Question

A circular loop of radius r carrying a current i is held at the centre of another circular loop of radius R(>>r) carrying a current I. The plane of the smaller loop makes an angle of 30° with that of the larger loop. If the smaller loop is held fixed in this position by applying a single force at a point on its periphery, what would be the minimum magnitude of this force? 

Answer 1

Given:
For the outer loop,
Magnitude of current = I
Radius of the loop = R 
Thus, the magnetic field at the centre due to the larger loop is given by

\[B = \frac{\mu_0 I}{2R}\]

Let A be the area of the smaller loop and let current i pass through it.
Now,
Angle between the area vector of the smaller loop and the magnetic field due to the larger loop = 30°
Thus, the torque on the smaller loop is given by

\[\Gamma = i( \vec{A} \times \vec{B} )\] 

    = iABsin 30°

\[= i\pi r^2 \frac{\mu_0 I}{4R}\]
\[ = \frac{\mu_0 \pi r^2 Ii}{4R}\]

If the smaller loop is held fixed in its position, then
Torque due to the magnetic field = Torque due to the external force at its periphery

\[\Rightarrow Fr = \frac{\mu_0 \pi r^2 Ii}{4R}\]
\[ \Rightarrow F = \frac{\mu_0 \pi Iir}{4R}\]

This is the minimum magnitude of force to balance the given condition.