# 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)

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