A conducting rod PQ of length L = 1.0, m is m | vedclass

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Question

(A) The motional electromotive force $(EMF)$ induced in the conducting rod $PQ$ is given by $e = BvL$.Substituting the given values: $e = 4.0\, T 	im

Vedclass · Accepted Answer

$q_A = 800\, \mu C, q_B = -800\, \mu C$

Vedclass · Answer

(A) The motional electromotive force $(EMF)$ induced in the conducting rod $PQ$ is given by $e = BvL$.Substituting the given values: $e = 4.0\, T 	imes 20\, m/s 	imes 1.0\, m = 80\, V$.According to the right-hand rule,the potential at $P$ will be higher than at $Q$ because the force on the free electrons in the rod is directed towards $Q$ (due to $\vec{F} = q(\vec{v} 	imes \vec{B})$).Thus,plate $A$ of the capacitor is connected to the higher potential $P$,and plate $B$ is connected to the lower potential $Q$.The charge on the capacitor is $q = CV$.Substituting the values: $q = 10\, \mu F 	imes 80\, V = 800\, \mu C$.Since plate $A$ is at higher potential,$q_A = +800\, \mu C$ and $q_B = -800\, \mu C$.

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with uniform speed $v = 20\, m/s$ in a uniform magnetic field $B = 4.0\, T$ directed into the paper. $A$ capacitor of capacity $C = 10\, \mu F$ is connected as shown in the figure. Then:

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