A rectangle loop of size `l xx b` carrying a steady current I is placed in a uniform magnetic field `vecB`. Prove that the torque `vec(tau)` acting on the loop is given by `vec(tau) = vecm xx vecB`, where `vecm` is the magnetic moment of the loop.

Explanation

Consider a rectangular coil PQRS of length l and breadth b, carrying current I, placed in a uniform magnetic field B such that a vectro normal to the plane of the coil subtends an angle `theta` from the direction of B.
In this situation forces `F_1 and F_2` having magnitude `I b B sin theta` are acting on arms PQ and RS. The forces are equal, opposite nad collinear, hence they cancel out.
Again forces `F_3 and F_4` having magnitude `I l B` are acting on each of the two arms QR and SP. These forces too are equal and opposite but these are non-collinear and form a couple whose torque is given as :
`tau = (I lB) xx b sin theta = I (lb) B sin theta = I A B sin theta`
where `A = I b` = area of the loop. If instead of a single loop, we have a rectangular coil having N turns, then
torque `tau = N I A B sin theta`
In vector notation `vec(tau) = N I (vecA xx vecB) = (vecm xx vecB) `, where `vecm = N I vecA` = magnetic moment of current carrying coil.