Two coils are placed close together in a physics lab to demonstrate Faraday's law of induction. A current of 8.00 A in one is switched off in 1.20 ms, inducing a 3.00 V emf in the other. What is their mutual inductance (in mH)

M = 0.45 mH

Explanation:

For the system described in the statement, you have that the induced emf of both coils are given by:

          (1)

ε1, ε2: induced emfs

M: mutual inductance

I1, I2: current on the wires

That is, the induced emf of one coil is generated by the change in the current of the other coil.

You can consider that the current of 8.0A belongs to the second coil, and the induced emf of 3.00V is generated in the first coil. The current of the second coil is switched off in 1.20ms.

With the previous information you have:

ε1 = 3.00V

ΔI2 = 0.00A - 8.00A = -8.00A

Δt = 1.20ms = 1.2*10^-3 s

You use the equation (1) to solve for M:

The mutual inductance of the coils is 0.45mH

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