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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 magnetic field B as shown in figure.
Due to this, an motional emf is induced across two ends of rod (e = vBd).
Since, switch S is closed at time t = 0 current starts growing in the inductor by the potential difference due to motional emf.
By applying KVL in the given circuit, we have
`- L (dI)/(dt) + vBd = IR` or `L (dI)/(dt) + IR = vBd`
This is the linear differential equation.
On solving, we get
`I = (vBd)/R + Ae^((- Rt)/2)`
At t = 0, I = 0
⇒ `A = - (vBd)/R`
⇒ `I = (vBd)/R (1 - e^((- Rt)/L))`
This is the required expression of current.
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