Question

When the current through an inductor is changed from 2 A to 6 A in time 2 s, emf induced in it is 3 V. Then the inductance of the inductor is

• A. \( 1.4 \) H
• B. \( 0.8 \) H
• C. \( 1.5 \) H
• D. \( 0.6 \) H

Hint:

The induced emf in an inductor is directly proportional to the rate of change of current through it. Use the formula \( |\varepsilon| = L \left| \frac{\Delta I}{\Delta t} \right| \) to find the inductance \( L \), where \( \Delta I \) is the change in current and \( \Delta t \) is the time interval over which the change occurs. Ensure the units are consistent (volts for emf, amperes for current, and seconds for time will yield inductance in Henrys).

Answer 1

The Correct Option is C

The induced electromotive force (emf) \( \varepsilon \) in an inductor is given by Faraday's law of induction, which states that the induced emf is proportional to the rate of change of current through the inductor: $$ \varepsilon = -L \frac{dI}{dt} $$ where \( L \) is the inductance of the inductor and \( \frac{dI}{dt} \) is the rate of change of current.
The negative sign indicates the direction of the induced emf (Lenz's law), but we are interested in the magnitude here.
Given: Initial current \( I_1 = 2 \) A Final current \( I_2 = 6 \) A Time interval \( \Delta t = 2 \) s Induced emf \( |\varepsilon| = 3 \) V The change in current is \( \Delta I = I_2 - I_1 = 6 - 2 = 4 \) A.
The rate of change of current is \( \frac{dI}{dt} \approx \frac{\Delta I}{\Delta t} = \frac{4 \text{ A}}{2 \text{ s}} = 2 \text{ As}^{-1} \).
Now, we can find the inductance \( L \) using the magnitude of the induced emf: $$ |\varepsilon| = L \left| \frac{dI}{dt} \right| $$ $$ 3 \text{ V} = L (2 \text{ As}^{-1}) $$ $$ L = \frac{3 \text{ V}}{2 \text{ As}^{-1}} = 1.
5 \text{ VsA}^{-1} $$ The unit of inductance is Henry (H), and 1 H = 1 VsA\(^{-1} \).
Therefore, the inductance of the inductor is \( L = 1.
5 \) H.