Question 234: Magnetism and Matter

Question

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 $'	heta '$ with the vertical. If wires have mass $\lambda $ per unit-length then the value of $I$ is :
($g=$ $gravitational$ $acceleration$ )
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l91f7cuv/3a7bb61e-6287-4a8c-a932-a655d2cc36cb/ce8d2580-47db-11ed-8284-6d7e98c66709/file-1l91f7cuw.png?format=png)

Accepted Answer

Let us consider $'\ell '$ length of current carrying wire,
At equilibrium
$T\cos 	heta  = \lambda g\ell $
and $T\sin 	heta  = {{{\mu _0}} \over {2\pi }}{{I 	imes Il} \over {2L\sin 	heta }}$
$\left[ {\,\,} ight.$ as $\left. {{{{F_B}} \over \ell } = {{{\mu _0}} \over {4\pi }}{{2I 	imes I} \over {2\ell \sin 	heta }}\,\,} ight]$
Therefore, $I = 2\sin 	heta \sqrt {{{\pi \lambda gL} \over {{u_0}\cos 	heta }}} $
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l91f8ury/49827386-97e5-4277-8921-73e46ec64f98/f833aee0-47db-11ed-8284-6d7e98c66709/file-1l91f8urz.png?format=png)

Answer

$2\sqrt {{{\pi gL} \over {{\mu _0}}}	an 	heta } $

Answer

$\sqrt {{{\pi \lambda gL} \over {{\mu _0}}}	an 	heta } $

Answer

$\sin 	heta \sqrt {{{\pi \lambda gL} \over {{\mu _0}\,\cos 	heta }}} $

Answer

$2\sin 	heta \sqrt {{{\pi \lambda gL} \over {{\mu _0}\,\cos 	heta }}} $

Magnetism and Matter

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