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. Calculate the charge on equivalent capacitance of the conbination shown in figure
2. A spherical body of emissivity e=0.6, placed inside a perfectly black body is maintained
3. When germanium is doped with phosphorus what type of semiconductor is produced?
4. Light emitted during the de excitation of electron from n=3 to n=2, when incident
5. Laser light is considered to be coherant because it consists of

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