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Question
Find the current in the sliding rod AB (resistance = R) for the arrangement shown in figure. B is constant and is out of the paper. Parallel wires have no resistance. v is constant. Switch S is closed at time t = 0.
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Solution
This is a similar problem as we discussed above. Here, a conductor of length d moves with speed v, perpendicular to the magnetic field B as shown in figure.
Due to this a motional emf is induced across two ends of rod (e = vBd). Since switch S is closed at time t = 0, capacitor is charged by this potential difference.
Let Q(t) be the charge on the capacitor and current flows from A to B.
Now, the induced current
`I = (dQ)/(dt) = (Bvd) - Q/(RC)`
`Q/(RC) + (dQ)/(dt) = (Bvd)/R`
`Q + RC (dQ)/(dt) = vBC` .....(Let `vBdC = A`)
`Q + RC Q/(dt) A`
`(dQ)/(A - Q) = 1/(RC) dt`
By integrating we have
`int_0^Q (dQ)/(A - Q) = 1/(RC) int_0^t dt - [In (A -Q) - In A] = t/(RC)`
In ``(A / Q)/A = - t/(RC)`
`(A - Q)/A = e^(-t/(RC))`
`Q = A(1 - e^(-t/(RC)))`
Current in the rod,
`I = (dQ)/(dt) = d/(dt) [A(1 - e^(-t/(RC)))]`
= `- A(e^-t/(RC))(- 1/(RC))`
`I = (vBd)/R e^(-t/(RC)`
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