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

Explanation:

To find: The minimum magnitude of the force under given condition

Step 1:

Given:

To the outside circle,

Current Magnitude = I

Loop Radius = R

The magnetic field at the center is thus supplied by the larger loop

$B=\frac{\mu_{0} I}{2 R}$

Let A be the smaller loop field, and let current pass through it.

Step 2:

Angle between the smaller loop area vector and the magnetic field because of the larger loop = 30 ° Thus the smaller torque  is given by

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

  $=\text { iABsin } 30^{\circ}$

  $=i \pi r^{2} \frac{\mu_{0} I}{4 R}$

  $=\frac{\mu_{0} I i \pi r^{2}}{4 R}$

If the smaller loop is fixed in its place, then  

Step 3:

Torque due to the magnetic field = Torque due to its peripheral external force

$F r=\frac{\mu_{0} \lim r^{2}}{4 R}$

$F=\frac{\mu_{0} I i \pi r^{2}}{4 R r}$

$F=\frac{\mu_{0} l i \pi r}{4 R}$

That is the minimum force magnitude to satisfy the condition given.