Question

A thin copper wire of length 100 m 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 :

• A. 4L
• B. 2L
• C. L/2
• D. L

Answer 1

Answer: (D)

$L=\mu N^2\frac{area}{length}$

$L^{\prime }=\mu N^{\prime 2}\frac{are{a}^{\prime }}{lengt{h}^{\prime }}$

$N=\frac{100}{2\pi r}$, $N^{\prime }=\frac{100}{2\pi r/2}=\frac{100}{\pi r}=2N$

$length=lengt{h}^{\prime }=l$

$area=\pi r^2$ , $are{a}^{\prime }=\pi (r/2{)}^2$

therefore,

$L^{\prime }=\mu (2N{)}^2\frac{\pi r^2/4}{l}=\mu N^2\frac{\pi r^2}{l}=L$