Question

An emf of 2.8 mV is induced in a rectangular loop of area 150 cm$^2$ when the current in the loop changes from 3 A to 8 A in a time of 0.2 s. Then the self-inductance of the loop is

• A. 112 $\mu$H
• B. 56 $\mu$H
• C. 28 $\mu$H
• D. 84 $\mu$H

Hint:

The formula for self-induced EMF is $\mathcal{E} = -L \frac{dI}{dt}$. Don't get confused by extra information provided in the problem, such as the area of the loop, which would be relevant for calculating mutual inductance or magnetic flux, but not self-inductance from EMF and current change.

Answer 1

The Correct Option is A

The induced emf ($\mathcal{E}$) in a loop due to a change in its own current is given by the formula for self-induction:
$\mathcal{E} = -L \frac{dI}{dt}$.
We are interested in the magnitude of the inductance, so we can use:
$|\mathcal{E}| = L \left|\frac{\Delta I}{\Delta t}\right|$.
We are given the following values:
Induced emf, $|\mathcal{E}| = 2.8$ mV = $2.8 \times 10^{-3}$ V.
Change in current, $\Delta I = I_{final} - I_{initial} = 8 \text{ A} - 3 \text{ A} = 5$ A.
Time interval, $\Delta t = 0.2$ s.
The area of the loop is extra information and not needed for this calculation.
Now, we rearrange the formula to solve for the self-inductance, $L$:
$L = \frac{|\mathcal{E}| \cdot \Delta t}{|\Delta I|}$.
$L = \frac{(2.8 \times 10^{-3} \text{ V}) \times (0.2 \text{ s})}{5 \text{ A}}$.
$L = \frac{0.56 \times 10^{-3}}{5} = 0.112 \times 10^{-3}$ H.
To express the answer in microhenries ($\mu$H), we convert the units. $1 \mu\text{H} = 10^{-6}$ H.
$L = 0.112 \times 10^{-3} \text{ H} = 112 \times 10^{-6}$ H = 112 $\mu$H.