiences a magnetic force in the positive y direction What is the direction of the magnetic field?
Solution:
An electron moving in the positive x-direction at right angles to a magnetic field, experiences a magnetic force in the positive y-direction The magnetic field is in the positive z-direction

Chapter 22 Magnetism Q.4CQ
Describe how the motion of a charged particle can be used to distinguish between an electric and a magnetic field
Solution:
In a uniform electric field, the force on a charged particle is always in the same direction, leading to parabolic trajectories In a uniform magnetic field, the force of charged particles is always right angles to the motion, resulting the circular paths (or) helical trajectories
Perhaps even more important. a charged particle experiences a force due to an electric field whether it is moving (or) at rest in the magnetic field, the particle must be moving to experience a force.

Chapter 22 Magnetism Q.4P
mastering-physics-solutions-chapter-22-magnetism4p
Solution:
mastering-physics-solutions-chapter-22-magnetism4ps
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Chapter 22 Magnetism Q.5CQ
Explain how a charged particle moving in a circle of small radius can take the same amount of time to complete an orbit as an identical particle orbiting in a circle of large radius.
Solution:
The radius of curvature is proportional to the speed of the par1icle It follows that the particle moving in a circle of large radius has a proportionally larger speed than the particle moving in a
circle of small radius. Therefore, the time required for an orbit (t=d/v) is the same for both particles

Chapter 22 Magnetism Q.5P
mastering-physics-solutions-chapter-22-magnetism5p
Solution:
Apply the concept of force on a moving charged particle in uniform magnetic field. Apply the right hand rule to find the charge of the deflecting particle in the uniform magnetic field.
From right hand rule, the direction of the particle in the uniform magnetic field depends on its sign. The force exerted on a negatively charged particle is opposite in direction to the force exerted on a positively charged particle.
For particle A: Initially place the fingers of right hand in the direction of motion of the particle. As the fingers of right hand curls in the direction of the magnetic field that is outward, the thumb points downward. The downward deflection force is for a positively charged particle. Since the deflecting force on the particle is directed up, the charged particle A must be negative.
For particle B: Initially place the fingers of right hand in the direction of motion of the particle. As the fingers of right hand curls in the direction of the magnetic field that is outward, the thumb points downward. The downward deflection force is for a positively charged particle. Since the deflecting force on the particle is directed up, the charged particle B must be negative.
For particle C: Initially place the fingers of right hand in the direction of motion of the particle. As the fingers of right hand curls in the direction of the magnetic field that is outward, the thumb points downward. The downward deflection force is for a positively charged particle. Since the deflecting force on the particle is directed down, the charged particle C must be positive.

Chapter 22 Magnetism Q.6CQ
A current-carrying wire is placed in a region with a uniform magnetic field. The wire experiences zero magnetic force. Explain.
Solution:
mastering-physics-solutions-chapter-22-magnetism6cqs

Chapter 22 Magnetism Q.6P
mastering-physics-solutions-chapter-22-magnetism6p
Solution:
mastering-physics-solutions-chapter-22-magnetism6ps

Chapter 22 Magnetism Q.7P
What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.6 T?
Solution:
mastering-physics-solutions-chapter-22-magnetism7ps

Chapter 22 Magnetism Q.8P
An electron moves at right angles to a magnetic field of 0.18 T. What is its speed if the force exerted on it is 8.9 × 1015 N?
Solution:
mastering-physics-solutions-chapter-22-magnetism8ps

Chapter 22 Magnetism Q.9P
A negatively charged ion moves due north with a speed of 1.5 × 106 m/s at the Earth’s equator. What is the magnetic force exerted on this ion?
Solution:
mastering-physics-solutions-chapter-22-magnetism9ps

Chapter 22 Magnetism Q.10P
A proton high above the equator approaches the Earth moving straight downward with a speed of 355 m/s. Find the acceleration of the proton, given that the magnetic field at its altitude is 4.05 × 10−5 T.
Solution:
mastering-physics-solutions-chapter-22-magnetism10ps

Chapter 22 Magnetism Q.11P
A 0.32-μ C particle moves with a speed of 16 m/s through a region where the magnetic field has a strength of 0.95 T. At what angle to the field is the particle moving if the force exerted on it is (a) 4.8 × 10−6 N, (b) 3.0 × 10−6, or (c) 1.0 × 10−7 N?
Solution:
mastering-physics-solutions-chapter-22-magnetism11ps
mastering-physics-solutions-chapter-22-magnetism11ps1

Chapter 22 Magnetism Q.12P
A particle with a charge of 14 μ C experiences a force of 2.2 × 10−4 N when it moves at right angles to a magnetic field with a speed of 27 m/s. What force does this particle experience when it moves with a speed of 6.3 m/s at an angle of 25° relative to the magnetic field?
Solution:
mastering-physics-solutions-chapter-22-magnetism12ps

Chapter 22 Magnetism Q.13P
An ion experiences a magnetic force of 6.2 × 10−16 N when moving in the positive × direction but no magnetic force when moving in the positive y direction. What is the magnitude of the magnetic force exerted on the ion when it moves in the x−y plane along the line × = y? Assume that the ion’s speed is the same in all cases.
Solution:
mastering-physics-solutions-chapter-22-magnetism13ps
mastering-physics-solutions-chapter-22-magnetism13ps1

