Question

A small magnet is placed on the axis, at a distance $\frac{3a}{4}$ from center of a fixed conducting ring of radius $a$ and resistance $R$. When it is moved from that point to infinity, charge $q$ flows through the ring then magnetic moment of magnet is

• A. $\frac{27aRq}{32\mu_0}$
• B. $\frac{32aRq}{27\mu_0}$
• C. $\frac{27\mu_0a}{32Rq}$
• D. $\frac{32\mu_0a}{27Rq}$

Answer 1

Answer: (A)

$\frac{27aRq}{32\mu_0}$

Answer 2

The total charge $q$ flowing through the ring is given by $q = \frac{\Delta \Phi_B}{R}$. The magnetic flux through the ring due to a magnetic dipole on its axis at distance $d$ is $\Phi_B = \frac{\mu_0 m a^2}{2 d^3}$. Initial flux at $d_1 = \frac{3a}{4}$ is $\Phi_{B1} = \frac{\mu_0 m a^2}{2 (\frac{3a}{4})^3} = \frac{32 \mu_0 m}{27a}$. Final flux at $d_2 = \infty$ is $\Phi_{B2} = 0$. So, $\Delta \Phi_B = \Phi_{B2} - \Phi_{B1} = -\frac{32 \mu_0 m}{27a}$. Therefore, $q = \frac{|\Delta \Phi_B|}{R} = \frac{32 \mu_0 m}{27aR}$. Solving for $m$, we get $m = \frac{27aRq}{32\mu_0}$.