Question

Two neighbouring coils $\text{A}$ and $\text{B}$ have a mutual inductance of $20\text{mH}$. The current flowing through $\text{A}$ is given by $i=3t^2-4t+6$. The induced emf at $t=2\text{s}$ in coil $\text{B}$ is-

• A. $160\text{mV}$
• B. $200\text{mV}$
• C. $260\text{mV}$
• D. $300\text{mV}$

Answer 1

The correct option is A $160\text{mV}$

Given, $M=20\text{mH};I_A=3t^2-4t+6$

From, the principle of mutual inductance,

${\varphi }_B=M{i}_A$

On differentiating both sides w.r.t time we get, induced emf as,

$\left|\begin{array}{c}{E}_B\end{array}\right|=\frac{d{\varphi }_B}{dt}=M\left(\frac{d{i}_A}{dt}\right)$

$\Rightarrow E_B=M\left(\frac{d{i}_A}{dt}\right)$

$=M\left[\frac{d}{dt}\right(3t^2-4t+6\left)\right]=M\times (6t-4)$

At, time $t=2\text{s}$

$E_B=20\times {10}^{-3}\times \left[6\right(2)-4]$

$=160\times {10}^{-3}\text{H}=160\text{mH}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.