When the current changes from +2 ~A to -2 ~A in 0.05 second, an e.m.f. of 8 ~V is induced in a coil. The coefficient of self-induction of the coil is

Correct answer: (D) 0.1 H. e=-L dl/dt ⇒ 8=(L) (2-(-2))/0.05 ⇒ L=0.1 H

Q187 / 225 · JEE Main 2003 · Physics

When the current changes from $+2 \mathrm{~A}$ to $-2 \mathrm{~A}$ in $0.05$ second, an e.m.f. of $8 \mathrm{~V}$ is induced in a coil. The coefficient of self-induction of the coil is

$e=-\mathrm{L} \frac{\mathrm{dl}}{\mathrm{dt}} \Rightarrow 8=(\mathrm{L}) \frac{2-(-2)}{0.05} \Rightarrow \mathrm{L}=0.1 \mathrm{H}$

Hence, the correct answer is option: 'D'.