Q115 / 125 · BITSAT 2010 · Physics
The back emf induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is
- A1 henry
- B4 henry
- C$10^{-3}$ henry
- Correct answerD$4 \times 10^{-3}$ henry
Solution
$\mathrm{e}=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}$
But $\mathrm{e}=4 \mathrm{~V}$ and $\frac{\mathrm{di}}{\mathrm{dt}}=\frac{0-1}{10^{-3}}=-1 / 10^{-3}$
$\therefore \frac{-1}{10^{-3}}(-\mathrm{L})=4 \Rightarrow \mathrm{L}=4 \times 10^{-3}$ henry
Hence, the correct answer is option: 'D'.