A (current vs time) graph of the current passing through a solenoid is shown in Fig. For which time is the back electromotive force (u) a maximum? If the back emf t = 3 s is e, find the back emf at t = 7 s, 15 s and 40 s OA, AB and BC are straight line segments.

Explanation

The back electromotive force in solenoid is (u) a maximum when there is maximum rate of change of current. This occurs is in AB part of the graph. So maximum back emf will be obtained between `5s lt tlt 10s`.
Since the back emf at t= 3s is e,
Also,
the rate of change of current at `t=3,s=` slope of OA from `t=0s` to `t=5s=1//5A//s`.
So, we have
If `u=L1//5(for t=3s,(dI)/(dt)=1//5)` (L is a constant). Applying `epsilon =-L(dI)/(dt)`
Similarly, we have for other values
For `5s lt t lt 10s`
`u_1=-L(3)/(5)=-(3)/(5)L=-3e`
Thus, at `t=7s,u_1=-3e`
For `10slttlt30s`
`u-2=L(2)/(20)=(L)/(10)=(1)/(2)e`
For `t gt 30s,u_2=0`
Thus, the back emf at `t=7s,15s` and `30s` are `-3e,e//2` and 0 respectively .