(a) Assume the solenoid carries current i(1). What is the magnitude of the magne
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Question
(a) Assume the solenoid carries current i(1). What is the magnitude of the magnetic field B(1) inside the solenoid? Express B(1) in terms of i(1), N(1), L, and any constants.
(b) What is the total magnetic flux (2) through the loop? Express it in terms of i(1), N(1), L, r, a, and any constants.
(c) What is the mutual inductance M? Express M in terms of N(1), L, r, a, and any constants.
(d) Now, assume that the loop carries current i(2). Show that the total flux (1) through all N1 turns of the solenoid is given by flux(1)=M*i(2) where M has the same value calculated in part c
Explanation / Answer
Calculate the mutual inductance of a solenoid 25 cm long of radius 1.5 cm with 500 turns, and a single loop of radius 3.0 cm centered on the solenoid, with its area perpendicular to the axis of the solenoid. __________µH The Mutual Inductance is 10.1µH ± 0.2µH Calculate self inductance of each coil L1 = µ0N1² A / l . . µ0=4p*10¯7, N1² = 500², A = p*0.015² m² , l = 0.25m = 0.888 mH L2 = µ0pa . . µ0=4p*10¯7, a = Loop radius = 0.03 m. Close result to more complex Eq ˜ 0.118 µH . . Close result to more complex Eq. just a bit too high, 0.115µH better = 0.115 µH Mutual inductance = v ( L1 L2 ) = v ( 0.888 mH * 0.115 µH ) = 10.1 µH = Answer Has this helped? Clarification of any point will be provided, if requested. Source(s): For L1 Inductance http://en.wikipedia.org/wiki/Inductor For L2 Single Loop Coil Inductance http://www.thompsonrd.com/induct2.pdf
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