Question

The number of turns in a coil of wire of fixed radius is $600$ and its self-inductance is $108\text{mH}$ .The self-inductance of a coil of $500$ turns will be_____(in $\text{mH}$)

• A. 75.0
• B. 75.00
• C. 75

Answer 1

Answer: (A), (B), (C)

Given:
$N_1=600,L_1=108\text{mH},N_2=500$

$\because$ self inductance of a coil is $L=\frac{{\mu }_0{N}^{2}A}{l}$

We can conclude that, $L\propto N^2$

Substituting the data gives, $\frac{108}{L_2}=(\frac{600}{500}{)}^2$

$\Rightarrow L_2=108\times \frac{25}{36}=75\text{mH}$