Question

A coil of $100$ turns and having radius $1\text{cm}$ is kept co-axially within a long solenoid of $8$ turns per $\text{cm}$ with radius $5\text{cm}.$ Find the mutual inductance between the coil and solenoid-

• A. $2.25\times {10}^{-5}\text{H}$
• B. $4.75\times {10}^{-5}\text{H}$
• C. $1.35\times {10}^{-5}\text{H}$
• D. $3.15\times {10}^{-5}\text{H}$

Answer 1

The correct option is D $3.15\times {10}^{-5}\text{H}$

The magnitude field $B$ inside the solenoid is,

$B={\mu }_0{n}_s{i}_s$

Where,

$n_s=\text{No. of turns per unit length of the solenoid}{i}_s=\text{current passing through the solenoid}$

Since, the coil is placed co-axially inside the solenoid, therefore, the field due to the solenoid perpendicularly intercepts the area of the coil. Hence, magnetic flux linked with the coil is,

${\varphi }_c=N_{c}B{A}_c=N_c\left({\mu }_0{n}_s{i}_s\right)A_c$

Where, $N_c=\text{No. of turns of the coil}{A}_c=\text{Cross sectional area of coil}$

The mutual inductance between solenoid and coil can be given as,

$M=\frac{{\varphi }_c}{i_s}={\mu }_0{n}_s{N}_c{A}_c.......\left(1\right)$

Here,

$n_s=\frac{N}{l}=\frac{8}{{10}^{-2}}=800$ and $N_c=100$

Substituting the given values in $\left(1\right)$ we get,

$M=4\pi \times {10}^{-7}\times 800\times 100\times \pi \times {10}^{-4}$

$\approx 315\times {10}^{-7}\text{H}=3.15\times {10}^{-5}\text{H}$

<!--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 between a solenoid and a loop.