A (current vs time) graph of the current passing through a solenoid is shown in figure. For which time is the back electromotive force (`epsilon`) a maximum. If the back emf at t = 3s is e, find the back emf at 1 = 7 s, 15 s and 40 s. OA, AB and BC are straight line segments.

Explanation

When rate of change of flux is maximum then induced emf is also t = 10 sec maximum. In graph at t = 5 sec. to slope is maximum, hence rate of change of flux is also maximum and induced emf will also be maximum.
For t=0 s to t=5 s , slope `(dI)/(dt)=1/5As^(-1)`
Back emf `epsilon=(-LdI)/(dt)`
So for t=0 to t=5s (At t=3 s also)
Back emf=e=`-L(1/5)=(-L)/5`
`therefore e=(-L)/5` ...(1)
(i)For 5s `lt t lt 10 s` slope `(dI)/(dt)=(-3)/5 As^(-1)`
Back emf for t=5 s to t=10 s,
(At t=7s as well)
`e_1=(-LdI)/(dt)=-L((-3)/5)`
`-3((-L)/5)`
`e_1=-3s [ because e=(-L)/5]`
(ii) For 10s to `t lt 30` s slope
`(dI)/(dt)=2/20=1/10 As^(-1)`
Back emf at t=15 s is ,
`e_2=-L xx (dI)/(dt)`
`-Lxx(+1/10)`
`=(-L)/10=1/2 e [ because e=(-L)/5]`
(iii)At `t ge 30s` , slope =0 ,
So, at t=40 s back emf `e_3=0`
Thus at t=7 s induced emf `e_1=-3e`
at t=15 s induced emf `e_2=e/2`
at t=40 s induced emf `e_3=0`