Chapter 22 Magnetism Q.14P
An electron moving with a speed of 4.2 × 105 m/s in the positive × direction experiences zero magnetic force. When it moves in the positive y direction, it experiences a force of 2.0 × 10−13 N that points in the negative z direction. What are the direction and magnitude of the magnetic field?
Solution:
mastering-physics-solutions-chapter-22-magnetism14ps

Chapter 22 Magnetism Q.15P
IP Two charged particles with different speeds move one at a time thro ugh a region of uni form magnetic field. The particles move in the same direction and experience equal magnetic forces, (a) If particle 1 has four tunes the charge of particle 2 which particle has the greater speed? Explain, (b) Find the ratio of the speeds, v1/v2.
Solution:
mastering-physics-solutions-chapter-22-magnetism15ps

Chapter 22 Magnetism Q.16P
A 6.60-μ C particle moves through a region of space where an electric field of magnitude 1250 N/C points in the positive x direction, and a magnetic field of magnitude 1.02 T points in the positive z direction. If the net force acting on the particle is 6.23 × 10−3 N in the positive x direction, find the magnitude and direction of the particle’s velocity. Assume the particle’s velocity is in the x-y plane.
Solution:
mastering-physics-solutions-chapter-22-magnetism16ps
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Chapter 22 Magnetism Q.17P
When at rest, a proton experiences a net electromagnetic force of magnitude 8.0 × 10−13 N pointing in the positive x direction. When the proton moves with a speed of 1.5 × 106 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.5 × 10−13 N, still pointing in the positive x direction. Find the magnitude and direction of (a) the electric field and (b) the magnetic field.
Solution:
mastering-physics-solutions-chapter-22-magnetism17ps
mastering-physics-solutions-chapter-22-magnetism17ps1
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Chapter 22 Magnetism Q.18P
CE A velocity selector is to be constructed using a magnetic field in the positive y direction. If positively charged particles move through the selector in the positive z direction, (a) what must be the direction of the electric field? (b) Repeat part (a) for the case of negatively charged particles.
Solution:
A positive charge always moves along the direction of electric field, but a negative charge moves opposite to the direction of electric field. Hence, from this we can say that the direction of electric field is opposite to the direction of motion of negative charge.
(a)
According to Right Hand Rule, if the direction of velocity of the positive charged particles is in positive z-direction and the direction of magnetic field is in positive y- direction then the direction of electric field is in positive x-direction.
(b)
According to Right Hand Rule, if the direction of velocity of the negative charged particles is in positive z-direction and the direction of magnetic field is in positive y- direction then the direction of electric field is in positive x-direction.

Chapter 22 Magnetism Q.19P
Find the radius of an electron’s orbit when it moves perpendicular to a magnetic field of 0.66 T with a speed of 6.27 × 105 m/s.
Solution:
mastering-physics-solutions-chapter-22-magnetism19ps

Chapter 22 Magnetism Q.20P
Find the radius of a proton’s orbit when it moves perpendicular to a magnetic field of 0.66 T with a speed of 6.27 × 105 m/s.
Solution:
mastering-physics-solutions-chapter-22-magnetism20ps

Chapter 22 Magnetism Q.21P
mastering-physics-solutions-chapter-22-magnetism21p
Solution:
mastering-physics-solutions-chapter-22-magnetism21ps

Chapter 22 Magnetism Q.22P
mastering-physics-solutions-chapter-22-magnetism22p
Solution:
mastering-physics-solutions-chapter-22-magnetism22ps
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Chapter 22 Magnetism Q.23P
IP BIO The artery in Figure 22–11 has an inside diameter of 2,75 mm and passes through a region where the magnetic field is 0.065 T. (a) If the voltage difference between the electrodes is 195 μ V, what is the speed of the blood? (b) Which electrode is at the higher potential? Does your answer depend on the sign of the ions in the blood? Explain.
Solution:
mastering-physics-solutions-chapter-22-magnetism23ps

Chapter 22 Magnetism Q.24P
An electron accelerated from rest through a voltage of 550 V enters a region of constant magnetic field, ff the electron follows a circularpath with a radius of 17 cm, what is the magnitude of the magnetic field?
Solution:
mastering-physics-solutions-chapter-22-magnetism24ps
mastering-physics-solutions-chapter-22-magnetism24ps1

Chapter 22 Magnetism Q.25P
A 12.5-μC particle with a mass of 2.80 × 10−5 kg moves perpendicular to a 1.01-T magnetic field in a circular path of radius 21.8 m. (a) How fast is the particle moving? (b) How long will it take the particle to complete one orbit?
Solution:
mastering-physics-solutions-chapter-22-magnetism25ps

Chapter 22 Magnetism Q.26P
mastering-physics-solutions-chapter-22-magnetism26p
Solution:
mastering-physics-solutions-chapter-22-magnetism26ps

Chapter 22 Magnetism Q.27P
A proton with a kinetic energy of 4.9 × 10 –16 J moves perpendicular to a magnetic field of 0.26 T. What is the radius of its circular path?
Solution:
mastering-physics-solutions-chapter-22-magnetism27ps

