Question

An ideal solenoid of cross-sectional area ${10}^{-4}{\text{m}}^2$ has $500$ turns per meter. At the centre of this solenoid, another coil of $100$ turns is wrapped closely around it. If the current in the coil changes from $0\text{A}$ to $2\text{A}$ in $3.14\text{ms}.$ The emf developed in the solenoid is

• A. $1\text{mV}$
• B. $2\text{mV}$
• C. $3\text{mV}$
• D. $4\text{mV}$

Answer 1

The correct option is D $4\text{mV}$

Let us consider, the given coil as coil-$1$ and the solenoid as coil-$2$

As we know, mutual inductance in case of solenoid is,

$M=\frac{{\varphi }_2}{i_1}={\mu }_0{n}_1{n}_2{A}_2{l}_2.......\left(1\right)$

Where, $n_1=\text{No. of turns per unit length of coil}{n}_2=\text{No. of turns per unit length of solenoid}{A}_2=\text{Area of the solenoid}{l}_2=\text{length of the solenoid}$

Here, $n_1=100;n_1=500A_2={10}^{-4}{\text{m}}^2;l_2=1\text{m}$

Putting this values in (1) we get,

$M=4\pi \times {10}^{-7}\times 500\times 100\times {10}^{-4}\times 1$

$M=20\pi \times {10}^{-7}\text{H}$

Now, induced emf in the solenoid is,

$\left|\begin{array}{c}{E}_2\end{array}\right|=\frac{\Delta {\varphi }_2}{\Delta t}=M\left(\frac{\Delta i_1}{\Delta t}\right)$

$=20\pi \times {10}^{-7}\times \left(\frac{2}{3.4\times {10}^{-3}}\right)$

$=40\times {10}^{-4}\text{V}\text{(or)}$

$=4\text{mV}$

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