Question

A long solenoid of cross-sectional area $5{\text{cm}}^2$ is wound with $25$ turns per $\text{cm}$. It is placed co-axially inside a closely wrapped coil of $10$ turns with radius of $25\text{cm}$. The magnitude of emf induced in the coil, when the current through the solenoid is decreasing at a rate of $0.2\text{A/s}$ is -

• A. $1.14\mu \text{V}$
• B. $2.14\mu \text{V}$
• C. $3.14\mu \text{V}$
• D. $4.14\mu \text{V}$

Answer 1

The correct option is C $3.14\mu \text{V}$

The magnetic field at any point inside the solenoid with $n$ turns per unit length carrying a current $i$ is given by the relation,

$B_1={\mu }_{0}ni$

The magnetic flux through the coil of total $N$ turns and through cross-sectional area $A$ of the solenoid is given as,

${\varphi }_{21}=N\left(B_{1}A\right)=N\left({\mu }_{0}niA\right)={\mu }_{0}niNA$

So, mutual inductance,

$M=\frac{{\varphi }_{21}}{i_1}=\frac{{\mu }_{0}niNA}{i}={\mu }_{0}nNA$

$\Rightarrow M=4\pi \times {10}^{-7}\times \frac{25}{{10}^{-2}}\times 10\times 5\times {10}^{-4}$

$\Rightarrow M=1.57\times {10}^{-5}\text{H}$

Now, the magnitude of emf induced in the coil,

$E=M\frac{di}{dt}=1.57\times {10}^{-5}\times 0.2=3.14\times {10}^{-6}\text{V}$

$\Rightarrow E=3.14\mu \text{V}$

Hence, option $\left(C\right)$ is the correct answer.