Question

A coil made of Copper wire having radius $r$, self-inductance $L$ is bent in two concentric turns each having radius $\frac{r}{2}$. The new self-inductance is

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

Answer 1

The correct option is A $2L$

Given,
$N_2=2N_1{r}_2=\frac{r}{2}or\frac{r_1}{2}$

Self inductance, $L\propto N^{2}r$
or $\frac{L_1}{L_2}={\left(\frac{N_1}{N_2}\right)}^2\times \frac{r_1}{r_2}$

$\Rightarrow \frac{L}{L_2}={\left(\frac{1}{2}\right)}^2\times \left(\frac{r}{\frac{r}{2}}\right)=\frac{1}{4}\times 2=\frac{1}{2}$

So, $L_2=2L$
Hence, correct answer is option (a).