Mutual inductance problem A square loop with side-length a is positioned at the

Question

Mutual inductance problem

A square loop with side-length a is positioned at the centre of a long thin solenoid, which has radius r (with r>a), length l and N turns. The plane of the loop is perpendicular to the 4 axis of the solenoid. A current I = I0 sin t flows through the loop. Derive an expression for the time-averaged power dissipated in a resistance R connected between the terminals of the solenoid. You may assume that R is much greater than the resistance of the solenoid and that the self-inductance of the solenoid is negligible.

Explanation / Answer

The magnetic field created by the current flowing through the loop is complex and the flux varies throughout the solenoid. By using the fact that the mutual inductance is the same for the coil and the solenoid we can find the emf generated in the solenoid. If we take the magnetic field produced by a solenoid as

B = 0nIsol ez where n = N/L

Then the flux through the square loop is = BA = 0nIsol*a^2

Now the mutual inductance is M = /Isol = 0na^2

The induced emf is V = -M(dIloop/dt) = -0na^2*Io*cost

power dissipated = P = V^2/R = (0na^2*Io*cost)^2/R

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