Chapter 22 Magnetism Q.28P
IP An alpha particle (the nucleus of a helium atom) consists of two protons and two neutrons, and has a mass of 6.64 × 10−27 kg. A horizontal beam of alpha particles is injected with a speed of 1.3 × 105 m/s into a region with a vertical magnetic field of magnitude 0.155 T. (a) How long does it take for an alpha particle to move halfway through a complete circle? (b) if the speed of the alpha particle is doubled, does the time found in part (a) increase, decrease, or stay the same? Explain, (c) Re-peat part (a) for alpha particles with a speed of 2.6 × 105 m/s.
Solution:
mastering-physics-solutions-chapter-22-magnetism28ps
mastering-physics-solutions-chapter-22-magnetism28ps1

Chapter 22 Magnetism Q.29P
mastering-physics-solutions-chapter-22-magnetism29ps
Solution:
mastering-physics-solutions-chapter-22-magnetism29ps1

Chapter 22 Magnetism Q.30P
What is the magnetic force exerted on a 2.15-m length of wire carrying a current of 0.899 A perpendicular to a magnetic field of 0.720 T?
Solution:
mastering-physics-solutions-chapter-22-magnetism30ps

Chapter 22 Magnetism Q.31P
A wire with a current of 2.8 Ais at an angle of 36.0° relative to a magnetic field of 0.88 T. Find the force exerted on a 2.25-m length of the wire.
Solution:
mastering-physics-solutions-chapter-22-magnetism31ps

Chapter 22 Magnetism Q.32P
The magnetic force exerted on a 1.2-m segment of straight wire is 1.6 N. The wire carries a current of 3.0 A in a region with a constant magnetic field of 0.50 T. What is the angle between the wire and the magnetic field?
Solution:
mastering-physics-solutions-chapter-22-magnetism32ps

Chapter 22 Magnetism Q.33P
A 0.45-m copper rod with a mass of 0.17 kg carries a current of 11 A in the positive × direction. What are the magnitude and direction of the minimum magnetic field needed to levitate the rod?
Solution:
mastering-physics-solutions-chapter-22-magnetism33ps
mastering-physics-solutions-chapter-22-magnetism33ps1

Chapter 22 Magnetism Q.34P
A wire with a length of 3.6 m and a mass of 0.75 kg is in a region of space with a magnetic field of 0.84 T. What is the minimum current needed to levitate the wire?
Solution:
mastering-physics-solutions-chapter-22-magnetism34p
mastering-physics-solutions-chapter-22-magnetism34ps
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Chapter 22 Magnetism Q.35P
A wire with a length of 3.6 m and a mass of 0.75 kg is in a region of space with a magnetic field of 0.84 T. What is the minimum current needed to levitate the wire?
Solution:
mastering-physics-solutions-chapter-22-magnetism35ps

Chapter 22 Magnetism Q.36P
A high-voltage power Une carries a current of 110 A at a location where the Earth’s magnetic field has a magnitude of 0.59 G and points to the north, 72° below the horizontal. Find the direction and magnitudeof the magnetic force exerted on a 250-m length of wire if the current in the wire flows (a) horizontally toward the east or (b) horizontally toward the south.
Solution:
mastering-physics-solutions-chapter-22-magnetism36ps
mastering-physics-solutions-chapter-22-magnetism36ps1

Chapter 22 Magnetism Q.37P
mastering-physics-solutions-chapter-22-magnetism37p
Solution:
mastering-physics-solutions-chapter-22-magnetism37ps
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Chapter 22 Magnetism Q.38P
mastering-physics-solutions-chapter-22-magnetism38p
Solution:
a) The force on the left vertical wire is out of the page, the force on the right vertical wire is into the page; the horizontal wire experiences zero force. As a result, the loop tends to rotate counter clockwise, as viewed from above.
b) This loop tends to rotate clockwise
c) This loop will not tend to rotate at all
mastering-physics-solutions-chapter-22-magnetism38ps

Chapter 22 Magnetism Q.39P
A rectangular loop of 260 turns is 33 cm wide and 16 cm high. What is the · tuent in this loop if the maximum torque in a field of 0.48 T is 23 N · m?
Solution:
mastering-physics-solutions-chapter-22-magnetism39ps

Chapter 22 Magnetism Q.40P
A single circular loop of radius 0.23 m carries a current of 2.6 Ain a magnetic field of 0.95 T. What is the maximum torque exerted on this loop?
Solution:
mastering-physics-solutions-chapter-22-magnetism40ps

Chapter 22 Magnetism Q.41P
In the previous problem, find the angle the plane of the loop must make with the field if the torque is to be half its maximum value.
Solution:
mastering-physics-solutions-chapter-22-magnetism41ps

Chapter 22 Magnetism Q.42P
mastering-physics-solutions-chapter-22-magnetism42p
Solution:
mastering-physics-solutions-chapter-22-magnetism42ps
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Chapter 22 Magnetism Q.43P
IP Two current loops, one square the other circular, have one turn made from wires of the same length. (a) If these loops carry the same current and are placed in magnetic fields of equal magnitude, is the maximum torque of the square loop greater than, less than, or the same as the maximum torque of the circular loop? Explain. (b) Calculate the ratio of the maximum torques, τsquare/τcirclc.
Solution:
mastering-physics-solutions-chapter-22-magnetism43ps

Chapter 22 Magnetism Q.44P
mastering-physics-solutions-chapter-22-magnetism44p
Solution:
mastering-physics-solutions-chapter-22-magnetism44pss
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Chapter 22 Magnetism Q.45P
· Find the magnetic field 6.25 cm from a long, straight wire that carries a current of 7.81 A.
Solution:
mastering-physics-solutions-chapter-22-magnetism45ps
mastering-physics-solutions-chapter-22-magnetism45ps1

