Question

The current in a coil decreases from 5 A to 0 in 0.1 sec. If the average e.m.f induced in the coil is 50 V, then the self-inductance of the coil is
A. 0.25 H
B. 0.5 H
C. 1 H
D. 2 H

Answer 1

Hint:Self-inductance is the characteristics of the current-carrying coil that opposes the change in current flowing through that coil. This occurs due to the self-induced emf produced in the coil itself. On the other hand, we can also say that self-inductance is a phenomenon where there is the induction(L) of a voltage in a current-carrying wire. It is always opposing the changing current and its SI unit is Henry.

Formula used:
The induced voltage in an inductor can be expressed in terms of the inductance and the rate of change of current as-
\[e = L\dfrac{{dI}}{{dt}}\]
Where, L is the inductance.
\[dI\] is the change in current.
\[dt\] is the change in time.

Complete step by step solution:
Given initial current=5 A
final current= 0
So change in current,
\[dI = (5 - 0) = 5A\]
Change in time,
\[dt = 0.1\sec \]
Average e.m.f induced in the coil, e=50 V
As we know the induced e.m.f is determined as,
\[e = L\dfrac{{dI}}{{dt}}\]
Substituting the given values, we get
\[50 = L\dfrac{5}{{0.1}}\]
\[\therefore L = 1\,H\]
Therefore, the self-inductance of the coil is 1 H.

Hence option C is the correct answer.

Note: Full form of e.m.f is Electromotive force. Electromotive force or emf is defined as the potential terminal variation when no electric current flows. Michael Faraday states that the magnitude of the electromotive force(EMF) generated is directly proportional to the rate of change of the magnetic field.