Torque on a bar magnet placed in a uniform magnetic field
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 (Fig.).
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 ? perpendicular distance between them
τ = F ? NA
= mB ? NA ...(1)
= mB ? 2l sin θ
∴ τ = MB sin θ ...(2)
Vec τ = Vec M ? Vec B
The direction of τ is perpendicular to the plane containing Vec M and Vec B.
If B = 1 and θ = 90o
Then from equation (2), τ = M
Hence, 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.