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 rod is given by $\varepsilon = B v L$.Substituting the given values: $\varepsilon = 4.0

Vedclass · Accepted Answer

$q_A = + 80\,\mu C$ and $q_B = - 80\,\mu C$

Vedclass · Answer

(A) The motional electromotive force $(EMF)$ induced in the rod is given by $\varepsilon = B v L$.Substituting the given values: $\varepsilon = 4.0 	imes 2 	imes 1.0 = 8.0\, V$.The charge stored in the capacitor is $Q = C \varepsilon$.$Q = 10 	imes 10^{-6}\, F 	imes 8.0\, V = 80 	imes 10^{-6}\, C = 80\,\mu C$.According to the right-hand rule for motional $EMF$,the force on positive charges in the rod $PQ$ is directed from $Q$ to $P$. Thus,$P$ acts as the positive terminal and $Q$ as the negative terminal.This makes plate $A$ of the capacitor positively charged and plate $B$ negatively charged.Therefore,$q_A = + 80\,\mu C$ and $q_B = - 80\,\mu C$.

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with a uniform speed $v = 2\, m/s$ 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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