Consider a bar magnet NS of length 2l and pole strength m placed in a uniform magnetic field of induction B at an angle θ with the direction of the field (Figure).
Due to the magnetic field B, a force mB acts on the North Pole along the direction of the field and a force mB acts on the South Pole along the direction opposite to the magnetic field.
These two forces are equal and opposite, hence constitute a couple.
The torque τ due to the couple is
τ = one of the forces x perpendicular distance between them
τ = F x NA
= mB x NA … …. (1)
= mB × 2l sin θ
so, τ = MB sin θ
Vectorially, τ = M x B
The direction of τ is perpendicular to the plane containing M and B.
If, B = 1 and θ = 90 θ
Then from equation (2), τ = M
Hence, the moment of the magnet M is equal to the torque necessary to keep the magnet at right angles to a magnetic field of unit magnetic induction.