6 Torque on Current Loop

Torque on Current Loop

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 =

= I(ab)B

τ = IAB

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:

∴ τ =

= Iab Bsinθ

τ = IAB sinθ

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

• Magnetic moment of the current, m = IA