The current in an inductor of self-inductance $L=40 \mathrm{mH}$ is to be increased uniformly from $2…
The current in an inductor of self-inductance $L=40 \mathrm{mH}$ is to be increased uniformly from $2 \mathrm{~A}$ to $12 \mathrm{~A}$ in $8 \mathrm{~ms}$. The emf induced in the inductor during this process is
$50 \mathrm{~V}$
$0.4 \mathrm{~V}$
$40 \mathrm{~V}$
$100 \mathrm{~V}$
Solution
Given,
Self-inductance of inductor, $L=40 \mathrm{mH}$
Initial current, $I_1=2 \mathrm{~A}$
Final current, $I_2=12 \mathrm{~A}$
Time interval, $d t=8 \mathrm{~ms}$
Using expression for induced emf,
$\varepsilon=-L \frac{d i}{d t}$
$\Rightarrow \quad|\varepsilon|=L \frac{d i}{d t}$
$=40 \times 10^{-3}\left(\frac{12-2}{8 \times 10^{-3}}\right)=50 \mathrm{~V}$