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 220 volt and 1000 watt bulb is connected across a 110 volt mains supply.
2. A student performs an experiment to determine the Young’s modulus of a wire
3. If a wire is stretched to four times its length, then the specific resistance
4. Work done by an external agent in separating the parallel plate capacitor is
5. In Young’s double slit experiment, the phase difference between the two waves

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