Two long current carrying thin wires, both with current $I$, are held by insulating threads oflength $L$ 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 $l$ is
($g =$ gravitational acceleration)
($g =$ gravitational acceleration)
JEE MAIN 2015, Diffcult
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Let us consider $'l'$ length of current carrying wire.
At equilibrium
$\mathrm{T} \cos \theta=\lambda \mathrm{g} \ell$
and $T \sin \theta=\frac{\mu_{0}}{2 \pi} \frac{I \times I l}{2 L \sin \theta}\left[\because \frac{F_{B}}{\ell}=\frac{\mu_{0}}{4 \pi} \frac{2 I \times I}{2 \ell \sin \theta}\right]$
Therefore, $I=2 \sin \theta \sqrt{\frac{\pi \lambda g L}{u_{0} \cos \theta}}$
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