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

⇦

⇨

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)]

Related Questions:

1. A bicyclist comes to a skidding stop in 10 m.During this process, the force
2. A charge of 40 µC is given to a capacitor having capacitance C=10 µF
3. A geostationary satellite is orbiting the earth at a height of 5R above that surface
4. A solid body rotates about a stationary axis, so that its angular velocity
5. A bottle full of water in a space ship at 30⁰C is carried on the moon.

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

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends