Question

A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field ${\displaystyle \overrightarrow{B}}$. Prove that the torque  ${\displaystyle \overrightarrow{\tau }}$acting on the loop is give by ${\displaystyle \overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B},}$where ${\displaystyle \overrightarrow{m}}$ is the magnetic moment of the loop.

Answer 1

Plane of the loop is at an angle with the direction of the magnetic field,

Let the angle between the field and the normal is θ. The forces on BC and DA are equal and opposite and they cancel each other as they are collinear.

Force on AB is F₁ and force on CD is F₂, thus

F₁ = F₂ = IbB

Magnitude of torque on the loop as in the figure:

${\displaystyle \because \tau =F_1\frac{1}{2}\sin \theta +F_2\frac{1}{2}\sin \theta =IlB\sin \theta }$

or, τ = IAB sin θ (Where area,A = lb)

If there are ‘n’ such turns the torque will be nIAB sinθ

Magnetic moment of the current, m = IA

${\displaystyle \overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}}$