Saved Bookmarks
| 1. |
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. |
|
Answer» 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 . Capacitor is charged by this potential difference. Let Q(t) is charge on the capacitor and current flows from A and B . Now , the induced current `I=(vBd)/(R)-(Q)/(RC)` On rearranging the terms, we have `(Q)/(RC)+(dQ)/(dt)=(vBd)/(R)` This is the linear differential equation. On solving we get `Q=vBdC+Ae^(-t//RC)` `rArrQ=vBdC[1-e^(-t//Rc)]` (At time `t =0,Q=0=A=-vBdc`). Differentiating, we get `I-(vBd)/(R)e^(-t//RC)` This is the required expression of current. |
|