A toroidal solenoid with mean radius r and cross-sectional area A is wound unifo

Question

A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N_1 turns. A second toroidal solenoid with N_2 turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius. What is the mutual inductance of the two solenoids? Assume that the magnetic field of the first solenoid is uniform across the cross section of the two solenoids. Express your answer in terms of the variables N_1, N_2, A, r, magnetic constant and others appropriate constants. mu_0 and other appropriate constants. If N_1 = 440 turns, N_2 = 250 turns, r = 13.0 cm and A = 0.800 cm^2, what is the value of the mutual inductance? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

Here ,

for the mutual inductance of solenoids

part A)

flux through the toroid

flux = u0 * N2 * I/(2pi * r) * N1 * A

as flux = M * I

M = u0 * N1 * N2*A/(2*r)

the mutual inductance is u0 * N1 * N2*A/(2*r)

part B)

M = 4pi *10^-7 * 440 * 2550 * 0.80 * 10^-4/(2 * 0.13)

M = 4.34 *10^-4 H

the mutual inductance is 4.34 *10^-4 H

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