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(B) As the rod moves down the rails with velocity $v$,the magnetic flux $\phi$ through the loop changes.According to Faraday's law of electromagnetic

Vedclass · Accepted Answer

$\frac{{mgR\sin 	heta }}{{{B^2}{l^2}}}$

Vedclass · Answer

(B) As the rod moves down the rails with velocity $v$,the magnetic flux $\phi$ through the loop changes.According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $e = B l v$.The induced current in the circuit is $i = \frac{e}{R} = \frac{B l v}{R}$.The magnetic force acting on the rod is $F_m = i l B = \left( \frac{B l v}{R} ight) l B = \frac{B^2 l^2 v}{R}$,which acts upwards along the rails.The component of the gravitational force acting down the rails is $F_g = mg \sin 	heta$.At terminal speed,the net force on the rod is zero,so $F_g = F_m$.$mg \sin 	heta = \frac{B^2 l^2 v}{R}$.Solving for $v$,we get the terminal speed $v = \frac{mgR \sin 	heta}{B^2 l^2}$.

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