RCW7%20solutions

RCW7%20solutions - Physics 204, morning summer 2008...

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Unformatted text preview: Physics 204, morning summer 2008 Recitation ClassWork 7 solutions Chapter 22 Conceptual Questions #4,12 Concepts & Calculations #76,77 Chapter 25 Conceptual Questions #9 Concepts & Calculations #46,47 4. SSM REASONING AND SOLUTION A magnetic field B is necessary if there is to be a magnetic flux passing through a coil of wire. Yet, just because there is a magnetic field does not mean that a magnetic flux will pass through a coil. The general expression for magnetic flux is given by Equation 22.2: Φ = ( B cos φ ) A , where B is the magnitude of the magnetic field, A is the cross‐sectional area of the coil, and is the angle between the magnetic field B and the normal to the surface of the coil. Equation 22.2 shows that the flux depends only on the component of the magnetic field that is perpendicular to the surface of the coil. As shown in Example 4 and in Figure 22.10, when the coil is oriented so that it is parallel to the field, = 90°, B has no component normal to the surface of the coil, and the magnetic flux through the coil is zero. Therefore, the magnetic flux through a coil can be zero even though there is a magnetic field present. 12. REASONING AND SOLUTION The figure Metal ring shows the set‐up. Iron core Switch a. When the switch is open, there is no current in the coil and no magnetic field is present. When the switch in the circuit is closed, a current is established in the coil. Coil From RHR‐2, the magnetic field asso‐ciated – + with the current in the coil leaves the bottom of the coil and enters the top of the coil (see Battery drawing below). In other words, when there is a current in the coil, the bottom of the coil acts like the north pole of a bar magnet, and the top of the coil acts like the south pole of a bar magnet. The magnetic field of the coil causes a magnetic flux through the metal ring. In the very short time that it takes for the current to rise from zero to its steady value, the magnetic field associated with the current also rises from zero to its steady value. As the magnetic field increases, the magnetic flux through the metal ring changes, and there is an induced voltage and an induced current around the ring. From Lenz's law, the polarity of the induced voltage will be such that it opposes N the changing flux through the metal ring. Induced The induced magnetic field of the metal ring magnetic field S will be directed upward through the ring to subtract from the magnetic field of the coil I S that points downward through the ring. This Coil is illustrated in the figure at the right. To – preserve clarity, the iron core and a portion + of the circuit have not been shown. The N induced magnetic field of the metal ring is I Magnetic field similar to that of a bar magnet, with the due to coil north pole above the ring, and the south pole below the ring. Thus, the situation is similar to that of two bar magnets with their south poles next to each other. The two south poles repel, and the ring "jumps" upward. b. If the polarity of the battery were reversed, the ring still "jumps" upward when the switch is closed. When the polarity of the battery is reversed, and the switch is closed, the current moves through the coil in the opposite direction. The magnetic field associated with the current points in a direction that is opposite S to the direction of the magnetic field in part Induced (a). In other words, the top of the coil acts magnetic field like the north pole of a bar magnet, and the N Magnetic field bottom of the coil acts like the south pole of due to coil a bar magnet. I N When the switch is closed, the magnetic Coil flux through the metal ring changes, and I – + there is an induced voltage and an induced S current around the ring. In accord with Lenz's law, the induced magnetic field of the metal ring will be directed downward through the ring to subtract from the magnetic field of the coil that points upward through the ring. The induced magnetic field of the metal ring is similar to that of a bar magnet, with the south pole above the ring, and the north pole below the ring. The two north poles repel, and the ring "jumps" upward. 76. CONCEPT QUESTIONS a. The induced emf is zero during the second interval, 3.0 – 6.0 s. According to Faraday’s law of electromagnetic induction, Equation 22.3, an induced emf arises only when the magnetic flux changes. During this interval, the magnetic field, the area of the loop, and the orientation of the field relative to the loop are constant. Thus, the magnetic flux does not change, so there is no induced emf. b. An emf is induced during the first and third intervals, because the magnetic field is changing in time. The time interval is the same (3.0 s) for the two cases. However, the magnitude of the field changes more during the first interval. Therefore, the magnetic flux is changing at a greater rate in that interval, which means that the magnitude of the induced emf is greatest during the first interval. c. During the first interval the magnetic