Chapter 22 Magnetism Q.46P
A long, straight wire carries a current of 7.2 A. How far from this wire is the magnetic field it produces equal to the Earth’s magnetic field, which is approximately 5.0 × 10−5 T?
Solution:
mastering-physics-solutions-chapter-22-magnetism46ps

Chapter 22 Magnetism Q.47P
You travel to the north magnetic pole of the Earth, where the magnetic field points vertically downward. There, you draw a circle on the ground. Applying Ampere’s law to this circle show that zero current passes through its area.
Solution:
mastering-physics-solutions-chapter-22-magnetism47ps

Chapter 22 Magnetism Q.48P
Two power lines, each 270 m in length, runparallel to each other with a separation of 25 cm. If the lines carry parallel currents of 110 A, what are the magnitude and direction of the magnetic force each exerts on the other?
Solution:
mastering-physics-solutions-chapter-22-magnetism48ps

Chapter 22 Magnetism Q.49P
· BIO Pacemaker Switches Some pacemakers employ magnetic reed switches to enable doctors to change their mode of operation without surgery. A typical reed switch can be switched from one position to another with a magnetic field of 5.0 × 10−4 T. What current must a wire carry if it is to produce a 5.0 × 10−4 T field at a distance of 0.50 m?
Solution:
mastering-physics-solutions-chapter-22-magnetism49ps

Chapter 22 Magnetism Q.50P
mastering-physics-solutions-chapter-22-magnetism50p
Solution:
mastering-physics-solutions-chapter-22-magnetism50ps
mastering-physics-solutions-chapter-22-magnetism50ps1

Chapter 22 Magnetism Q.51P
mastering-physics-solutions-chapter-22-magnetism51p
Solution:
mastering-physics-solutions-chapter-22-magnetism51ps
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Chapter 22 Magnetism Q.52P
In Oersted’s experimen t, suppose that the compass was 0.25 m from the current-carrying wire. If a magnetic field of half the Earth’s magnetic field of 5.0 × 10−5 T was required to give a noticeable deflection of the compass needle, what current must the wire have carried?
Solution:
mastering-physics-solutions-chapter-22-magnetism52ps

Chapter 22 Magnetism Q.53P
IP Two long, straight wires are separated by a distance of 9.25 cm. One wire carries a current of 2.75 A, the other carries a current of 4.33 A. (a) Find the force per meter exerted on the 2.75-A wire, (b) Is the force per meter exerted on the 4.33-A wire greater than, less than, or the same as the force per meter exerted on the 2.75-A wire? Explain.
Solution:
mastering-physics-solutions-chapter-22-magnetism53ps

Chapter 22 Magnetism Q.54P
mastering-physics-solutions-chapter-22-magnetism54p
Solution:
mastering-physics-solutions-chapter-22-magnetism54ps
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Chapter 22 Magnetism Q.55P
mastering-physics-solutions-chapter-22-magnetism55p
Solution:
Use the concept of the right hand rule to find the direction of the electric current passing through the loop. The conventional current direction is always from the positive terminal to the negative terminal.
The loop creates the North Pole towards the south pole of the bar magnet as the loop attracts the bar magnet. From the right hand rule, the current in the loop is in a counter clockwise direction. Since the conventional current is always directed from the positive terminal, the terminal A must be positive of the battery.

Chapter 22 Magnetism Q.56P
· CE Predict/Explain The number of turns in a solenoid is doubled, and at the same time its length is doubled. Does the magnetic field within the solenoid increase, decrease, or stay the same? (b) Choose the best explanation from among the following:
I. Doubling the number of turns in a solenoid doubles its magnetic field, and hence the field increases.
II. Making a solenoid longer decreases its magnetic field, and therefore the field decreases.
III. The magnetic field remains the same because the number of turns per length is unchanged.
Solution:
mastering-physics-solutions-chapter-22-magnetism56ps

Chapter 22 Magnetism Q.57P
It is desired that a solenoid 38 cm long and with 430 turns produce a magnetic field within it equal to the Earth’s magnetic field (5.0 × 10−5 T). What current is required?
Solution:
mastering-physics-solutions-chapter-22-magnetism57ps

Chapter 22 Magnetism Q.58P
A solenoid that is 62 cm long produces a magnetic field of 1.3 T within its core when it carries a current of 8.4 A. How many turns of wire are contained in this solenoid?
Solution:
mastering-physics-solutions-chapter-22-magnetism58ps

Chapter 22 Magnetism Q.59P
The maximum current in a superconducting solenoid can be as large as 3.75 kA. If the number of turns per meter in such a solenoid is 3650, what is the magnitude of the magnetic field it produces?
Solution:
mastering-physics-solutions-chapter-22-magnetism59ps

Chapter 22 Magnetism Q.60P
To construct a solenoid, you wrap insulated wire uniformly around a plastic tube 12 cm in diameter and 55 cm in length. You would like a 2.0-A current to produce a 2.5-kG magnetic field inside your solenoid. What is the total length of wire you will need to meet these specifications?
Solution:
mastering-physics-solutions-chapter-22-magnetism60ps
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Chapter 22 Magnetism Q.61GP
CE At a point near the equator, the Earth’s magnetic field is horizontal and points to the north. If an electron is moving vertically upward at this point, does the magnetic force acting on it point north, south, east, west, upward, or downward? Explain.
Solution:
At a point near the equator, the Earth’s magnetic field is horizontal and points to the north. If the electron is moving vertically upward at this point, the magnetic force acting on the electron points to the east.

