Q189 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
Average e.m.f. of 2.4 V is induced in a coil when a current in it is changed from 10 A in one direction to 10 A in opposite direction in 0.4 second. The self inductance of the coil is
- A$50 \text{ mH}$
- Correct answerB$48 \text{ mH}$
- C$45 \text{ mH}$
- D$42 \text{ mH}$
Solution
Given $e = 2.4 \text{ V}$, $I_1 = 10 \text{ A}$, $I_2 = -10 \text{ A}$, and $\Delta t = 0.4 \text{ s}$.
The magnitude of the average induced e.m.f. is given by:
$|e| = L \dfrac{|\Delta I|}{\Delta t}$
Substituting the given values:
$2.4 = L \dfrac{|-10 - 10|}{0.4}$
$2.4 = L \dfrac{20}{0.4}$
$2.4 = 50 L$
$L = \dfrac{2.4}{50} = 0.048 \text{ H}$
$L = 48 \text{ mH}$
Answer: $48 \text{ mH}$
The magnitude of the average induced e.m.f. is given by:
$|e| = L \dfrac{|\Delta I|}{\Delta t}$
Substituting the given values:
$2.4 = L \dfrac{|-10 - 10|}{0.4}$
$2.4 = L \dfrac{20}{0.4}$
$2.4 = 50 L$
$L = \dfrac{2.4}{50} = 0.048 \text{ H}$
$L = 48 \text{ mH}$
Answer: $48 \text{ mH}$
Hence, the correct answer is option: 'B'.