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Question
A (current vs time) graph of the current passing through a solenoid is shown in figure. 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 = 7s, 15s and 40s. OA, AB and BC are straight line segments.
Solution
Whenever the electric current passing through a coil or circuit changes, the magnetic flux linked with it will also change. As a result of this, in accordance with Farraday's laws of electromagnetic induction, an emf is induced in the coil or the circuit which opposes the change that causes this induced emf is called back emf, the current so produced in the coil is called induced current. The induced emf is given by
`ε = - (d(Nphi_B))/(dt)`
`ε = - L (dl)/(dt)`
Thus, negative sign indicates that induced emf (e) opposes any change (increase or decrease) of current in the coil.
When the rate of change of current is maximum, then back emf in the solenoid is (u) a maximum. This occurs in AB part of the graph. So maximum back emf will be obtained between 5s < t < 10s.
Since, the back emf at t = 3s is e.
Also, the rate of change of current at t = 3
And slope(s) of OA (from t = 0s to t = 5s) = 1/5 A/s
So, we have
If u = L 1/5 (for t = 3s, `(dI)/(dt)` = 1/5)
Where L is a constant (coefficient of self-induction)
And emf is ε = `- L (dI)/(dt)`
Similarly, we have for other values
For 5s < t < 10s, u1 = `- L 3/5 = - 3/5 L = - 3e`
Thus, at t = 7s, u1 = – 3e
For 10s < t < 30s
u2 = `L 2/20 = L/10 = 1/2 e`
For t > 30s, u2 = 0
Thus, the back emf at t = 7s, 15s and 40s are – 3e e/2 and 0 respectively.
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