A circular loop of radius r carrying a current i is maintained at...

The magnetic field produced by the larger loop at its centre is given by:

$B = \dfrac{\mu_0 I}{2 R}$

The magnetic moment of the smaller loop is:

$M = i A = i \pi r^2$

The angle between the plane of the smaller loop and the plane of the larger loop is 30^{}^{{}^{\circ}}. Therefore, the angle between their area vectors (or between the magnetic moment $\vec{M}$ and the magnetic field $\vec{B}$) is also \theta = 30^{}^{{}^{\circ}}.

The magnitude of the torque exerted by the magnetic field on the smaller loop is:

$\tau = M B \sin \theta$

Substituting the values, we get:

\tau = (i \pi r^2) \left(\dfrac{\mu_0 I}{2 R}\right) \sin 30^{}^{{}^{\circ}}

$\tau = \dfrac{\mu_0 \pi i I r^2}{2 R} \times \dfrac{1}{2} = \dfrac{\mu_0 \pi i I r^2}{4 R}$

To hold the smaller loop fixed, an external torque of equal magnitude must be applied. This torque is provided by a single force $F$ applied at a point on the periphery of the smaller loop.

The torque due to this force about the centre is $\tau = F \times d$, where $d$ is the perpendicular distance from the centre to the line of action of the force. To minimize the required force, the distance $d$ must be maximized. The maximum possible distance for a point on the periphery is the radius of the smaller loop, $r$.

Thus, F_\min r = \tau

F_\min r = \dfrac{\mu_0 \pi i I r^2}{4 R}

F_\min = \dfrac{\mu_0 \pi i I r}{4 R}