Physics Solution Manual for 1100 and 2101

0 v solution solving equation 233 for xl we obtain xl

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Unformatted text preview: ed by the picture tube means that we know the power provided by the secondary. Since power is current times voltage (Equation 20.15a), we know that P = ISVS . This expression can be solved for VS, and the result can be substituted into Equation 22.12 to give the turns ratio: NS NP = VS VP = P / IS VP = P ISVP (1) Chapter 22 Problems 1233 SOLUTION Using Equation (1), we find that the turns ratio is NS NP = P 91 W = = 140 −3 ISVP 5.5 ×10 A (120 V ) ( ) 69. REASONING AND SOLUTION Ohm’s law written for the secondary is Vs = IsR2. We know that Vs = (Ns/Np)Vp Is = (Np/Ns)Ip and Substituting these expressions for Vs and Is into Vs = Is R2 and recognizing that R1 = VP/IP, we find that 2 Np R1 = R N 2 s ______________________________________________________________________________ 70. REASONING According to Ohm’s law (see Section 20.2), the resistance of the wire is equal to the emf divided by the current. The emf can be obtained from Faraday’s law of electromagnetic induction. SOLUTION The resistance R of the wire is R= ξ I (20.2) According to Faraday’s law of electromagnetic induction, the induced emf is ξ = −N Φ − Φ0 ∆Φ = −N ∆t t − t0 (22.3) where N is the number of loops in the coil, Φ and Φ0 are, respectively, the final and initial fluxes, and t − t0 is the elapsed time. Substituting Equation 22.3 into Equation 20.2 yields Φ − Φ0 −N ( ) 4.0 Wb − 9.0 Wb t − t0 − 12 ξ 0.050 s = 5.2 Ω R= = = I I 230 A ______________________________________________________________________________ 1234 ELECTROMAGNETIC INDUCTION 71. SSM REASONING We can use the information given in the problem statement to determine the area of the coil A. Since it is square, the length of one side is l = A. SOLUTION According to Equation 22.4, the maximum emf ξ 0 induced in the coil is ξ0 = NABω . Therefore, the length of one side of the coil is l= A= ξ0 75.0 V = = 0.150 m NBω (248)(0.170 T)(79.1 rad/s) ______________________________________________________________________________ 72. REASONING The magnitude ξ of the emf induced in the loop can be found using Faraday’s law of electromagnetic induction: ξ = −N Φ − Φ0 t − t0 (22.3) where N is the number of turns, Φ and Φ0 are, respectively, the final and initial fluxes, and t − t0 is the elapsed time. The magnetic flux is given by Φ = BA cos φ (Equation 22.2), where B is the magnitude of the magnetic field, A is the area of the surface, and φ is the angle between the direction of the magnetic field and the normal to the surface. 2 SOLUTION Setting N = 1 since there is only one turn, noting that the final area is A = 0 m and the initial area is A0 = 0.20 m × 0.35 m, and noting that the angle φ between the magnetic field and the normal to the surface is 0°, we find that the magnitude of the emf induced in the coil is ξ = −N BA cos φ − BA0 cos φ t − t0 ( 0.65 T ) ( 0 m 2 ) cos 0° − ( 0.65 T )( 0.20 m × 0.35 m ) cos 0° = − (1) = 0.25 V 0.18 s ______________________________________________________________________________ 73. SSM REASONING The current I produces a magnetic field, and hence a magnetic flux, that passes through the loops A and B. Since the current decreases to zero when the switch is opened, the magnetic flux also decreases to zero. According to Lenz’s law, the current induced in each coil will have a direction such that the induced magnetic field will oppose the original flux change. Chapter 22 Problems 1235 SOLUTION a. The drawing in the text shows that the magnetic field at coil A is perpendicular to the plane of the coil and points down (when viewed from above the table top). When the switch is opened, the magnetic flux through coil A decreases to zero. According to Lenz's law, the induced magnetic field produced by coil A must oppose this change in flux. Since the magnetic field points down and is decreasing, the induced magnetic field must also point down. According to Right-Hand Rule No. 2 (RHR-2), the induced current must be clockwise around loop A. b. The drawing in the text shows that the magnetic field at coil B is perpendicular to the plane of the coil and points up (when viewed from above the table top). When the switch is opened, the magnetic flux through coil B decreases to zero. According to Lenz's law, the induced magnetic field produced by coil B must oppose this change in flux. Since the magnetic field points up and is decreasing, the induced magnetic field must also point up. According to RHR-2, the induced current must be counterclockwise around loop B. ____________________________________________________________________________________________ 74. REASONING AND SOLUTION The step down transformer reduces the voltage by a factor of 13: 1 V2 = 13 (120 V ) = 9.2 V ______________________________________________________________________________ 75. SSM REASONING AND SOLUTION The emf ξ induced in the coil of an ac generator is given by Equation 22.4 as ξ = NABω sinω t = (500)(1.2 ×10 –2 m 2 )(0.13 T)(3...
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