Derive an expression for the torque expereinced by a amgnetic dipole suspened in a uniform magneic field.

Explanation

Let a bar magnet NS is placed in a uniform magnetic field of strength B. Let m= Pole strength of each pole of magnet and NS= 2I= Magnetic dipole length
Suppose `theta`= Angle between the plane of magnet and direction of magnetic field
Force acting on North Pole `= F_(1) = m vec(B)` (in the direction of `vec(B)`)
Force acting on South Pole `=F_(2)= m vec(B)` (in a direction opposite to `vec(B)`)

These two forces `vec(F)_(1) and F_(2)` are equal and opposite and are having different lines of action. They constitute a couple and rotate the dipole in clockwise direction as shown in fig above. Torque acting on dipole is given by the product of magnitude of either force and perpendicular distance between them,
`therefore tau= m B xx` distance NA
`=m B xx 2l sin theta`
or `tau= (m.2l ) B sin theta [(because (NA)/(NS)= sin theta),(therefore NA= 2l sin theta)]`
`rArr tau = MB sin theta [ because m.2l= M]`
In vector form, `vec(tau) = vec(M) xx vec(B)`
which is the required expression for the torque acting on a magnetic dipole placed in a uniform magnetic field.