Electromagnetic Induction - Result Question 38

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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

======= ####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

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/electromagnetic-induction/electromagnetic-induction—result-question-38.md (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:

  1. (a) $e=L \frac{d i}{d t}$

Given that $L=40 \times 10^{-3} H$,

$di=11 A-1 A=10 A$

and $dt=4 \times 10^{-3} s$

$\therefore e=40 \times 10^{-3} \times(\frac{10}{4 \times 10^{-3}})=100 V$

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