Two long current-carrying thin wires,both wit | vedclass

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Question

(A) Let us consider a length $l$ of the current-carrying wire.At equilibrium,the forces acting on the wire are tension $T$,gravitational force $(\lamb

Vedclass · Accepted Answer

$2 \sin 	heta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos 	heta}}$

Vedclass · Answer

(A) Let us consider a length $l$ of the current-carrying wire.At equilibrium,the forces acting on the wire are tension $T$,gravitational force $(\lambda l)g$,and magnetic force $F_B$.The distance between the two wires is $r = 2L \sin 	heta$.The magnetic force per unit length is $\frac{F_B}{l} = \frac{\mu_0 I^2}{2 \pi r} = \frac{\mu_0 I^2}{2 \pi (2L \sin 	heta)} = \frac{\mu_0 I^2}{4 \pi L \sin 	heta}$.Resolving forces in the vertical and horizontal directions:$T \cos 	heta = \lambda l g$$T \sin 	heta = F_B = \frac{\mu_0 I^2 l}{4 \pi L \sin 	heta}$Dividing the two equations:$	an 	heta = \frac{F_B}{\lambda l g} = \frac{\mu_0 I^2 l}{4 \pi L \sin 	heta \cdot \lambda l g} = \frac{\mu_0 I^2}{4 \pi \lambda g L \sin 	heta}$Solving for $I$:$I^2 = \frac{4 \pi \lambda g L \sin 	heta 	an 	heta}{\mu_0} = \frac{4 \pi \lambda g L \sin^2 	heta}{\mu_0 \cos 	heta}$$I = 2 \sin 	heta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos 	heta}}$.

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