Work Done Bringing Dipole Near Current Loop

To find the work required to bring the magnetic dipole from infinity to a distance $r$ from the center of the loop, we analyze the interaction between the dipole and the induced current in the loop.

1. Magnetic Flux through the Loop

Consider a circular loop of radius $a$ and a small magnetic dipole of magnetic moment $m$ located on its axis at a distance $r$ ($r \gg a$) from its center.

The magnetic field generated along the axis by a loop carrying a unit current $I = 1$ at distance $r$ is: $B(r) = \frac{\mu_0 a^2}{2(a^2 + r^2)^{3/2}} \approx \frac{\mu_0 a^2}{2 r^3}$

Using the reciprocity property of mutual inductance, the magnetic flux $\Phi$ passing through the loop due to the dipole $m$ located at distance $r$ is: $\Phi = B(r) \cdot m = \frac{\mu_0 a^2 m}{2 r^3}$

2. Induced Current in the Loop

During the process of bringing the dipole from infinity to position $r$, the change in magnetic flux induces a current $I$ in the loop. By flux conservation (or self-inductance relation $L I + \Phi = 0$): $I = -\frac{\Phi}{L} = -\frac{\mu_0 a^2 m}{2 L r^3}$

3. Magnetic Force on the Dipole

The magnetic force acting on the dipole along the axis of the loop is given by: $F_{\text{mag}} = I \frac{\partial \Phi}{\partial r}$

Substituting $I$ and $\frac{\partial \Phi}{\partial r}$: $\frac{\partial \Phi}{\partial r} = \frac{\partial}{\partial r}\left(\frac{\mu_0 a^2 m}{2 r^3}\right) = -\frac{3 \mu_0 a^2 m}{2 r^4}$

$F_{\text{mag}} = \left(-\frac{\mu_0 a^2 m}{2 L r^3}\right) \left(-\frac{3 \mu_0 a^2 m}{2 r^4}\right) = \frac{3 \mu_0^2 a^4 m^2}{4 L r^7}$

4. Work Done by External Agent

The work required by an external agent to bring the dipole quasi-statically from infinity to distance $r$ against the magnetic force is: $W = \int_{\infty}^r -F_{\text{mag}} \, dr'$

$W = -\int_{\infty}^r \frac{3 \mu_0^2 a^4 m^2}{4 L r'^7} \, dr' = \left[ \frac{\mu_0^2 a^4 m^2}{8 L r'^6} \right]_{\infty}^r = \frac{\mu_0^2 a^4 m^2}{8 L r^6}$

Thus, the work done $W$ is proportional to: $W \propto \frac{m^2}{r^6}$

Correct Option: (C)