Electromagnetic Induction Ques 32
32. The current in self inductance $L=40 $ $mH$ is to be increased uniformly from $1 $ $amp$ to $11$ $ amp$ in $4 $ milliseconds. The e.m.f. induced in the inductor during the process is
[1990]
(a) $100$ volt
(b) $0.4$ volt
(c) $4.0$ volt
(d) $440$ volt
Show Answer
Answer:
Correct Answer: 32.(a)
Solution:
- (a) $e=L \frac{d i}{d t}$
Given that $L=40 \times 10^{-3}$ $ H$,
$di=11 A-1 A=10 $ $A$
and $dt=4 \times 10^{-3} $ $s$
$\therefore e=40 \times 10^{-3} \times(\frac{10}{4 \times 10^{-3}})=100$ $ V$
BETA
