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 now

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

Answer 1

The correct option is A 2 L

$\because L\infty N^{2}r;\frac{L_1}{L_2}={\left(\frac{N_1}{N_2}\right)}^2\times \frac{r_1}{r_2}$
$\Rightarrow \frac{L}{L_2}={\left(\frac{1}{2}\right)}^2\times \left(\frac{r}{\frac{r}{2}}\right)=\frac{1}{2};L_2=2L$