Chapter 22 Magnetism Q.62GP
CE A proton is to orbit the Earth at the equator using the Earth’s magnetic field to supply part of the necessary centripetal force. Should the proton move eastward or westward? Explain.
Solution:
The Earth’s magnetic field will be directed towards north at the equator. The necessary centripetal force to the proton has to be provided by the magnetic field of the earth. As the centripetal force is to be in the outward direction, the direction of the magnetic force will be in the inward direction. Therefore, from Fleming’s left hand rule, the proton has to move in the westward direction in order to orbit the earth at the equator.

Chapter 22 Magnetism Q.63GP
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Solution:
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Chapter 22 Magnetism Q.64GP
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Solution:
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Chapter 22 Magnetism Q.65GP
CE Each of the current-carrying wires in Figure 22–42 is long and straight, and carries the current I either into or out of the page, as shown. What is the direction of the net magnetic field produced by these three wires at the center of the triangle?
Solution:
mastering-physics-solutions-chapter-22-magnetism65ps
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Chapter 22 Magnetism Q.66GP
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Solution:
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Chapter 22 Magnetism Q.67GP
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Solution:
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Chapter 22 Magnetism Q.68GP
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Solution:
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Chapter 22 Magnetism Q.69GP
A stationary proton (q = 1.60 × 10−19 C) is Located between the poles of a horseshoe magnet, where the magnetic field is 0.35 T. What is the magnitude of the magnetic force acting on the proton?
Solution:
mastering-physics-solutions-chapter-22-magnetism69ps

Chapter 22 Magnetism Q.70GP
BIO Brain Function and Magnetic Fields Experiments have shown that thought processes in the brain can be affected if the parietal lobe is exposed to a magnetic field with a strength of 1.0 T. How much current must a long, straight wire carry if it is to produce a 1.0-T magnetic field at a distance of 0.50 m? (For comparison, a typical lightning bolt carries a current of about 20,000 A, which would melt most wires.)
Solution:
mastering-physics-solutions-chapter-22-magnetism70ps

Chapter 22 Magnetism Q.71GP
A mixture of two isotopes is injected into a mass spectrometer. One isotope follows a curved path of radius R1 = 48.9 cm; the other follows a curved path of radius R2 = 51.7 cm. Find the mass ratio, m1/m2, assuming that the two isotopes have the same charge and speed.
Solution:
mastering-physics-solutions-chapter-22-magnetism71ps

Chapter 22 Magnetism Q.72GP
High above the surface of the Earth, charged particles (such as electrons and protons) can become trapped in the Earth’s magnetic field in regions known as Van Alien belts. A typical electron in a Van Allen belt has an energy of 45 keV and travels in a roughly circular orbit with an average radius of 220 m. What is the magnitude of the Earth’s magnetic field where such an electron orbits?
Solution:
mastering-physics-solutions-chapter-22-magnetism72ps
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Chapter 22 Magnetism Q.73GP
Credit-Card Magnetic Strips Experiments carried out on the television show Mythbusters determined that a magnetic field of 1000 gauss is needed to corrupt the information on a credit card’s magnetic strip. (They also busted the myth that a credit card can be demagnetized by an electric eel or an eelskin wallet.) Suppose a long, straight wire carries a current of 3.5 A. How close can a credit card be held to this wire without damaging its magnetic strip?
Solution:
mastering-physics-solutions-chapter-22-magnetism73ps

Chapter 22 Magnetism Q.74GP
Superconducting Solenoid Cryomagnetics, Inc., advertises a high-field, superconducting solenoid that produces a magnetic field of 17 T with a current of 105 A. What is the number of turns per meter in this solenoid?
Solution:
mastering-physics-solutions-chapter-22-magnetism74ps

Chapter 22 Magnetism Q.75GP
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Solution:
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Chapter 22 Magnetism Q.76GP
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Solution:
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Chapter 22 Magnetism Q.77GP
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Solution:
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Chapter 22 Magnetism Q.78GP
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Solution:
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Chapter 22 Magnetism Q.79GP
mastering-physics-solutions-chapter-22-magnetism79p
Solution:
mastering-physics-solutions-chapter-22-magnetism79ps

Chapter 22 Magnetism Q.80GP
BIO Magnetic Resonance Imaging An MR1 (magnetic resonance imaging) solenoid produces a magnetic field of 1.5 T. The solenoid is 2.5 m long, 1.0 m in diameter, and wound with insulated wires 2.2 mm in diameter. Find the current that flows in the solenoid. (Your answer should be rather large. A typical MR1 solenoid uses niobium-titanium wire kept at liquid helium temperatures, where it is superconducting.)
Solution:
mastering-physics-solutions-chapter-22-magnetism80ps

Chapter 22 Magnetism Q.81GP
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Solution:
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Chapter 22 Magnetism Q.82GP
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Solution:
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Chapter 22 Magnetism Q.83GP
mastering-physics-solutions-chapter-22-magnetism83p
Solution:
mastering-physics-solutions-chapter-22-magnetism83pss

