Question

Two coils are at fixed locations. When coil-$1$ has no current and current in the coil-$2$ increases at the rate $15.0{\text{As}}^{-1}.$ The e.m.f. in the coil-$1$ is $25.0\text{mV}$. When coil-$2$ has no current and coil-$1$ has a current of $3.6\text{A}.$, the flux linkage in coil-$2$ is-

• A. $16\text{mWb}$
• B. $10\text{mWb}$
• C. $4\text{mWb}$
• D. $6\text{mWb}$

Answer 1

The correct option is D $6\text{mWb}$

Case-$\left(I\right)$

When current in the coil-$1$ is zero and current in the coil-$2$ increases at a rate of $\left(\frac{d{i}_2}{dt}\right)=15{\text{As}}^{-1}$ and emf in the coil-$1$ is $E_1=25\text{mV}.$

Using, the principle of mutual inductance,

${\varphi }_1=M{i}_2$

On differentiating both sides we get,

$\left|\begin{array}{c}{E}_1\end{array}\right|=\frac{d{\varphi }_1}{dt}=M\frac{d{i}_2}{dt}.......\left(1\right)$

Where, ${\varphi }_1=\text{Flux linked with the coil-}1M=\text{Co-efficient of mutual inductance}{i}_2=\text{Current in the coil-}2$

Putting the values in (1) we get,

$25\times {10}^{-3}=M\times 15$

$\Rightarrow M=\frac{5}{3}\times {10}^{-3}\text{H}$

Case-$\left(II\right)$

Now, flux linked with coil-$2$ when current in coil-$1$ is $3.6\text{A}$ is,

$=M{i}_1$

$=\frac{5}{3}\times {10}^{-3}\times 3.6$

$=6\times {10}^{-3}\text{Wb}=6\text{mWb}$

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

Why this question ? This question tests your basic understanding of mutual inductance.