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
$B=\frac{{\mathrm{\mu }}_{0}I}{2R}$
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
$\Gamma =i(\overrightarrow{A}\times \overrightarrow{B})$
= iABsin 30°
$$\begin{gathered} =i\mathrm{\pi }{r}^2\frac{{\mathrm{\mu }}_{0}I}{4R} \\ =\frac{{\mathrm{\mu }}_0\mathrm{\pi }{r}^{2}Ii}{4R} \end{gathered}$$
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
$$\begin{gathered} \Rightarrow Fr=\frac{{\mathrm{\mu }}_0\mathrm{\pi }{r}^{2}Ii}{4R} \\ \Rightarrow F=\frac{{\mathrm{\mu }}_0\mathrm{\pi }Iir}{4R} \end{gathered}$$
This is the minimum magnitude of force to balance the given condition.