Electromagnetic Induction - Result Question 38
38. 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
(a) 100 volt
(b) 0.4 volt
(c) 4.0 volt
(d) 440 volt
[1990]
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Answer:
Correct Answer: 38. (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 $dt=4 \times 10^{-3} s$
$\therefore e=40 \times 10^{-3} \times(\frac{10}{4 \times 10^{-3}})=100 V$
