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 bimetallic strip consists of metals X and Y. It is mounted rigidly at the base as shown
2. A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane
3. The number of NAND gates required to form an AND gate is
4. The resistance of a wire is 5 ohm at 50° C and 6 ohm at 100° C. The resistance
5. A block of mass m is moving on a wedge with the acceleration aᵒ

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