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

E = - 12V

Explanation:

Given

L = self inductance = 6.0H

di/dt = 2.0A/s

E = - L * di/dt

= - 6.0 * 2.0

= - 12V.