A conductor of length I and mass m can slide without any friction along the two vertical conductors connected at the top through a capacitor (figure). A uniform magnetic field B is setup⊥to the plane of paper. The acceleration of the conductor
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a
is constant
b
increases
c
decreases
d
cannot say
answer is A.
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Detailed Solution
Let v is the velocity of conductor at any time,then induced emf : e=Blv……..(i)Charge on capacitor : q=Ce=CBlvCurrent in circuit : i=dqdt=CBldvdtfor conductor: mg−iBl=mdvdt⇒mg−CB2l2dvdt=mdvdt ⇒dvdt=mgm+CB2l2This is the acceleration of conductor which is constant.