# Torque Experienced by a Rectangular Coil in a Uniform Magnetic Field

Consider a rectangular coil, PQRS is carrying current ‘I’ as shown in figure. Let PQ=RS=L and QR=SP=B

Suppose the coil is placed in a uniform magnetic field of strength ‘B’ (magnetic flux density) as shown in figure. Let θ be the angle between the magnetic field and plane of coil.

Now,

Force experienced by PQ, F_{1}=BI.PQ.sin 90º

∴ F_{1 }= BI.l [perpendicularly outward]

Force experienced by QR, F_{2}=BI.QR.sinθ

∴ F_{2 }= BIb.sinθ (downward)

Force experienced by RS, F_{3}=BI.RS.sin 90º

∴ F_{3 }= BI.l [perpendicularly inward]

Force experienced by SP, F_4=BI.SP.sin(π-θ)

∴ F_{4 }=BI.b.sinθ (upward)

Here, F_{1} and F_{3} are equal, opposite and acting at a certain separation,thus they form couple. F_{2} and F_{4} are equal, opposite and acting in a straight line. Their effect will be cancelled by the rigidity of the coil.

**Torque produced by the couple**

In figure,

τ = F_{1} or F_{3} × XZ

τ =BI.lb.cosθ

τ =BI.A.cosθ

If there are ‘N’ turns of coil,

τ =BI.A.cosθ.N

∴ τ =BINAcosθ

This is the required relation.

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