field in increasing with time. During the third interval, the field is decreasing with time. As a result, the induced emfs will have opposite polarities during these intervals. If the direction of the induced current is clockwise during the first interval, it will be counterclockwise during the third interval. SOLUTION a. The induced emf is given by Equations 22.3 and 22.3: 0–3.0 s: ξ =−N ⎛ BA cos φ − B0 A cos φ ΔΦ =−N⎜ ⎜ t − t0 Δt ⎝ ⎛ B − B0 = − NA cos φ ⎜ ⎜ t −t 0 ⎝ 3.0–6.0 s: ⎞ ⎟ ⎟ ⎠ ⎞ ⎛ 0.40 T − 0 T ⎞ 2 ⎟ = − ( 50 ) 0.15 m ( cos 0° ) ⎜ ⎟ = −1.0 V ⎟ ⎝ 3.0 s − 0 s ⎠ ⎠ ( ) ⎛ B − B0 ⎞ ⎛ 0.40 T − 0.40 T ⎞ 2 ⎟ = − ( 50 ) 0.15 m ( cos 0° ) ⎜ ⎟= 0V t − t0 ⎟ ⎝ 6.0 s − 3.0 s ⎠ ⎝ ⎠ ( ξ = − NA cos φ ⎜ ⎜ ) 6.0–9.0 s: ⎛ B − B0 ⎞ ⎛ 0.20 T − 0.40 T ⎞ 2 ⎟ = − ( 50 ) 0.15 m ( cos 0° ) ⎜ ⎟ = +0.50 V t − t0 ⎟ ⎝ 9.0 s − 6.0 s ⎠ ⎝ ⎠ ( ξ = − NA cos φ ⎜ ⎜ ) b. The induced current is given by Equation 20.2 as I = (Emf)/R. 0–3.0 s: I= ξ R = −1.0 V = −2.0 A 0.50 Ω 6.0–9.0 s: I= ξ R = +0.50 V = +1.0 A 0.50 Ω As expected, the currents are in opposite directions. 77. CONCEPT QUESTIONS a. The magnetic field produced by I extends throughout the space surrounding the loop. Using RHR‐2, it can be shown that the magnetic field is parallel to the normal to the loop. Thus, the magnetic field penetrates the loop and generates a magnetic flux. b. According to Faraday’s law of electromagnetic induction, an emf is induced when the magnetic flux through the loop is changing in time. If the current I is constant, the magnetic flux is constant, and no emf is induced in the loop. However, if the current is decreasing in time, the magnetic flux is decreasing and an induced current exists in the loop. c. No. Lenz’s law states that the induced magnetic field opposes the change in the magnetic field produced by the current I. The induced magnetic field does not necessarily oppose the magnetic field itself. Thus, the induced magnetic field does not always have a direction that is opposite to the direction of the field produced by I. SOLUTION At the location of the loop, the magnetic field produced by the current I is directed into the page (this can be verified by using RHR‐2). The current is decreasing, so the magnetic field is decreasing. Therefore, the magnetic flux that penetrates the loop is decreasing. According to Lenz’s law, the induced emf has a polarity that leads to an induced current whose direction is such that the induced magnetic field opposes this flux change. The induced magnetic field will oppose this decrease in flux by pointing into the page, in the same direction as the field produced by I. According to RHR‐2, the induced current must flow clockwise around the loop in order to produce such an induced field. The current then flows from left-to-right through the resistor. 78. CONCEPT QUESTIONS a. An emf is induced in the coil because the magnetic flux through the coil is changing in time. The flux is changing because the angle between the normal to the coil and the magnetic field is changing. b. The amount of induced current is equal to the induced emf divided by the resistance of the coil (see Equation 20.2). c. According to Equation 20.1, the amount of charge q that flows is equal to the induced current I multiplied by the time interval t = t − t0 during which the coil rotates, or q = I t − t0). SOLUTION According to Equation 20.1, the amount of charge that flows is q = I t. The current is related to the emf ξ in the coil and the resistance R by Equation 20.2 as I = ξ /R. The amount of charge that flows can, therefore, be written as ⎛ξ ⎞ Δq = I Δt = ⎜ ⎟ Δt ⎝R⎠ The emf is given by Faraday’s law of electromagnetic induction as ⎛ BA cos φ − BA cos φ0 ⎞ ⎛ ΔΦ ⎞ ⎟ ⎟=−N⎜ Δt ⎝ Δt ⎠ ⎝ ⎠ ξ =−N⎜ where we have also used Equation 22.2, which gives the definition of magnetic flux as Φ = BA cos φ . With this emf, the expression for the amount of charge becomes ⎡ ⎛ BA cos φ − BA cos φ0 ⎢ −N⎜ Δt ⎝ Δq = I Δt = ⎢ ⎢ R ⎣ Solving for the magnitude of the magnetic field yields ⎞⎤ ⎟⎥ ⎠ ⎥ Δt = − NBA ( cos φ − cos φ0 ) ⎥ R ⎦ ( ) − (140 Ω ) 8.5 × 10−5 C − R Δq = B= = 0.16 T NA ( cos φ − cos φ0 ) ( 50 ) 1.5 × 10−3 m 2 ( cos 90° − cos 0° ) ( ) 3. REASONING AND SOLUTION When parallel rays of light strike a concave mirror, they are reflected; these reflected rays converge at the focal point of the mirror. When parallel rays of light strike a convex mirror, they are also reflected; these reflected rays diverge from the mirror's surface and appear to originate from the focal point located behind the mirror. a. The earth‐sun distance is very large; therefore, when rays of light from the sun reach the earth, they are essentially parallel. If it is desired to start a fire with sunlight, it is necessary to focus the parallel light rays from the sun on a very small area, preferably a point, on the piece of paper. Since a concave mirror reflects parallel rays so that they converge in front of the mirror, a concave mirror, rather than a convex mirror, should be used. b. For best results, the piece of paper should be placed at the focal point of the mirror since this is the location where the rays converge to a point and the heating would be greatest. 