A wire ab of length l,mass m,and resistance R | vedclass

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Question

(C) The component of the magnetic field $B$ perpendicular to the area of the loop is $B \cos 	heta$. The motional $emf$ induced in the wire is $\vare

Vedclass · Accepted Answer

$B = \sqrt {\frac{{mgR\,\sin \,	heta }}{{v{l^2}\,{{\cos }^2}\,	heta }}} $

Vedclass · Answer

(C) The component of the magnetic field $B$ perpendicular to the area of the loop is $B \cos 	heta$. The motional $emf$ induced in the wire is $\varepsilon = B l v \cos 	heta$. The induced current in the circuit is $I = \frac{\varepsilon}{R} = \frac{B l v \cos 	heta}{R}$. The magnetic force acting on the wire is $F_m = I l B$. This force acts horizontally. The component of this magnetic force along the inclined plane is $F_{m, 	ext{parallel}} = F_m \cos 	heta = (I l B) \cos 	heta = \frac{B^2 l^2 v \cos^2 	heta}{R}$. For the wire to slide at a constant speed,the net force along the incline must be zero. The component of gravity along the incline is $mg \sin 	heta$. Equating the forces: $mg \sin 	heta = \frac{B^2 l^2 v \cos^2 	heta}{R}$. Solving for $B$: $B^2 = \frac{mgR \sin 	heta}{v l^2 \cos^2 	heta} \implies B = \sqrt{\frac{mgR \sin 	heta}{v l^2 \cos^2 	heta}}$.

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