JEE Main 2015 · 4 Apr · Shift 1 · Physics Q20

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PhysicsSingle correct+4 / -1Question 20 of 90

Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle 'θ' with the vertical. If wires have mass λ per unit length then the value of I is :

(g = gravitational acceleration)

Two wires seen end-on, each hanging from a common point by a thread of length L at angle θ to the vertical, carrying current I.

\sin\theta\sqrt{\dfrac{\pi\lambda gL}{\mu_0\cos\theta}}

2\sin\theta\sqrt{\dfrac{\pi\lambda gL}{\mu_0\cos\theta}}

2\sqrt{\dfrac{\pi gL}{\mu_0}\tan\theta}

\sqrt{\dfrac{\pi\lambda gL}{\mu_0}\tan\theta}

Answer and worked solution

Official answer: B

The wires repel. Their separation is $d=2L\sin\theta$ . Per unit length: $T\cos\theta=\lambda g$ and $T\sin\theta=\dfrac{\mu_0I^2}{2\pi d}$ . So
$\frac{\mu_0I^2}{4\pi L\sin\theta}=\lambda g\tan\theta\implies I=2\sin\theta\sqrt{\frac{\pi\lambda gL}{\mu_0\cos\theta}}.$

Topics:Force on a current-carrying conductor Free body diagrams

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 20