Question

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

Answer 1

Consider a magnet of length 21 of pole strength q_m kept in a uniform magnetic field `vec"B"` Each pole experiences a force of magnitude q_mB but acts in 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 `vec"B"`.

The force experienced by north pole, `vec"F"_"N" = "q"_"m" vec"B"`    ....(1)

The force experienced by south pole, `vec"F"_"S" = "q"_"m" vec"B"`    ....(2)

Adding equations (1) and (2), we get the net force acting on the dipole as

`vec"F" = vec"F"_"N" + vec"F"_"S" = vec"O"`

Magnetic dipole kept in a uniform magnetic field

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 `vec"B"`.

The moment of force or torque experienced by north and south pole about point O is

`vectau = vec"ON" xx vec"F"_"N" + vec"OS" xx vec"F"_"S"`

`vectau = vec"ON" xx "q"_"m" vec"B" + vec"OS" xx (- "q"_"m" vec"B")`

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" vec"B"| = |- "q"_"m" vec"B"|`,

The magnitude of total torque about point O

τ = l × q_m B sin θ +l × q_m B sin θ

τ = 2l xq_m B sin θ

τ = P_m B sin θ

(∴ q_m × 2l = P_m )

In vector notation, τ = p_m × `vec"B"`