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Q.

A parallel plate capacitor is moving with a velocity of 25 ms−1 through a uniform magnetic field of 4.0 T as shown in figure. If the electric field within the capacitor plates is 400 NC−1 and plate area is  25×10−7m2, then the magnetic force experienced by the positive charge plate is

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a

8.85×10−13 N directed out of the plane of the paper

b

Zero

c

8.85×10−15 N directed out of the plane of the paper

d

none of the above

answer is A.

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Detailed Solution

Electric filed in between the capacitor plates is given by E=Qε0AWhere Q is the charge on capacitor. Q=ε0A×E=8.85×10−12×25×10−7×400 =8.85×10−15CMagnetic force experienced by +ve plate is, Fm=QvB=8.85×10−15×25×4=8.85×10−13N in direction out of the plane of paper.
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