Q. 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
NTA AbhyasNTA Abhyas 2022
Solution:
As the current is increasing uniformly so the rate of increase in current is
$\frac{d i}{d t}=\frac{\Delta i}{\Delta t}=\frac{11 - 1}{4 \times 10^{- 3}}=\frac{5}{2}\times 10^{3}As^{- 1}$
So the magnitude of induced emf in the coil is
$Ε=\left|L \frac{d I}{d t}\right|=40\times 10^{- 3}\times \frac{5}{2}\times 10^{3}=100V$
$\frac{d i}{d t}=\frac{\Delta i}{\Delta t}=\frac{11 - 1}{4 \times 10^{- 3}}=\frac{5}{2}\times 10^{3}As^{- 1}$
So the magnitude of induced emf in the coil is
$Ε=\left|L \frac{d I}{d t}\right|=40\times 10^{- 3}\times \frac{5}{2}\times 10^{3}=100V$