Question.

An inductor consists of 500 turns of wire of resistance 6.0 Ω wound tightly and uniformly on a toroidal ring of an insulating magnetic material with relative permeability μ = 40. The material is linear so μ is independent of the magnetic field. The mean radius of the toroid is 15 cm and the cross sectional area is 1.0 cm2.

a) Starting from Ampere's law, calculate the self inductance of the coil, assuming that the cross sectional area of the coil is sufficiently small that we can assume that the magnetic field has constant strength inside the coil.

b) Use two methods to calculate the energy that would be stored in the inductor when 5.0V is applied to it, one method involving the inductance and one method involving the fields within the inductor.
A second coil consisting of a single loop of wire is now wound around the toroidal ring.

c) Determine the mutual inductance M between the two coils.

d) Suppose now that a current I1 = I0 sin of amplitude I = 40A flows through the 500 turn coil. Find an expression for the induced current in the single turn coil in terms of I0, M, , t and the resistance R of the single turn coil. You may assume that the resistance of this second coil is sufficiently large that you can neglect its self-inductance. If the resistance is R = 20 Ω and the angular frequency of the current is =100s-1, calculate the amplitude of the induced current.