Question

If radius of long solenoid is reduced to half of original without changing other physical factor,then its self inductance will change-

• A. $\frac{1}{3}times$
• B. $\frac{1}{2}times$
• C. $\frac{1}{5}times$
• D. $\frac{1}{4}times$

Answer 1

The correct option is D $\frac{1}{4}times$

$N$=total number of turns in solenoid

$l$=length of solenoid

$R$=radius of cross section of solenoid

Magnetic field inside solenoid =$B={\mu }_o\frac{N}{l}i$ where, $\frac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$⟹E=N\times {\mu }_o\frac{N}{l}i\times \pi R^2$

$⟹E=\frac{{\mu }_o{N}^2\pi R^2}{l}i=Li$

where $L$=self inductance

Thus, $L=\frac{{\mu }_o{N}^2\pi R^2}{l}$

Hence, $L\propto R^2$

Hence, on reducing the radius to half, $L$ will becomes $\frac{1}{4}$ times.

Answer-(D)