The current in a self-inductance L=40 mH is to be increased uniformly from 1A to 11 A

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The current in a self-inductance L=40 mH is to be increased uniformly from 1A to 11A in 4 ms. The emf indued in the inductor during the process is

Options

(a) 100 V
(b) 0.4 V
(c) 40 V
(d) 440 V

Correct Answer:

100 V

Explanation:

Consider the inductor of inductance L.

The current flowing through the inductor is i.

Now, we can write ɸ = Li

where, ɸ is magnetic flux linked with the inductor dɸ / dt = L (di / dt)

Given, L = 40 mH

dt = change in time = t₂ – t₁ = 4 ms = Δi

So, dɸ / dt = Δɸ / dt = L (Δi / dt) = (40 mH) [10 /4 ms] = 10 × 10 = 100 —-(i)

According to Faraday’s law of electromagnetic induction, Emf induced, e = -(dɸ / dt)

|e| = Δɸ / dt = 100 V [from equation (i)]

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Topics: Electromagnetic Induction (76)
Subject: Physics (2479)

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