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

Question

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

Infinity Learn · Accepted Answer

Electric filed in between the capacitor plates is given by $$E=Q$$ε$$0AWhere$$ Q is the charge on capacitor. $$Q=$$ε$$0A$$×$$E=8.85$$×$$10$$−$$12$$×$$25$$×$$10$$−$$7$$×$$400$$ $$=8.85$$×$$10$$−$$15CMagnetic$$ 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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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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