Find the current in the sliding rod AB (resistance = R) for the arranged shown in Fig. B is constant and is out of the paper. Paralllel wires have no resistance. `upsilon` is constant. Switch is closed at time t = 0.

Explanation

The conductor of length d moves with speed `v_1` perpendicular to magnetic field B as shown in figure. The produces motional emf across two ends of rod. Which is given by =vBd. Since, switch S is closed at time t=0 current start growing in inductor by the potential difference due to motional emf.
Applying Kichhoff's voltage rule, we have
`-L(dI)/(dt)+vBd=IRor L(dI)/(dt)+IR=vBd`
This is the linear difference equation. On solving, we get
`l=(vBd)/(R)+Ae^(RT//2)`
At `t=0l=0`
`rArrA=-(vBd)/(R)rArrI(vBd)/(R)(1-e^(-RT//L))`
This is the required expression of current.