Question

The current in a self-inductance $L = 40 \,mH$ increased uniformly from $0$ ampere to $10\, A$ in $ 4\times 10^{-3} s. $ The induced emf produced in during this process will be :

• A. 40 V
• B. 400 V
• C. 0.4 V
• D. 100 V

Answer 1

Answer: (D)

From Faradays law of electromagnetic induction, the induced emf is given by
$e=-L \frac{d i}{d t}$ where $L$ is coefficient of self induction,
$\frac{d i}{d t}$ is rate of change of current.
Given, $L=40\, m H=40 \times 10^{-3} H$,
$ \frac{d i}{d t}=\frac{10}{4 \times 10^{-3}} $
$\therefore $ $e=-40 \times 10^{-3} \times \frac{10}{4 \times 10^{-3}}$
Negative sign signifies Lenzs law.
$\Rightarrow e=100 \,V$