Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. `vecB` is constant and is out of the paper. Parallel wires have no resistance, v is constant. Switch S is closed at time t = 0.

Explanation

A rod of length d is moving with velocity v perpendicular to magnetic field B which produces emf between two ends of rod induced emf `epsilon=Bvd`
Now when switch is on at t = 0, capacitor start charging, suppose charge on capacitor at t time is Q(t).
Now according to Kirchhoff.s second law,
`epsilon=V_R+V_C`
`Bvd=IR+Q/C`
`therefore R=(dQ)/(dt)+Q/C=Bvd`
`therefore (dQ)/(dt)+1/(RC)Q=(Bvd)/R`
Now solution of this linear different equation,
`Q=(Bvd//R)/(1//RC)+A "exp"(-1/"RC"t)`
`Q=BvdC+Ae^(-t//RC)`...(1)
where A is constant that can be found by initial condition.
At t=0, Q=0
`therefore ` 0=BvdC+A(1)
`therefore` A=-BvdC
From equation (1)
`Q=BvdC-BvdC e^(-(t/(RC)))`
`Q=BvdC-(1-e^(-(t/(RC)))`
Induced charge
`I=(dQ)/(dt)=BvdC [ 0-("-1"/"RC")e^(-(t/"RC"))]`
`I=(BvdC)/(RC)e^(-(t)/(RC))`
`I=(Bvd)/R e^(-(t/"RC"))` which is required equation of current.