A conducting rod PQ of length 1 m is moving | vedclass

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Question

(A) The motional electromotive force $(EMF)$ induced in the conducting rod $PQ$ is given by $V = Bvl$.Given: $B = 4 \ T$,$v = 2 \ ms^{-1}$,$l = 1 \ m$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The motional electromotive force $(EMF)$ induced in the conducting rod $PQ$ is given by $V = Bvl$.Given: $B = 4 \ T$,$v = 2 \ ms^{-1}$,$l = 1 \ m$.Therefore,$V = 4 	imes 2 	imes 1 = 8 \ V$.According to Fleming's Right-Hand Rule,the potential at $P$ is higher than at $Q$. Thus,the upper plate of the capacitor (connected to $P$) becomes positively charged and the lower plate (connected to $Q$) becomes negatively charged.The charge $q$ on the capacitor plates is given by $q = CV$.Given: $C = 10 \ \mu F = 10 	imes 10^{-6} \ F$.$q = (10 	imes 10^{-6} \ F) 	imes (8 \ V) = 80 	imes 10^{-6} \ C = 80 \ \mu C$.Thus,$q_A = +80 \ \mu C$ and $q_B = -80 \ \mu C$.

$A$ conducting rod $PQ$ of length $1 \ m$ is moving with a uniform speed $2 \ ms^{-1}$ in a uniform magnetic field of $4 \ T$ which is directed into the paper. $A$ capacitor of capacity $10 \ \mu F$ is connected as shown in the figure. Then,the charge on the plates of the capacitor are

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