Question

A wire of fixed length is wound on a solenoid of length l and radius r. Its self inductance is found to be L. Now, if same wire is wound on a solenoid of length $\frac{l}{2}$ and radius $\frac{r}{2}$, then self inductance will be

• A. 2L
• B. $\frac{L}{2}$
• C. 3L
• D. 4L

Answer 1

The correct option is A 2L

$L={\mu }_0\times n^2\times V$ (n $\to$ Number of turns per unit length , V $\to$ Volume of cylinder)

$={\mu }_0\times \frac{N^2}{l^2}\times \pi r^2\times l$

$L=\frac{{\mu }_0{N}^2\pi r^2}{l}$

Let length of wire be $l_1$

$l_1=\left(2\pi r\right)\times N\Rightarrow N=\left(\frac{l_1}{2\pi r}\right)$

$L=\frac{{\mu }_0\pi r^2}{l}\times \frac{l_1^2}{4{\pi }^2{r}^2}=\frac{{\mu }_0{l}_1^2}{4\pi l}$

$\therefore L\propto \frac{1}{l}\Rightarrow \frac{L_2}{L_1}=\frac{l_1}{l_2}=\frac{l}{\frac{l}{2}}=2$

$L_2=2L_1$