JEE Main 2015 · 11 Apr · Shift 1 · Physics Q20

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PhysicsSingle correct+4 / -1Question 20 of 90

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 centre of the circle (of which it forms an arch) then the tension in the wire is :

A wire arching upward from P to Q in a field B into the page; radii of length R from the centre below make angles θ₀ either side of the vertical.

IBR

\dfrac{IBR}{\sin\theta_0}

\dfrac{IBR}{2\sin\theta_0}

\dfrac{IBR\theta_0}{\sin\theta_0}

Answer and worked solution

Official answer: A

A small element subtending $d\theta$ feels an outward magnetic force $IB\,R\,d\theta$ , balanced by the two tensions: $2T\sin\frac{d\theta}2=T\,d\theta$ . So $T=IBR$ .

Topics:Force on a current-carrying conductor

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JEE Main 2015 · 11 Apr · Shift 1, Physics question 20