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Electromagnetic Induction — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction5 Marks
Question

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 at t = 3s is e, find the back emf at t = 7s, 15s and 40s. OA, AB and BC are straight line segments.
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Answer

Key concept: 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 'Faraday'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,
$\epsilon=-\frac{\text{d}(\text{N}\phi_\text{B})}{\text{dt}}$
$\epsilon=-\text{L}\frac{\text{dl}}{\text{dt}}$
Thus, negative sign indicates that induced emf (e) apposes any change (increase of decrease) of current in the coil.
When the rate of change of current is maximum, then back emf in 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 $(\text{from t}=0\text{s to t}=5\text{s})=\frac{1}{5}\frac{\text{A}}{\text{S}}$.
So, we have
If $\text{u}=\text{L}\frac{1}{5}\Big(\text{for t}=3\text{s},\frac{\text{dI}}{\text{dt}}=\frac{1}{5}\Big).$
where, L is a constant (coefficient of self-induction).
And emf is $\epsilon=-\text{L}\frac{\text{dI}}{\text{dt}}$
Similarly, we have for other values.
For $5\text{s}<\text{t}<10\text{s},\text{u}_1=-\text{L}\frac{3}{5}=-\frac{3}{5}\text{L}=-3\text{e}$
Thus, at $\text{t}=7\text{s},\text{u}_1=-3\text{e}$
For $10\text{s}<\text{t}<30\text{s}$
$\text{u}_2=\text{L}\frac{2}{20}=\frac{\text{L}}{10}=\frac{1}{2}\text{e}$
for $\text{t}>30\text{s},\text{u}_2=0$
Thus, the back emf ar t = 7s, 15s and 40s are $-3\text{e},\frac{\text{e}}{2}$ and 0 respectively.

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