Question

A rectangular loop of sides a and b carrying current I is placed in a magnetic field `vec B` such that its area vector `vec A` makes an angle θ with `vec B`. With the help of a suitable diagram, show that the torque `vec tau` acting on the loop is given by `vec tau = vec m xx vec B`, where `vec m(= I vec A)` is the magnetic dipole moment of the loop.

Answer 1

A current-carrying loop in a magnetic field experiences torque due to magnetic forces on its sides.

Magnetic force on a current element:

`vec F = I(vec l xx vec B)`

Magnitude of force for side of length a:

F = IaB sin θ

Torque = force × perpendicular distance between forces

τ = F × b

= (IaB sin θ)b

= IabB sin θ    ...[Since Area of rectangular loop (A) = ab]

= IAB sin θ

Magnetic dipole moment of loop `(vec m) = I vec A`

So the magnitude of torque (τ) = mB sin θ

The direction of torque is perpendicular to both `vec m` and `vec B`.

Thus:

`vec tau = vec m xx vec B`