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A conducting rod PQ of length L = 1.0m is moving with a uniform speed v= 2.0 m/s in a uniform magnetic field B = 4.0 T directed into the paper. A capacitor of capacity C = 10 mu Fis connected as shown in the figure. Then what are the charges on the plates A and B of the capacitor. |
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Answer» Solution : The motional emf is `e = Blv` `therefore ` p.d across the capacitor ` = Blv = 4 xx 1 xx 2 = 8V` `q = CV= 10 xx 8 = 80 MU C` A is + Ve w.r.t B (from fleming right hand rule) The CHARGE on PLATE A is `q_A = + 80 mu C` The charge on plate B is `q_B = -80muC` |
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