A thin copper wire of length 100m is wound as a solenoid of length l and radius r. Its self-inductance is found to be L. Now, if the same length of wire is wound as a solenoid of length l but of radius r/2, then its self-inductance will be

Given, $\mathrm{l}=100\mathrm{m}$, radius r , then self inductance of thin copper wire,

$\mathrm{L}=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^2\mathrm{A}}{\mathrm{l}}$  i.e., $2\mathrm{\pi rN}=100$

Similarly same length of wire is wound as a solenoid of length l of radius r/2, then its self inductance will be $\mathrm{L}\text{'}=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^{\text{'}2}(\mathrm{A}/4)}{\mathrm{l}}$ So, $\frac{2\mathrm{\pi r}}{2}\mathrm{N}\text{'}=100\Rightarrow \mathrm{N}\text{'}=2\mathrm{N}$

$\mathrm{L}\text{'}=\frac{{\mathrm{\mu }}_0{\left(2\mathrm{N}\right)}^2(\mathrm{A}/4)}{\mathrm{l}}\_\_\_\_\left(\mathrm{ii}\right)$

On comparing self-inductance of equations(i) and (ii), we get L' = L