Torque on a Bar Magnet in Uniform Magnetic Field:
Consider a bar magnet NS of dipole length 2l placed in a uniform magnetic field \(\vec B\). Let m be the pole strength of its each pole. Let the magnetic axis of the bar magnet make an angle θ with the magnetic field \(\vec B\), as shown in figure
Force on N-pole = mB, along \(\vec B\)
Force on S-pole = mB, opposite to \(\vec B\)
The force on the two poles are equal and opposite. They form a couple, moment of couple or torque is given by relation
τ = force × perpendicular distance from the axis of rotation
= mB × 2lsinθ
= (m × 2l)Bsinθ
or τ = MBsinθ ……………. (1)
where, M = m × 2l is the magnetic dipole moment of the bar magnet.
In vector notaion, \(\vec{\tau}=\vec{M} \times \vec{B}\) ……………. (2)
Special Cases:
(i) When bar magnet is parallel to magnetic field (θ = 0), then
τ = MBsin0 = 0
(ii) When bar magnet is perpendicular to magnetic field (θ = 90°), then
τmax = MB sin 90°= MB
The direction of the torque 
is given by the right hand screw rule as indicated in the figure 8.18. The effect of the torque is to make the magnet align itself parallel to the field \(\vec B\). That is why a freely suspended magnet aligns itself in the N-S direction because the earth has its own magnetic field which exerts a torque on the magnet tending it to align along the field.
Definition of Magnetic Dipole Moment: If in equation (1), B = 1, θ = 90° , then τ = M
Hence the magnetic dipole moment may be defined as the torque acting on a magnetic dipole placed perpendicular to a uniform magnetic field of unit strength,
SI Unit of Magnetic Moment
