Explanation Given Info: The current i in the large circuit is 5.00 A . The emf in the circuit is 90.0 V . The small circuit has a resistance of 1.0 Ω / m with 25 loops. The dimensions of the small circuit a is 10.0 cm and b is 20.0 cm . The distance c between the large circuit and small circuit is 5.0 cm . The resistance R 0 is 10.0 Ω . The capacitance C of the capacitor is 20 μF . Sketch the circuit as given in problem 29.47 as follows with an emf is placed between the capacitor C and switch S , Refer figure.1. The long wire nearest to the small loop produces the magnetic flux through the small loop is given by, Φ B = ∫ c c + a μ 0 i b 2 π r d r = μ 0 i b 2 π ∫ c c + a 1 r d r Take integration for the above equation, Φ B = μ 0 i b 2 π ( ln ( c + a ) − ln c ) = μ 0 i b 2 π ln ( c + a c ) = μ 0 i b 2 π ln ( 1 + a c ) Take differentiation on both sides with respect to time, d Φ B d t = μ 0 b 2 π ln ( 1 + a c ) d i d t (I) Formula to calculate the charging current on the RC circuit is, i ( t ) = ε R e − t / R C (II) ε is the emf of the circuit, R is the resistance in the circuit, C is the capacitance in the circuit. Take differentiation with respect to time for the equation (II), d i d t = d d t ( ε R e − t / R C ) = ε R d d t ( e − t / R C ) = ε R ( − 1 R C e − t / R C ) = − ε R 2 C e − t / R C (III) Substitute equation (III) in equation (I), d Φ B d t = ( μ 0 b 2 π ln ( 1 + a c ) ) ( − ε R 2 C e − t / R C ) (IV) Formula to calculate the magnitude of the induced emf by Faraday’s law is, | ε ind | = N | d Φ B d t | (V) d Φ B d t is the rate of change of magnetic flux with respect to time, N is the number of turns in the circuit. Substitute equation (IV) in equation (V), | ε ind | = N | ( μ 0 b 2 π ln ( 1 + a c ) ) ( − ε R 2 C e − t / R C ) | = N μ 0 b 2 π ln ( 1 + a c ) ε R 2 C e − t / R C (VI) Rearrange the equation (II) as follows, i R = ε e − t / R C ε = i R e − t / R C (VII) Substitute equation (VII) in equation (VI), | ε ind | = N μ 0 b 2 π ln ( 1 + a c ) ( i R e − t / R C ) R 2 C e − t / R C = N μ 0 b 2 π ln ( 1 + a c ) 1 R C i (VIII) Refer figure 1. The total length L of the small loop is, L = a + b + a + b = 2 a + 2 b Substitute 10.0 cm for a and 20.0 cm for b to find L . L = 2 ( 10.0 cm ) + 2 ( 20.0 cm ) = 60.0 cm = 60.0 × 10 − 2 m = 0

University Physics with Modern Physics (14th Edition)

14th Edition

ISBN: 9780321973610

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Chapter 29, Problem 29.48P

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The magnitude and direction of the induced current in the small circuit.

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Chapter 29, Problem 29.47P

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University Physics with Modern Physics (14th Edition)

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The magnitude and direction of the induced current in the small circuit.

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The magnitude and direction of the induced current in the small circuit.

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