Question

A thin copper wire of length 100 meters is wound as a solenoid of length $\lambda$ 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 $\lambda$ but the radius r/2, then its self inductance will be:

• A. 4 L
• B. 2 L
• C. L
• D. L/2

Answer 1

The correct option is C L

As we know that, self inductance = ${\mu }_{\circ }{n}^{2}lA$

where n= number of turns per unit length

l=length of solenoid

A=cross section

So, self inductance(L) =${\mu }_{\circ }{n}^{2}l\pi R^2$

$\Rightarrow L={\mu }_{\circ }{\left(\frac{N}{l}\right)}^{2}l\pi R^2$ (Where N= Total number of turns)

$\Rightarrow L=\frac{{\mu }_{\circ }{N}^2\pi R^2}{l}$

Now, it is given that when length is $\lambda$ and radius is'r' then ,

$L={\mu }_o{\left(\frac{100}{2\pi r}\right)}^2\frac{\pi r^2}{\lambda }$

Now, in second case length is $\lambda$ and radius is 'r/2' then,

$L^{\prime }={\mu }_o{\left(\frac{100}{2\pi \frac{r}{2}}\right)}^2\frac{\pi }{\lambda }{\left(\frac{r}{2}\right)}^2$

$\Rightarrow L^{\prime }={\mu }_o{\left(\frac{100}{2\pi r}\right)}^2\frac{\pi r^2}{\lambda }=L$

$\Rightarrow L^{\prime }=L$