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 $\varepsilon = BvL$.Substituting the given values: $\varepsilon = 4.0 	im

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The motional electromotive force $(EMF)$ induced in the rod is given by $\varepsilon = BvL$.Substituting the given values: $\varepsilon = 4.0 	imes 20 	imes 1.0 = 80\, V$.According to Fleming's right-hand rule,for a rod moving to the right in a magnetic field directed into the paper,the induced current flows from $Q$ to $P$ inside the rod.Thus,point $P$ acts as the positive terminal and point $Q$ acts as the negative terminal.Since plate $A$ is connected to $P$ and plate $B$ is connected to $Q$,plate $A$ becomes positively charged and plate $B$ becomes negatively charged.The charge on the capacitor is $q = C\varepsilon$.$q = 10 	imes 10^{-6} 	imes 80 = 800 	imes 10^{-6}\, C = 800\, \mu C$.Therefore,$q_A = + 800\, \mu C$ and $q_B = - 800\, \mu C$.

$A$ conducting rod $PQ$ of length $L = 1.0\, m$ is moving with a 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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