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Question:

As shown in figure, two vertical conducting rails separated by distance 1.0 m are placed parallel to z-axis. At z=0, a capacitor of 0.15 F is connected between the rails and a metal rod of mass 100 g placed across the rails slides down along the rails. If a constant magnetic field of 2.0 T exists perpendicular to the plane of the rails, what is the acceleration of the rod?

1.4 m/s2

2.5 m/s2

9.8 m/s2

0

Solution:

Due to motion of rod, emf induced across capacitor, ε=Blv
Therefore, Charge stored in capacitor, Q=C(Blv)
I=dQ/dt=CBl(dv/dt)=CBla
Force opposing the downward motion, Fm=BIl
Therefore, Fm=B(CBla)l=B²l²Ca
Net force on rod, Fnet=W−Fm=mg−B²l²Ca
Therefore, ma=mg−B²l²Ca
or a=mg/(m+B²l²C)
So, a=(0.1×9.8)/(0.1+2²×1²×0.15)=1.4 m/s².