A wire carrying current I is tied between points P and Q and is in… | JEE Main 2015 (11 Apr, Shift 1)

Q54JEE Main 2015 · 11 APR · SHIFT 1Difficult

Force on Current-Carrying Conductors

A wire carrying current I is tied between points P and Q and is in the shape of a circular arch of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle 2 $\theta$ 0 at the center of the circle (of which it forms an arch) then the tension in the wire is:

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Answer: (a)

Take a small element $dl = R\,d\phi$ . Magnetic force on it is radial: $dF = IB R\,d\phi$ .
Tension at its two ends gives a net radial pull $2T\sin\frac{d\phi}{2} = T\,d\phi$ .
Balance: $T\,d\phi = IBR\,d\phi \Rightarrow T = IBR$ (independent of $\theta_0$ ).
So the answer is (a) IBR.

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