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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)

 

Option: 1

\sin \Theta \sqrt{\frac{\pi \lambda gL}{\mu _{0}\cos \Theta }}


Option: 2

2\sin \Theta \sqrt{\frac{\pi \lambda gL}{\mu _{0}\cos \Theta }}


Option: 3

2 \sqrt{\frac{\pi \ gL}{\mu _{0}}\tan \Theta }


Option: 4

\sqrt{\frac{\pi \lambda gL}{\mu _{0}}\tan \Theta }


Answers (1)

best_answer

The wires repel each other with some force.

The magnitude of this force can be given by

F=\overrightarrow{I l} \times \vec{B}$ where I is length of wire \\ $B=\frac{\mu_{0} I}{2 \pi r}$ where $r$ is distance between two wires \\ $r=2 L \sin \theta$\\ so \ $F=\frac{\mu_{o} I^{2} l}{2 \pi 2 L \sin \theta}

Now the tension in the thread is keeping both wires from moving to sideways

From FBD of the wire 

T \sin \theta=F_{\dots}(\mathrm{ii})$\\ $T \cos \theta=m g_{\cdots \cdot}(\mathrm{iii})$\\ dividing equation (ii) by (iii)\\ $\tan \theta=\frac{F}{m g}

and we know m=\lambda l

on solving we get

I=2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_{0} \cos \theta}}

Posted by

Pankaj

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