Question

A solenoid has $600$ turns per meter and has a self-inductance of $108\text{mH}.$ What will be the self inductance of the solenoid if the number of turns per unit length is changed to $500$ turns per meter keeping length and area of the solenoid same.

• A. $74\text{mH}$
• B. $75\text{mH}$
• C. $76\text{mH}$
• D. $77\text{mH}$

Answer 1

The correct option is B $75\text{mH}$

We know that self inductance of solenoid is,

$L={\mu }_0{n}^{2}Al$

$\Rightarrow L\propto n^2$

$\therefore \frac{L_1}{L_2}=\frac{n_1^2}{n_2^2}$

Here, $L_1=108\text{mH},n_1=600\text{turns/meter}$

$L_2=?,n_2=500\text{turns/meter}$

$\Rightarrow \frac{108}{L_2}=\frac{(600{)}^2}{(500{)}^2}$

$\Rightarrow L_2=\frac{25}{36}\times 108$

$\therefore L_2=75\text{mH}$

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