72) A coil with a self-inductance of 6.0 H is connected to a dc battery through a switch. As soon as the switch is closed, the rate of change of current is 2.0 A/s. What is the emf induced in this coil at this instant

12.0 V

Explanation:

The E.M.F induced in a coil is given as,

E = LΔ(i/t) Equation 1

Where E = E.M.F induced in the coil, L = inductance of the coil, Δ(i/t) = rate of change of current when the switch is closed

Given: L = 6.0 H, Δ(i/t) = 2.0 A/s.

Substitute into equation 1

E = 6.0(2.0)

E = 12.0 V.

Hence the emf induced in the coil = 12.0 V

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