Question

A coil of wire of radius $r$ has $600$ turns and a self-inductance of $108mH$. The self-inductance of a coil with the same radius and $500$ turns is

• A. $80mH$
• B. $75mH$
• C. $108mH$
• D. $90mH$

Answer 1

The correct option is D

$90mH$

Step1: Given data.

Radius of the coil $=$ $r$

Number of turns, $N_1=600$

Self-inductance, $L_1=108mH$

Changed number of turns, $N_2=500$

Step2: Finding self-inductance of another coil with same radius.

Formula used:

$L=\frac{{\mu }_0{N}^{2}A}{l}$

Where, ${\mu }_0$$=$permeability of Free Space, $N=$number of turns, $A=$Inner Core Area, $\mathit{l}=$length of the coil.

According to question, length of the Coil and Inner Core Area are constant.

Therefore,

$L_1=\frac{{\mu }_0{N_1}^{2}A}{l}$ …..$\left(i\right)$

$L_2=\frac{{\mu }_0{N_2}^{2}A}{l}$ …..$\left(ii\right)$

Dividing equation $\left(ii\right)$ by $\left(i\right)$ we get.

$\Rightarrow$$\frac{L_2}{L_1}=\frac{N_2^2}{N_1^2}$

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

$\Rightarrow$ $L_2=90mH$

Hence, option D is correct.