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

Q: 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)

a 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, $I=2 \sin \theta \sqrt{\frac{\pi \lambda g L}{u_{0} \cos \theta}}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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)

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

JEE MAIN 2015, Diffcult

STD 11 Science
NEET
STD 12 - 4. Moving charges and magnetism
Physics

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