Physics Solution Manual for 1100 and 2101

# 40 t 0 t 2 50 015 m cos 0 10 v 30 s 0

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Unformatted text preview: e 1190 ELECTROMAGNETIC INDUCTION ∆tA = ∆tB = ∆t , ∆tC = 2∆t . Dividing gives the following results for the magnitudes of the emfs: ξ A = BL2 B 2 L2 B 2 L2 BL2 . = , ξB = , ξC = ∆t ∆t 2∆t ∆t 8. (a) An induced current appears only when there is an induced emf to drive it around the loop. According to Faraday’s law, an induced emf exists only when the magnetic flux through the loop changes as time passes. Here, however, there is no magnetic flux through the loop. The magnetic field lines produced by the current are circular and centered on the wire, with the planes of the circles perpendicular to the wire. Therefore, the magnetic field is always parallel to the plane of the loop as the loop falls and never penetrates the loop. In other words, no magnetic flux passes through the loop. No magnetic flux, no induced emf, no induced current. 9. (d) When the switch is closed, current begins to flow counterclockwise in the larger coil, and the field that it creates appears inside the smaller coil. Using RHR-2 reveals that this field points out of the screen toward you. According to Lenz’s law, the induced current in the smaller coil flows in such a direction that it creates an induced field that opposes the growth of the field from the larger coil. Thus, the induced field must point into the screen away from you. Using RHR-2 reveals that the induced current must, then, flow clockwise. The induced current exists only for the short period following the closing of the switch, when the field from the larger coil is growing from zero to its equilibrium value. Once the field from the larger coil reaches its equilibrium value and ceases to change, the induced current in the smaller coil becomes zero. 10. (b) The peak emf is proportional to the area A of the coil, according to Equation 22.4. Thus, we need to consider the areas of the coils. The length of the wire is L and is the same for each of the coil shapes. For the circle, the circumference is 2π r = L, so that the area is 2 2 L2 L2 L L . For the square, the area is Asquare = = . For the Acircle = π r 2 = π = 4π 16 2π 4 L 2 L L2 rectangle, the perimeter is 2(D + 2D) = L, so that the area is Arectangle = = . The 6 6 18 circle has the largest area, while the rectangle has the smallest area, corresponding to answer b. 11. 5.3 cm 12. (d) The back emf is proportional to the motor speed, so it decreases when the speed V −ξ decreases. The current I drawn by the motor is given by Equation 22.5 as I = , where R V is the voltage at the socket, ξ is the back emf, and R is the resistance of the motor coil. As ξ decreases, I increases. Chapter 22 Answers to Focus on Concepts Questions 13. (c) According to Equation 22.7, the mutual inductance is M = 1191 −ξS∆t . If the time interval is ∆I P cut in half and the change in the primary current is doubled, while the induced emf remains the same, the mutual inductance must be reduced by a factor of four. 14. (b) The energy stored in an inductor is given by Equation 22.10 as Energy = 1 LI 2 . Since 2 the two inductors store the same amount of energy, we have I1 I2 = L2 L1 = L2 L2 / 2 1 L I2 2 11 2 = 1 L2 I 2 . 2 Thus, = 2 = 1.414 . LI . Since Ф is the same for each coil, the Φ number of turns is proportional to the product LI of the inductance and the current. For the coils specified in the table, this product is (LI)A = L0I0, (LI)B = L0I0/2, (LI)C = 4L0I0. 15. (e) According to Equation 22.8, we have N = 16. (c) The current in the primary is proportional to the current in the secondary according to Equation 22.13: I P = IS NS / NP . The current in the secondary is the secondary voltage divided by the resistance, according to Ohm’s law. Thus, when the resistance increases, the current in the secondary decreases and so does the current in the primary. The wall socket delivers to the primary the same power that the secondary delivers to the resistance, assuming that no power is lost within the transformer. The power delivered to the resistance is given by Equation 20.15c as the square of the secondary voltage divided by the resistance. When the resistance increases, the power decreases. Hence, the power delivered to the primary by the wall socket also decreases. 17. 0.31 W 18. (a) The current in resistor 2 (without the transformer) is the same as the current in resistor 1 (with the transformer). In either event, the current I is I = V/R, where V is the voltage across the resistance R. Since the transformer is a step-up transformer, the voltage applied across resistor 2 is smaller than the voltage applied across resistor 1. The smaller voltage across resistor 2 can lead to the same current as does the greater voltage across resistor 1 only if R2 is less than R1. 1192 ELECTROMAGNETIC INDUCTION CHAPTER 22 ELECTROMAGNETIC INDUCTION PROBLEMS ______________________________________________________________________________ 1. REASONING AND SOLUTION Using Equation 22.1, we find ξ = vBL = (220 m/s)(5.0 × 10–6 T)(59 m) = 0.065 V _______________________________...
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