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 '$\theta$' with the vertical. If wires have mass $\lambda$ per unit length then the value of I is :
(g = gravitational acceleration)

• A. sin$\theta \sqrt{\frac{\pi \lambda gL}{{\mu }_{0}cos\theta }}$
• B. 2sin$\theta \sqrt{\frac{\pi \lambda gL}{{\mu }_{0}cos\theta }}$
• C. 2$\sqrt{\frac{\pi gL}{{\mu }_0}tan\theta }$
• D. $\sqrt{\frac{\pi \lambda gL}{{\mu }_0}tan\theta }$

Answer 1

Answer: (B)

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

The wires repel each other with some force.

The magnitude of this force can be given by

$F=\overrightarrow{Il}\times \overrightarrow{B}$ where l is length of wire

$B=\frac{{\mu }_{\circ }I}{2\pi r}$

where r is distance between two wires

$r=2L\sin \theta$
$F=\frac{{\mu }_{\circ }{I}^{2}l}{2\pi 2L\sin \theta }$...(i)

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

From FBD of the wire

$T\sin \theta =F$...(ii)

$T\cos \theta =mg$...(iii)

dividing equation (ii) by (iii)

$\tan \theta =\frac{F}{mg}$

$m=\lambda L$

$I=2\sin \theta \sqrt{\frac{\pi \lambda gL}{{\mu }_{\circ }\cos \theta }}$