Question

A straight solenoid has $50$ turns in the primary coil and $200$ turns in the secondary coil. The area of cross-section of the solenoid is $4{\text{cm}}^2$ and length is $4\pi \text{cm}$. The mutual inductance, if primary coil is tightly kept co-axial inside the secondary coil, is

Assume almost equal cross-sectional area and length.

• A. $0.01\text{mH}$
• B. $0.02\text{mH}$
• C. $0.04\text{mH}$
• D. $0.06\text{mH}$

Answer 1

The correct option is C $0.04\text{mH}$

The magnetic field at any point inside the straight solenoid of the primary coil 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 secondary coil of total $N$ turns each of cross-sectional area $A$ is given as,

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

So, mutual inductance,

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

$\Rightarrow M=4\pi \times {10}^{-7}\times \frac{50}{4\pi \times {10}^{-2}}\times 200\times 4\times {10}^{-4}$

$\Rightarrow M=4\times {10}^{-5}\text{H}=0.04\text{mH}$

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