Question

Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field.

Answer 1

 Consider a rectangular loop ABCD carrying current I.

Case I - The rectangular loop is placed such that the uniform magnetic field B is in the plane of loop.

No force is exerted by the magnetic field on the arms AD and BC.

Magnetic field exerts a force F₁ on arm AB.

∴F₁ = IbB

Magnetic field exerts a force F₂ on arm CD.

∴F₂ = IbB = F₁

Net force on the loop is zero.

The torque on the loop rotates the loop in anti-clockwise direction.

Torque, τ = `f_1  a/2 + f_2  a/2`

`= IbBa/2 +IbBa/2`

= I(ab)B

τ = BIA

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

where, b → Breadth of the rectangular coil

a → Length of the rectangular coil

A = ab → Area of the coil

Case II - Plane of the loop is not along the magnetic field, but makes angle with it.

Angle between the field and the normal is θ.

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.

F₁ = F₂ = IbB

Magnitude of torque on the loop as in the figure: