Derivation of Expression for Torque on a Rectangular Coil

Let us consider a rectangular coil PQRS of length ‘l’ and breadth ‘b’ suspended in uniform horizontal magnetic field ‘B’. Let the plane of the coil make an angle ‘θ’ with the direction of magnetic field.

When current ‘I’ is passed through the coil, the magnetic force produced on the various arms of the coil are:

i) Force on arm PQ,

F₁ = IlBsinθ
     = IlBsin90^o
     = IlB
∴ F₁ is perpendicular to both l and B and is directed outwards.

ii) Force on arm QR,

F₂ = IbBsinθ
∴ F₂ is perpendicular to both b and B and is directed downwards.

iii) Force on arm RS,

F₃ = IlBsinθ
     = IlBsin90^o
     = IlB
∴ F₃ is perpendicular to both l and B and is directed inwards.

iv) Force on arm SP,

F₄ = IbBsinθ
∴ F₄ is perpendicular to both b and B is directed upwards.

The forces F₂ and F₄ are equal and opposite and they pass through same line of action, so they cancel each other resulting to no torque or force on the coil.
The forces F₁ and F₃ are also equal and opposite but they pass through different lines of action, so they constitute couple or torque which is given by:

τ = (F₁ or F₃) × Arm of couple
   = IlB × bcosθ
   = IBAcosθ [∵ A = l × b]

If there are ‘N’ number of turns in the coil, then the total torque is:
   τ = N × IBAcosθ
∴ τ = BINAcosθ

Special Cases:
i) If θ = 0^o, τ = BINA (maximum value)
When the plane of the coil is kept parallel to the magnetic field, torque on the coil is maximum.
ii) If θ = 90^o, τ = 0 (minimum value)
When the plane of the coil is kept perpendicular to the magnetic field, torque on the coil is minimum.