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 set up ⊥ 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 be 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.
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 set up ⊥ to the plane of paper. The acceleration of the conductor