Derive the expression for the torque acting on a current carrying loop placed in a magnetic field.

Explanation

Consider a rectangular current carrying loop placed in a uniform magnetic field `vec B.` Let l and b be the length and breadth of the loop respectively. Let `theta ` be the angle between the normal to the plane and the magnetic field `vec B.`

Force on a current carrying cunductor of length l is given by, `vec F = l ( vec l xx vec B)....(i)`
Force on arm PQ
`vecF _(1) = l (vecl xx vec B) (bot r ` to the arm PQ and direction outwords)....(ii)
Force on arm RS
`vec F _(2)= l ( vec l xx vec B) (bot r` to the arm RS & and direction outwards).....(iii)
Force acting on arm QR and SP are equal in magnitude and opposite in direction. Therefore they cancel each other.
Force `vec F _(1) and vec F _(2)` constitute a couple.
Hence, loop experience a torque which is given by
`tau =` Magnitude of either force `xx` Lever arm `= F _(1) xx SM`
`= (IIB sin 90^(@)) xx SM " "[because theta= 90 ^(@)]`
`= Il B xx b sin theta " "[SM = b sin theta]`
`= I l B xx b sin theta `
`= I ( lb ) B sin theta = I AB sin theta `
For n turns `tau = nLAB sin theta = MB sin theta or vec tau = vec M xx vec B [because M = n lA]`