Question

If two coils placed next to one another have a mutual inductance of 3.00 mH, what voltage (in V) is induced in one when the 2.50 A current in the other is switched off in 40.0 ms

Answer 1

Answer:

-0.1875 V.

Explanation:

Using

E₂ = MdI₁/dt........................ Equation 1

Where E₂ = Voltage induced in the second coil, M = mutual inductance of both coil, dI₁ = change in current in the first coil, dt = change in time.

Given: M = 3.00 mH = 0.003 H, dI₁ = (0-2.50) = -2.5 A, dt = 40 ms = 0.04 s.

Substitute into equation 1

E₂ = 0.003(-2.5)/0.04

E₂ = -0.1875 V.

Hence the induced emf = -0.1875 V.