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Q.Arrive at an expression for torque on a rectangular current loop in a uniform magnetic field.

Kerala DhseKerala DHSE Plus Two Board 2026Subjective· 2mImportance★★★★★
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A current loop in a uniform field B experiences a torque τ=NIABsin⁡θ\tau = NIAB\sin\theta, arising from a pair of equal and opposite forces (a couple) on the two sides of the loop perpendicular to B.

Consider a single-turn rectangular loop PQRS carrying current I, with sides PQ = RS = aa and QR = SP = bb, placed in a uniform magnetic field B lying in the plane of the page. Let the normal n^\hat{n} to the loop make an angle θ with B (equivalently, the loop is tilted by θ from the position where its plane contains B).

Forces on QR and SP (length bb, always perpendicular to B):

Each carries current I in a direction perpendicular to B, so each experiences a force of magnitude

F=BIbF = BIb

These two forces are equal in magnitude, opposite in direction, and act along the two parallel sides QR and SP — a classic force couple.

Perpendicular distance between the two forces:

Because the loop is tilted by θ from the plane containing B, the perpendicular distance between the lines of action of these two forces is

asin⁡θa\sin\theta

Torque of the couple:

τ=F×(perpendicular distance)=(BIb)(asin⁡θ)=BIAsin⁡θ\tau = F\times(\text{perpendicular distance}) = (BIb)(a\sin\theta) = BIA\sin\theta …

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