A (current vs time) graph of the current passing through a solenoid is shown in Fig 6.9. For which time is the back electromotive force (u) a maximum. If the back emf at t = 3s is e, find the back emf at t = 7 s, 15s and 40s. OA, AB and BC are straight line segments.
From the graph, we can say that when there is a maximum rate of change of magnetic flux, the electromagnetic force will be maximum which is proportional to the rate
of change of current. The rate of current will be maximum when the time axis makes a maximum angle at AB.
Therefore, when t = 3 seconds, s = slope of OA
Therefore, the rate of change of current at t = 3 = ¼ A/s
Therefore, the electromotive force is L/5.
When the emf is between 5 to 10 seconds, it is -3e
When the emf is between 10 to 30 seconds, it is +1/2e
When the emf is at 40 seconds, dI/dt = 0