Question

A coil of $Cu$ wire (radius $-r$, self-inductance-$L$) is bent in two concentric turns each having radius $\frac{r}{2}$. The self-inductance is now

• A. $2L$
• B. $L$
• C. $4L$
• D. $\frac{L}{2}$

Answer 1

The correct option is A $2L$

Given,

Let self inductance before bent is $L_1$

Let self inductance after bent is $L_2$

Radius of inductance before bent is $r$

Radius of inductance after bent is $r/2$

Number of turn is constant.

We know that

$L=N^2{\mu }_0{\mu }_r\left(\frac{D}{2}\right)L\propto N^{2}r\frac{L_1}{L_2}=\frac{r_1}{r_2}\frac{L}{L_2}=\frac{r}{r/2}{L}_2=2L$

Correct option is A.