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Exercises · 6.7

Q.Current in a circuit falls from 5.0 A5.0\ \text{A} to 0.0 A0.0\ \text{A} in 0.1 s0.1\ \text{s}. If an average emf of 200 V200\ \text{V} induced, give an estimate of the self-inductance of the circuit.

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The self-inductance is found using Faraday’s law for a changing current: L=∣E∣∣ΔI/Δt∣L = \frac{|\mathcal{E}|}{|\Delta I / \Delta t|}. With E=200 V\mathcal{E} = 200\ \text{V}, ΔI=−5.0 A\Delta I = -5.0\ \text{A}, and Δt=0.1 s\Delta t = 0.1\ \text{s}, we get L=4.0 HL = 4.0\ \text{H}.

The key idea here is self-inductance — a circuit’s property that opposes a change in current by inducing an emf. When the current changes, the magnetic flux through the circuit itself changes, and that induces an emf (back emf) given by:

E=−LdIdt\mathcal{E} = -L \frac{dI}{dt}

The negative sign is Lenz’s law: the induced emf opposes the change. But for magnitude, we drop the sign and use the average values.

Since the current falls uniformly from 5.0 A5.0\ \text{A} to 0.0 A0.0\ \text{A} in 0.1 s0.1\ \text{s}, the average rate of change is:

ΔIΔt=0.0−5.00.1=−5.00.1=−50 A/s\frac{\Delta I}{\Delta t} = \frac{0.0 - 5.0}{0.1} = \frac{-5.0}{0.1} = -50\ \text{A/s}

The magnitude of this rate is 50 A/s50\ \text{A/s}.

The average induced emf is given as 200 V200\ \text{V}. Using the magnitude form of Faraday’s law:

∣E∣=L∣ΔIΔt∣|\mathcal{E}| = L \left| \frac{\Delta I}{\Delta t} \right|

So:

200=L×50200 = L \times 50

Therefore:

L=20050=4.0 HL = \frac{200}{50} = 4.0\ \text{H} …

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