A conducting rod with resistance r per unit l | vedclass

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Question

(C) The length of the rod between the rails is $L = \frac{d}{\sin 	heta}$.The induced electromotive force $(EMF)$ in the rod is $\varepsilon = B v d$

Vedclass · Accepted Answer

$F=\frac{B^2d^2v/\sin 	heta}{(R+dr/\sin 	heta)}$

Vedclass · Answer

(C) The length of the rod between the rails is $L = \frac{d}{\sin 	heta}$.The induced electromotive force $(EMF)$ in the rod is $\varepsilon = B v d$.The resistance of the rod is $R_{rod} = L \cdot r = \frac{d}{\sin 	heta} \cdot r$.The total resistance of the circuit is $R_{eq} = R + R_{rod} = R + \frac{dr}{\sin 	heta}$.The induced current in the circuit is $i = \frac{\varepsilon}{R_{eq}} = \frac{Bvd}{R + \frac{dr}{\sin 	heta}}$.The magnetic force on the rod is $F_m = i L B = i \left(\frac{d}{\sin 	heta}ight) B$.Substituting the value of $i$,we get $F_m = \left(\frac{Bvd}{R + \frac{dr}{\sin 	heta}}ight) \left(\frac{d}{\sin 	heta}ight) B = \frac{B^2 d^2 v / \sin 	heta}{R + \frac{dr}{\sin 	heta}}$.Since the rod moves at a constant speed,the external force $F$ must balance the magnetic force,so $F = F_m = \frac{B^2 d^2 v / \sin 	heta}{R + \frac{dr}{\sin 	heta}}$.

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$	$0$

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