A conducting rod of length L = 0.1 , m is mov | vedclass

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Question

(C) The motional electromotive force $(EMF)$ induced in the rod is given by $\varepsilon = B v L$.Given: $B = 0.5 \, T$,$v = 0.2 \, m/s$,$L = 0.1 \, m

Vedclass · Accepted Answer

$q_A = +0.2 \, \mu C, q_B = -0.2 \, \mu C$

Vedclass · Answer

(C) The motional electromotive force $(EMF)$ induced in the rod is given by $\varepsilon = B v L$.Given: $B = 0.5 \, T$,$v = 0.2 \, m/s$,$L = 0.1 \, m$.$\varepsilon = (0.5 \, T) 	imes (0.2 \, m/s) 	imes (0.1 \, m) = 0.01 \, V$.Using the right-hand rule for the motional $EMF$,the upper end of the rod becomes positive and the lower end becomes negative. Consequently,plate $A$ of the capacitor becomes positively charged and plate $B$ becomes negatively charged.The charge on the capacitor is $q = C \varepsilon$.$q = (20 	imes 10^{-6} \, F) 	imes (0.01 \, V) = 0.2 	imes 10^{-6} \, C = 0.2 \, \mu C$.Thus,$q_A = +0.2 \, \mu C$ and $q_B = -0.2 \, \mu C$.

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