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. $2L$
• B. $4L$
• C. $6L$
• D. $\frac{1}{2}L$

Answer 1

The correct option is A $2L$

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