Chapter 22 Magnetism Q.84GP
IP Medical X-rays An electron hi a medical X-ray machine is accelerated from rest through a vol tage of 10.0 kV. (a) Find the maximum force a magnetic field of 0.957 T can exert on this electron. (b) If the voltage of the X-ray machine is increased, does the maximumforce found in part (a) increase, decrease, or stay the same? Explain. (c) Repeat part (a) for an electron accelerated through a potential of 25.0 kV.
Solution:
mastering-physics-solutions-chapter-22-magnetism84ps
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Chapter 22 Magnetism Q.85GP
A particle with a charge of 34 μC moves with a speed of 73 m/s in the positive x direction. The magnetic field in this region of space has a component of 0.40 T inthe positive y direction, and a component of 0.85 T in the positive z direction. What are the magnitude and direction of the magnetic force on the particle?
Solution:
mastering-physics-solutions-chapter-22-magnetism85ps
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Chapter 22 Magnetism Q.86GP
mastering-physics-solutions-chapter-22-magnetism86p
Solution:
mastering-physics-solutions-chapter-22-magnetism86ps
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Chapter 22 Magnetism Q.87GP
IP A charged particle moves in a horizontal plane with a speed of 8.70 × 106 m/s. When this particle encounters a uniform magnetic field in the vertical direction it begins to move on a circular path of radius 15.9 cm. (a) If the magnitude of the magnetic field is 1.21 T, what is the charge-to-mass ratio (q/m)of this particle? (b) If the radius of the circular path were greater than 15.9 cm, would the corresponding charge-to-mass ratio be greater than, less than, or the same as that found in part (a)? Explain. (Assume that the magnetic field remains the same.)
Solution:
mastering-physics-solutions-chapter-22-magnetism87ps

Chapter 22 Magnetism Q.88GP
mastering-physics-solutions-chapter-22-magnetism88p
Solution:
mastering-physics-solutions-chapter-22-magnetism88ps
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mastering-physics-solutions-chapter-22-magnetism88ps2
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Chapter 22 Magnetism Q.89GP
Repeat Problem 88 for the case where the current in wire 1 is reversed in direction.
Solution:
mastering-physics-solutions-chapter-22-magnetism89ps
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Chapter 22 Magnetism Q.90GP
Lightning Bolts A powerful bolt of lightningcan carry a current of 225 kA. (a) Treating a lightning bolt as along, thin wire, calculate the magnitude of the magnetic field produced by such a bolt of lightning at a distance of 35 m. (b) ff two such bolts strike simultaneously at a distance of 35 m from each other, what is the magnetic force per meter exerted by one bolt on the other?
Solution:
mastering-physics-solutions-chapter-22-magnetism90ps

Chapter 22 Magnetism Q.91GP
mastering-physics-solutions-chapter-22-magnetism91p
Solution:
mastering-physics-solutions-chapter-22-magnetism91ps
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Chapter 22 Magnetism Q.92GP
IP Consider the physical system shown in Figure 22—49. which consists of two current-carrying wires each with a length of 71 cm. (a) If the net magnetic field at the point A is out of the page. is the force between the wires attractive or repulsive? ExpIain (b) Calculate the magnitude of the force exerted by each wire on the other wire, given that the magnetic field at point A is out of the page with a magnitude of 2.1 x 1.0—6 T
Solution:
mastering-physics-solutions-chapter-22-magnetism92ps
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Chapter 22 Magnetism Q.93GP
Magnetars The astronomical object 4U014+61 has the distinction of creating the most powerful magnetic field ever observed. This object is referred to as a “magnetar” (a subclass of puisars), and its magnetic field is 1.3 × 1015 times greater than the Earth’s magnetic field. (a) Suppose a 2.5-m straight wire carrying a current of 1.1 A is placed in this magnetic field at an angle of 65° to the field lines. What force does this wire experience? (b) A field this strong can significantly change the behavior of an atom. To see this, consider an electron moving with a speed of 2.2 × 106 m/s. Compare the maximum magnetic force exerted on the electron to the electric force a proton exerts on an electron in a hydrogen atom. The radius of the hydrogen atom is 5.29 × 10−11 m.
Solution:
mastering-physics-solutions-chapter-22-magnetism93ps
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Chapter 22 Magnetism Q.94GP
mastering-physics-solutions-chapter-22-magnetism94p
Solution:
mastering-physics-solutions-chapter-22-magnetism94ps
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Chapter 22 Magnetism Q.95GP
IP A long, straight wire on the x axis carries a current of 3.12 A in the positive x direction. The magnetic field produced by the wive combines with a uniform magnetic field of 1.45 × 10−6 T that points in the positive z direction. (a) Is the net magnetic field of this system equal to zero at a point on the positive y axis or at a point on the negative y axis? Explain. (b) Find the distance from the wire to the point where the field vanishes.
Solution:
mastering-physics-solutions-chapter-22-magnetism95ps

Chapter 22 Magnetism Q.96GP
Find the angle between the plane of a loop and the magnetic field for which the magnetic torque acting on the loop is equal to × times its maximum value, where 0 ≤ x ≤ 1.
Solution:
mastering-physics-solutions-chapter-22-magnetism96ps

