Question

Two current carrying loops having $N_1$ turns and $N_2$ turns respectively both carrying a current equal to $I$ in the same direction, are placed inside a magnetic field $B$. If the radii of both loops are in the ratio $1:3$ then what will be the ratio of the potential energy of loops in that magnetic field?

• A. $\frac{N_1}{N_2}$
• B. $\frac{2N_1}{N_2}$
• C. $\frac{N_1}{3N_2}$
• D. $\frac{N_1}{9N_2}$

Answer 1

The correct option is D $\frac{N_1}{9N_2}$

Given : $\frac{R_1}{R_2}=\frac{1}{3}$

Magnetic moment of the loop $\mu =NIA$

where $A=\pi R^2$

$\therefore$ $\mu =\pi NI{R}^2$

Potential energy of loop in the magnetic field $P.E=\mu B$

$\therefore$ $P.E=\pi NI{R}^{2}B$

We get $P.E\propto N{R}^2$

$⟹$ $\frac{P.E_1}{P.E_2}=\frac{N_1}{N_2}\frac{R_1^2}{R_2^2}$

Or $\frac{P.E_1}{P.E_2}=\frac{N_1}{N_2}\frac{1}{9}=\frac{N_1}{9N_2}$