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. $2\sin\theta\sqrt{\frac{\pi \lambda gL}{\mu_0 \cos\theta}}$
• B. $\sin\theta\sqrt{\frac{\pi \lambda gL}{\mu_0 \cos\theta}}$
• C. $2\sqrt{\frac{\pi gL}{\mu_0}\tan \theta}$
• D. $\sqrt{\frac{\pi \lambda gL}{\mu_0}\tan \theta}$

Answer 1

Answer: (A)

forces per unit length are taken
$\tan \theta=\frac{\frac{\mu_{0} i^{2}}{2 \pi d }}{\lambda_{ g .}}$
$i ^{2}=\frac{\lambda .g .\sin \theta}{\mu_{0} .\cos \theta}(2 \pi)\, d \,\,\,\,\,\,[ d =2 L \sin \theta]$
$i =2 \sin \theta \sqrt{\frac{\lambda g \pi L }{\mu_{0} \cos \theta}}$