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

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 moving 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 $Q = C \varepsilon = 10 	imes 10^{-6} 	imes 8.0 = 80 	imes 10^{-6}\, C = 80\, \mu C$.According to Fleming's right-hand rule, for a rod moving to the right in a magnetic field directed into the paper, the induced $EMF$ makes the upper end (connected to plate $A$) at a higher potential and the lower end (connected to plate $B$) at a lower potential.Therefore, plate $A$ acquires a positive charge $(+80\, \mu C)$ and plate $B$ acquires a negative charge $(-80\, \mu C)$.

$A$ 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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