Chapter 22 Magnetism Q.97GP
Solenoids produce magnetic fields that arc relatively intense for the amount of current they carry. To make a direct comparison, consider a solenoid with 55 turns per centimeter, a radius of 1.0S cm, and a current of 0.622 A. (a) Find the magnetic field at the center of the solenoid. (b) What current must a long, straight wire carry to have the same magnetic field as that found in part (a)? Let the distance from the wire be the same as the radius of the solenoid, 1.05 cm.
Solution:
mastering-physics-solutions-chapter-22-magnetism97ps

Chapter 22 Magnetism Q.98GP
The current in a solenoid with 22 turns per centimeter is 0.50 A. The solenoid has a radius of 1.5 cm. A long, straight wire runs along the axis of the solenoid, carrying a current of 13 A. Find the magnitude of the net magnetic field a radial distance of 0.75 cm from the straight wire.
Solution:
mastering-physics-solutions-chapter-22-magnetism98ps

Chapter 22 Magnetism Q.99GP
IP BIO Transcranial Magnetic Stimulation A recently developed method to study brain function is to produce a rapidly changing magnetic field within the brain. When this technique, known as transcranial magnetic stimulation. (TMS), is applied to the prefrontal cortex, for example, it can reduce a person’s ability to conjugate verbs, though other thought processes are unaffected. The rapidly varying magnetic field is produced with a circular coil of 21 turns and a radius of 6.0 cm placed directly on the head. The current in this loop increases at the rate of 1.2 × 107 A/s (by discharging a capacitor). (a) At what rate does the magnetic field at the center of the coil increase? (b) Suppose a second coil with half the area of the first coil is used instead. Would your answer to part (a) increase, decrease, or stay the same? By what factor?
Solution:
mastering-physics-solutions-chapter-22-magnetism99ps
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Chapter 22 Magnetism Q.100GP
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Solution:
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Chapter 22 Magnetism Q.101GP
A thin ring of radius R and charge per length A rotates with an angular speed ω about an axis perpendicular to its plane and passing through its center. Find the magnitude of the magnetic field at the center of the ring.
Solution:
mastering-physics-solutions-chapter-22-magnetism101ps

Chapter 22 Magnetism Q.102GP
A solenoid is made from a 25-m length of wire of resistivity 2.3 × 10−8 Ω · m. The wire, whose radius is 2.1 mm, is wrapped uniformly onto a plastic tube 4.5 cm in diameter and 1.65 m iong. Find the emf to which the ends of the wire must be connected to produce a magnetic field of 0.015 T within the solenoid.
Solution:
mastering-physics-solutions-chapter-22-magnetism102ps
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Chapter 22 Magnetism Q.103GP
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Solution:
mastering-physics-solutions-chapter-22-magnetism103pss
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Chapter 22 Magnetism Q.104GP
Magnetic Fields in the Bohr Model hi the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius 5.29 × 10−11 m about the nucleus. Given that the charge on the electron is −1.60 × 10−19 C, and that its speed is 2.2 × 106 m/s, find the magnitude of the magnetic field the electron produces at the nucleus of the atom.
Solution:
mastering-physics-solutions-chapter-22-magnetism104ps

Chapter 22 Magnetism Q.105GP
A single-turn square loop carries a current of 18 A. The loop is 15 cm on a side and has a mass of 0.035 kg. Initially the loop lies flat on a horizontal tabletop. When a horizontal magnetic field is turned on, it is found that only one side of the loop experiences an upward force. Find the minimum magnetic field, Bmin, necessary to start tipping the loop up from the table.
Solution:
mastering-physics-solutions-chapter-22-magnetism105ps
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Chapter 22 Magnetism Q.106GP
mastering-physics-solutions-chapter-22-magnetism106p
Solution:
mastering-physics-solutions-chapter-22-magnetism106ps
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mastering-physics-solutions-chapter-22-magnetism106ps2
mastering-physics-solutions-chapter-22-magnetism106ps3

Chapter 22 Magnetism Q.107PP
To read and understand this sentence your brain must process visual input from your eyes and translate it into words and thoughts. As you do so, minute electric currents flow through the neurons in your visual cortex. These currents, like any electric current, produce magnetic fields. In fact, even your innermost thoughts and dreams produce magnetic fields that can be detected outside your head.
Magnetoencephalography (MEG) is the study of magnetic fields produced by electrical activity in the brain. Though completely noninvasive, MEG can provide detailed information on spontaneous brain function—like alpha waves and pathological epileptic spikes—as well as brain activity that is evoked by visual, auditory, and tactile stimuli.
The magnetic fields produced by brain activity are incredibly weak—roughly 100 million times smaller than the Earth’s magnetic field. Even so, sensitive detectors called SQUIDS (superconducting quantum interference devices), which were invented by physicists as a research tool, can detect fields as small as 1.0 × 10″15 T. Coupled with sophisticated electronics and software, and operating at liquid helium temperatures (−269 °C), SQUIDS can localize the source of brain activity to within millimeters. When the information from MEG is overlaid with the anatomical data from an MRI scan, the result is a richly detailed “map” of the electrical activity within the brain.
mastering-physics-solutions-chapter-22-magnetism107p
Solution:
mastering-physics-solutions-chapter-22-magnetism107ps

