JEE MAIN - Physics (2015 (Offline) - No. 12)
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 $$'\theta '$$ with the vertical. If wires have mass $$\lambda $$ per unit-length then the value of $$I$$ is :
($$g=$$ $$gravitational$$ $$acceleration$$ )
_en_12_1.png)
($$g=$$ $$gravitational$$ $$acceleration$$ )
$$2\sqrt {{{\pi gL} \over {{\mu _0}}}\tan \theta } $$
$$\sqrt {{{\pi \lambda gL} \over {{\mu _0}}}\tan \theta } $$
$$\sin \theta \sqrt {{{\pi \lambda gL} \over {{\mu _0}\,\cos \theta }}} $$
$$2\sin \theta \sqrt {{{\pi \lambda gL} \over {{\mu _0}\,\cos \theta }}} $$
Explanation
Let us consider $$'\ell '$$ length of current carrying wire,
At equilibrium
$$T\cos \theta = \lambda g\ell $$
_en_12_2.png)
and $$T\sin \theta = {{{\mu _0}} \over {2\pi }}{{I \times Il} \over {2L\sin \theta }}$$
$$\left[ {\,\,} \right.$$ as $$\left. {{{{F_B}} \over \ell } = {{{\mu _0}} \over {4\pi }}{{2I \times I} \over {2\ell \sin \theta }}\,\,} \right]$$
Therefore, $$I = 2\sin \theta \sqrt {{{\pi \lambda gL} \over {{u_0}\cos \theta }}} $$
At equilibrium
$$T\cos \theta = \lambda g\ell $$
and $$T\sin \theta = {{{\mu _0}} \over {2\pi }}{{I \times Il} \over {2L\sin \theta }}$$
$$\left[ {\,\,} \right.$$ as $$\left. {{{{F_B}} \over \ell } = {{{\mu _0}} \over {4\pi }}{{2I \times I} \over {2\ell \sin \theta }}\,\,} \right]$$
Therefore, $$I = 2\sin \theta \sqrt {{{\pi \lambda gL} \over {{u_0}\cos \theta }}} $$
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