
Consider any small element dl of arc of angle dθ . Then magnetic force dF=Idl.B outwards

For equilibrium of element
IdlB=2Tsin2dθ
For small angle sinθ≃θ
IdlB=2T2dθ
IdlB=TRdl
T=IBR

JEE Main 2015 — Physics Electromagnetism
A wire carrying current I is tied between points P and Q and is in the shape of a circular arc of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, as shown in the figure) in the vicinity of the wire. If the wire subtends an angle 2θo at the center of the circle (of which it forms an arch) then the tension in the wire is:

Held on 11 Apr 2015 · Verified 6 Jul 2026.
IBR
sinθ0IBR
2sinθ0IBR
sinθ0IBRθ0
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