Question

The current in a coil of $L=40mH$ is to be increased uniformly from $1A$ to $11A$ in $4$ milliseconds. The induced e.m.f. will be

• A. $100V$
• B. $0.4V$
• C. $440V$
• D. $40V$

Answer 1

Answer: (A)

As the current is increasing uniformly so the rate of increase in current is
$\frac{di}{dt}=\frac{\Delta i}{\Delta t}=\frac{11-1}{4\times 10^{-3}}=\frac{5}{2}\times 10^{3}A{s}^{-1}$
So the magnitude of induced emf in the coil is
$E=\left|L\frac{dI}{dt}\right|=40\times 10^{-3}\times \frac{5}{2}\times 10^3=100V$