A coil of 100 turns is wound on a magnetic circuit of reluctance 1000 AT/mWb. The current of 1A flowing in the coil is reversed in 10 ms. The average EMF induced in the coil is ________ V.

This question was previously asked in
MPPGCL JE Electrical 28 April 2023 Shift 3 Official Paper
View all MPPGCL Junior Engineer Papers >
  1. 0.1 V
  2. 2 V
  3. 0.2 V
  4. 1 V

Answer (Detailed Solution Below)

Option 2 : 2 V
Free
MPPGCL JE Electrical Full Test 1
3.9 K Users
100 Questions 100 Marks 120 Mins

Detailed Solution

Download Solution PDF

Concept

The average EMF induced in the coil is given by:

\(E=-N{Δ ϕ \over Δ t}\)

The magnetic flux is given by:

\(ϕ = {NI\over R}\)

where, E = EMF

N = No. of turns

Δϕ = Change in flux

Δt = Change in time

I = Current

R = Reluctance

Calculation

Given, N = 100

I = 1 A

R = 1000 AT/Wb

When the current reverses, the flux changes from  Wb to  Wb.

The change in flux is given by:

Δϕ = ϕfinal - ϕinitial

Δϕ = (−0.1) − (0.1) = −0.2 Wb

\(E=-(100)× {-0.2 \over 10× 10^{-3}}\)

E = 100 × 20 × 10-3 V

E = 2 V

Latest MPPGCL Junior Engineer Updates

Last updated on May 29, 2025

-> MPPGCL Junior Engineer result PDF has been released at the offiical website.

-> The MPPGCL Junior Engineer Exam Date has been announced.

-> The MPPGCL Junior Engineer Notification was released for 284 vacancies.

-> Candidates can apply online from 23rd December 2024 to 24th January 2025.

-> The selection process includes a Computer Based Test and Document Verification.

-> Candidates can check the MPPGCL JE Previous Year Papers which helps to understand the difficulty level of the exam.

More Faraday's Law Questions

More Maxwell's Equations Questions

Get Free Access Now
Hot Links: teen patti circle teen patti pro teen patti joy 51 bonus teen patti game - 3patti poker