A circular loop of radius r carrying a current i is held at the center of another circular loop of radius R (>>r) carrying a current I. The plane of the smaller loop makes and 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 the force?

Given:

Current carried by circular loop of radius r = i

Current carried by circular loop of radius R(>>r) = I

Angle between the planes of the two loops = 30⁰

Diagram:

Formula used:

Magnetic field at the center of a circular loop of radius r

M= ,

where

I = current through the loop,

r = radius of the circle,

𝛍₀ = 4π x 10^-7 kgm s^-2 A^-2

Magnetic torque acting on a loop of radius r

where

I = current carried by the loop,

A = area of loop,

B = magnetic field,

F = force on periphery

Therefore, magnetic field at the center of the loops due to the bigger loop(B) =

Area of the smaller loop(A) = πr²

Therefore, torque on the smaller loop = i (A X B), since the smaller loop carries current i.

A = area of smaller loop, B = magnetic field due to bigger loop.

Now, we know that A X B = |A||B|sinθ,

Where

θ is the angle between the area vector of the smaller loop and the magnetic field of the larger loop.

|A| = magnitude of area of smaller loop,

|B| = magnitude of magnetic field due to larger loop.

Given:

The angle between the area vector of the smaller loop and the magnetic field of the larger loop is θ = 30⁰.

Hence, torque acting on the smaller loop = i(|A||B|sin 30⁰) =

We know, |A| = πr², |B| =

Hence,

Torque on smaller loop due to magnetic field,

Torque due to external force on its periphery,

Therefore

Where

F = external force,

r = perpendicular distance between the line of action of force and the point at which the torque is acting.

Here, F and r are perpendicular to each other, since the force is being applied to the periphery.

Here r = radius of smaller loop.

Hence,

which gives us

(Answer)