Question

The current in a self-inductance $L=40mH$ increased uniformly from $0$ ampere to $10A$ 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
$e=-L\frac{di}{dt}$ where $L$ is coefficient of self induction,
$\frac{di}{dt}$ is rate of change of current.
Given, $L=40mH=40\times 10^{-3}H$,
$\frac{di}{dt}=\frac{10}{4\times 10^{-3}}$
$\therefore$ $e=-40\times 10^{-3}\times \frac{10}{4\times 10^{-3}}$
Negative sign signifies Lenzs law.
$\Rightarrow e=100V$