Chapter 22 Magnetism Q.108PP
To read and understand this sentence your brain must process visual input from your eyes and translate it into words and thoughts. As you do so, minute electric currents flow through the neurons in your visual cortex. These currents, like any electric current, produce magnetic fields. In fact, even your innermost thoughts and dreams produce magnetic fields that can be detected outside your head.
Magnetoencephalography (MEG) is the study of magnetic fields produced by electrical activity in the brain. Though completely noninvasive, MEG can provide detailed information on spontaneous brain function—like alpha waves and pathological epileptic spikes—as well as brain activity that is evoked by visual, auditory, and tactile stimuli.
The magnetic fields produced by brain activity are incredibly weak—roughly 100 million times smaller than the Earth’s magnetic field. Even so, sensitive detectors called SQUIDS (superconducting quantum interference devices), which were invented by physicists as a research tool, can detect fields as small as 1.0 × 10″15 T. Coupled with sophisticated electronics and software, and operating at liquid helium temperatures (−269 °C), SQUIDS can localize the source of brain activity to within millimeters. When the information from MEG is overlaid with the anatomical data from an MRI scan, the result is a richly detailed “map” of the electrical activity within the brain.
mastering-physics-solutions-chapter-22-magnetism108p
Solution:
mastering-physics-solutions-chapter-22-magnetism108ps

Chapter 22 Magnetism Q.109PP
To read and understand this sentence your brain must process visual input from your eyes and translate it into words and thoughts. As you do so, minute electric currents flow through the neurons in your visual cortex. These currents, like any electric current, produce magnetic fields. In fact, even your innermost thoughts and dreams produce magnetic fields that can be detected outside your head.
Magnetoencephalography (MEG) is the study of magnetic fields produced by electrical activity in the brain. Though completely noninvasive, MEG can provide detailed information on spontaneous brain function—like alpha waves and pathological epileptic spikes—as well as brain activity that is evoked by visual, auditory, and tactile stimuli.
The magnetic fields produced by brain activity are incredibly weak—roughly 100 million times smaller than the Earth’s magnetic field. Even so, sensitive detectors called SQUIDS (superconducting quantum interference devices), which were invented by physicists as a research tool, can detect fields as small as 1.0 × 10″15 T. Coupled with sophisticated electronics and software, and operating at liquid helium temperatures (−269 °C), SQUIDS can localize the source of brain activity to within millimeters. When the information from MEG is overlaid with the anatomical data from an MRI scan, the result is a richly detailed “map” of the electrical activity within the brain.
mastering-physics-solutions-chapter-22-magnetism109p
Solution:
mastering-physics-solutions-chapter-22-magnetism109ps

Chapter 22 Magnetism Q.110PP
To read and understand this sentence your brain must process visual input from your eyes and translate it into words and thoughts. As you do so, minute electric currents flow through the neurons in your visual cortex. These currents, like any electric current, produce magnetic fields. In fact, even your innermost thoughts and dreams produce magnetic fields that can be detected outside your head.
Magnetoencephalography (MEG) is the study of magnetic fields produced by electrical activity in the brain. Though completely noninvasive, MEG can provide detailed information on spontaneous brain function—like alpha waves and pathological epileptic spikes—as well as brain activity that is evoked by visual, auditory, and tactile stimuli.
The magnetic fields produced by brain activity are incredibly weak—roughly 100 million times smaller than the Earth’s magnetic field. Even so, sensitive detectors called SQUIDS (superconducting quantum interference devices), which were invented by physicists as a research tool, can detect fields as small as 1.0 × 10″15 T. Coupled with sophisticated electronics and software, and operating at liquid helium temperatures (−269 °C), SQUIDS can localize the source of brain activity to within millimeters. When the information from MEG is overlaid with the anatomical data from an MRI scan, the result is a richly detailed “map” of the electrical activity within the brain.
mastering-physics-solutions-chapter-22-magnetism110p
Solution:
mastering-physics-solutions-chapter-22-magnetism110ps
mastering-physics-solutions-chapter-22-magnetism110ps1

Chapter 22 Magnetism Q.111IP
IP Referring to Example 22–3 Suppose the speed of the isotopes is doubled. (a) Does the separation distance, d, increase, decrease, or stay the same? Explain. (b) Find the separation distance for this case.
Solution:
mastering-physics-solutions-chapter-22-magnetism111ps
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Chapter 22 Magnetism Q.112IP
· · IP Referring to Example 22–3 Suppose we change the initial speed of 238 U, leaving everything else the same. (a) If we want the separation distance to be zero, should the initial speed of U be increased or decreased? Explain. (b) Find the required A initial speed.
Solution:
mastering-physics-solutions-chapter-22-magnetism112ps

Chapter 22 Magnetism Q.113IP
Referring to Active Example 22-2 The current lx is adjusted until the magnetic field halfway between the wires has a magnitude of 2.5 × 10−6 T and points into the page. Everything else in the system remains the same as in Active Example 22–2. Find the magnitude and direction of I1.
Solution:
mastering-physics-solutions-chapter-22-magnetism113ps

Chapter 22 Magnetism Q.114IP
Referring to Active Example 22-2 The current I2 is adjusted until the magnetic field 5,5 cm below wire 2 has a magnitude of 2.5 × 10−6 T and points out of page. Everything else in the system remains the same as in Active Example 22–2. Find the magnitude and direction of I2
Solution:
mastering-physics-solutions-chapter-22-magnetism114ps
mastering-physics-solutions-chapter-22-magnetism114ps1