7. REASONING AND SOLUTION a. The back side of a shiny teaspoon acts like a convex mirror. When an object is placed in front of a convex mirror, a virtual image is produced that is reduced in size and upright, as shown in Figure 25.22. Thus, when you look at the back side of a shiny teaspoon held at arms length, you will see yourself upright. b. The concave side of a shiny teaspoon acts like a concave mirror. If the teaspoon is held at arm's length, then you (the object) are farther from the reflecting surface than the center of curvature of the surface. The situation is similar to that of an object located beyond the center of curvature of a concave mirror. A real image is formed that is reduced in size and inverted with respect to the object, as shown in Figure 25.19b. Therefore, when you look at the concave side of a shiny teaspoon, you will see yourself upside down. 9. REASONING AND SOLUTION The microphone arrangement shown in the figure is used to pick up weak sounds. It consists of a "hollowed‐out" shell behind the mike. The shell acts like a mirror for sound waves. Therefore, when parallel rays of sound hit the inside surface of the shell, they will be reflected from it. Since the shell is concave, the reflected rays will converge at the focal point of the shell. Presumably, weak sounds will originate far from the microphone; therefore, when the they reach the microphone arrangement, these rays will be essentially parallel to the principal axis of the shell. Since the reflected rays will converge at the focal point, the microphone should be located at the focal point to detect them optimally. 46. CONCEPT QUESTIONS a. According to the discussion about relative velocity in Section 3.4, vIY = vIM + vMY. b. As you walk toward the stationary mirror, you perceive the mirror moving toward you in the opposite direction. Thus, vMY = –vYM. c. The velocity vYM has the components vYMx and vYMy, while the velocity vIM has the components vIMx and vIMy. The x direction is perpendicular to the mirror, and the two x components have the same magnitude. This is because the image in a plane mirror is always just as far behind the mirror as the object is in front of it. For instance, if an object moves 1 meter perpendicularly toward the mirror in 1 second, the magnitude of its velocity relative to the mirror is 1 m/s. But the image also moves 1 meter toward the mirror in the same time interval, so that the magnitude of its velocity relative to the mirror is also 1 m/s. The two x components, however, have opposite directions. The two y components have the same magnitude and the same direction. This is because an object moving parallel to a plane mirror has an image that remains at the same distance behind the mirror as the object is in front of it and moves in the same direction as the object. SOLUTION According to the discussion in Section 3.4, we have vIY = vIM + vMY = vIM – vYM This vector equation is equivalent to two equations, one for the x components and one for the y components. For the x direction, we note that vYMx = –vIMx. vIYx = vIMx − vYMx = − ( 0.90 m/s ) cos 50.0° − ( 0.90 m/s ) cos 50.0° = −1.2 m/s For the y direction, we note that vYMy = vIMy. vIYy = vIMy − vYMy = ( 0.90 m/s ) sin 50.0° − ( 0.90 m/s ) sin 50.0° = 0 m/s Since the y component of the velocity vIY is zero, the velocity of your image relative to you points in the –x direction and has a magnitude of 1.2 m/s . 47. CONCEPT QUESTIONS a. You will need a concave mirror. Since you must measure the image distance and image height of the tree, the image must be a real image. Only a concave mirror can produce a real image. b. The image distance of the sun is equal to the focal length of the mirror. The sun is extremely far away, so the light rays from it are nearly parallel when they reach the mirror. According to the discussion in Section 25.5, light rays near and parallel to the principal axis are reflected from a concave mirror and converge at the focal point. c. With numerical values for the focal length f of the mirror and the image distance di, the mirror equation (Equation 25.3) can be used to find the object distance do: 1 1 1 = − do f di The height ho of the tree is related to the height hi of its image by the magnification m; ho = hi/m. However, the magnification is given by the magnification equation (Equation 25.4) as m = −di/do, where di is the image distance and do is the object distance, both of which we know. SOLUTION a. The distance to the tree is given by the mirror equation as 1 1 1 1 1 = − = − d o f di 0.9000 m 0.9100 m so d o = 82 m b. Since ho = hi/m and m = −di/do, we have that ho = hi hi = m ⎛ di ⎜− d ⎝ o ⎛ d = hi ⎜ − o ⎞ ⎝ di ⎟ ⎠ ⎞ ⎟ ⎠ Now hi = −0.12 m, where the minus sign has been used since the image is inverted relative to the tree. Thus, the height of the tree is ⎛ d ho = hi ⎜ − o ⎝ di ⎞ 82 m ⎞ ⎛ ⎟ = 11 m ⎟ = ( −0.12 m ) ⎜ − ⎝ 0.9100 m ⎠ ⎠ ...
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This note was uploaded on 10/04/2009 for the course PHYS 750:204 taught by Professor Rollino during the Spring '09 term at Rutgers.

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