Compute the torque experimenced by a magnetic needle in a uniform magnetic field.

Explanation

Consider a magnet of length 2l of pole strength `q_(m)` kept in a uniform magnetic field `vecB` as shown in Figure. Each pole experiences a force of magnitude `q_(m)B` but acts opposite direction.
Therefore, the net force exerted on the magnet is zero, so that there is no translatory motion. These two forces constitute a couple ( about midpoint of bar magnet ) which will rotate and try to align in the direction of the magnetic field `vecB`.

The force experienced by north pole,
`vecF_(N) = q_(m) vecB` .....(1)
The force experienced by south pole,
`vecF_(S) = - q_(m) vecB` ...(2)
Adding equations (1) and (2) , we get the net force acting on the dipole as
`vecF = vecF_(N) + vecF_(S) = vec0`
This implies, that the net force acting on the dipole is zero , but forms a couple which tends to rotate the bar magnet clockwise ( here ) in order to align it along `vecB`.
The moment of force or torque experienced by north and south pole about point O is
`vec tau = vec(ON) xx vecF_(N) + vec(OS) xx vecF_(S) `
` vectau = vec(ON) xx q_(m) vecB + vec(OS) xx (-q_(m) vecB)`
By using right hand cork screw rule, we conclude that the total torque is pointing into the paper . Since the magnitudes
`|vec(ON)|= |vec(OS)| = l " and " |q_(m) vecB| = |- q_(m) vecB|`, magnitude of total torque about point O
`tau = l xx q_(m) B sin theta + l xx q_(m) B sin theta`
` tau = 2l xx q_(m)B sin theta`
` tau = p_(m) Bsin theta" " (.:. q_(m) xx 2l = p_(m))`
In vector notation , `vectau = vecp_(m) xx vecB`