Physics Solution Manual for 1100 and 2101

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Unformatted text preview: mf. When the motor is just turned on, there is no back emf, so the current is just the applied voltage divided by the resistance of the wire. To limit the starting current to 15.0 A, a resistor R1 is placed in series with the resistance of the wire, so the equivalent resistance is R + R1. The current to the motor is equal to the applied voltage divided by the equivalent resistance. SOLUTION a. According to Equation 22.5, the back emf ξ generated by the motor is ξ = V − I R = 120.0 V − ( 7.00 A ) ( 0.720 Ω ) = 115 V b. When the motor has been just turned on, the back emf is zero, so the current is I= V − ξ 120.0 V − 0 V = = 167 A R 0.720 Ω c. When a resistance R1 is placed in series with the resistance R of the wire, the equivalent resistance is R1 + R. The current to the motor is I= V −ξ R1 + R Chapter 22 Problems 1217 Solving this expression for R1 gives V −ξ 120.0 V − 0 V −R= − 0.720 Ω = 7.28 Ω I 15.0 A ______________________________________________________________________________ R1 = 41. SSM WWW REASONING AND SOLUTION a. From the drawing, we see that the period of the generator (the time for one full cycle) is 0.42 s; therefore, the frequency of the generator is f= 1 1 = = 2.4 Hz T 0.42 s b. The angular speed of the generator is related to its frequency by ω = 2π f , so the angular speed is ω = 2π (2.4 Hz) = 15 rad/s c. The maximum emf ξ 0 induced in a rotating planar coil is given by ξ 0 = NABω (see Equation 22.4). The magnitude of the magnetic field can be found by solving this expression for B and noting from the drawing that ξ 0 = 28 V: B= ξ0 28 V = = 0.62 T NAω (150)(0.020 m 2 )(15 rad/s) ______________________________________________________________________________ 42. REASONING The peak emf ξ0 of a generator is found from ξ 0 = NABω (Equation 22.4), where N is the number of turns in the generator coil, A is the coil’s cross-sectional area, B is the magnitude of the uniform magnetic field in the generator, and ω is the angular frequency of rotation of the coil. In terms of the frequency f (in Hz), the angular frequency is given by ω = 2π f (Equation 10.6). Substituting Equation 10.6 into Equation 22.4, we obtain ξ0 = NAB ( 2π f ) = 2π NABf (1) When the rotational frequency f of the coil changes, the peak emf ξ0 also changes. The quantities N, A, and B remain constant, however, because they depend on how the generator is constructed, not on how rapidly the coil rotates. We know the peak emf of the generator at one frequency, so we will use Equation (1) to determine the peak emf for a different frequency in part (a), and the frequency needed for a different peak emf in part (b). 1218 ELECTROMAGNETIC INDUCTION SOLUTION a. Solving Equation (1) for the quantities that do not change with frequency, we find that ξ0 f = 2π24 1 NAB 43 (2) Same for all frequencies The peak emf is ξ0,1 = 75 V when the frequency is f1 = 280 Hz. We wish to find the peak emf ξ0,2 when the frequency is f2 = 45 Hz. From Equation (2), we have that ξ0,2 f2 ξ0,1 = 2π24 = 1 NAB 43 f1 Same for all (3) frequencies Solving Equation (3) for ξ0,2, we obtain f2 45 Hz ξ0,1 = ( 75 V ) = 12 V 280 Hz f1 ξ0,2 = b. Letting ξ0,3 = 180 V, Equation (2) yields ξ0,3 f3 = ξ0,1 f1 (4) Solving Equation (4) for f3, we find that ξ0,3 180 V f3 = f1 = ( 280 Hz ) = 670 Hz ξ0,1 75 V 43. REASONING The number N of turns in the coil of a generator is given by N = ξ0 ABω (Equation 22.4), where ξ 0 is the peak emf, A is the area per turn, B is the magnitude of the magnetic field, and ω is the angular speed in rad/s. We have values for A, and B. Although we are not given the peak emf ξ 0 , we know that it is related to the rms emf, which is known: ξ 0 = 2 ξ rms (Equation 20.13). We are also not given the angular speed ω, but we know that it is related to the frequency f in hertz according to ω = 2π f (Equation 10.6). Chapter 22 Problems 1219 SOLUTION Substituting Equation 20.13 for ξ 0 and Equation 10.6 for ω into Equation 22.4 gives N= ξ0 2 ξ rms 2 (120 V ) = = = 3.0 ×105 2 −5 ABω AB 2π f 0.022 m 6.9 × 10 T 2π ( 60.0 Hz ) ( )( ) 44. REASONING AND SOLUTION Using Equation 22.5 to take the back emf into account, we find V – ξ (120.0 V ) – ( 72.0 V ) = = 16 Ω R= I 3.0 A ______________________________________________________________________________ 45. REASONING The length of the wire is the number N of turns times the length per turn. Since each turn is a square that is a length L on a side, the length per turn is 4L, and the total length is N(4L). Since the area of a square is A = L2, the length of a side of the square can be ( ) obtained as L = A , so that the total length is N 4 A . The area can be found from the peak emf, which is ξ 0 = NABω , according to Equation 22.4. Solving this expression for A and substituting the result into the expression for the total length gives ξ0 Nξ 0 Total length = N 4 A = N 4 =4 NBω Bω ( ) (1) Although we are not given the peak emf ξ 0 , we know that it is related to the...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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