Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • A small flat search coil of area  with 140 closely wound turns is placed between the poles of a powerful magnet producing magnetic field

A small flat search coil of area 5cm^{2} with 140 closely wound turns is placed between the poles of a powerful magnet producing magnetic field 0.09\; T and then quickly removed out of the field region. Calculate

(a) change of magnetic flux through the coil, and

(b) emf induced in the coil.

 

 
 
 
 
 

Answers (1)

(a) The change of magnetic flux through the coil will be given as :

\Delta \phi =\phi _{2}-\phi _{1}

\Delta \phi =0-\phi _{1}\; \; \; (\because \phi _{2}=0)

\Delta \phi =0-NBA\cos \theta

Where, N = 140 = number of turns

            B = 0.09 = magnetic field

            A = S = Area

Then,

\Delta \phi =-140\times 0.09\times5\times 10^{-4}\cos \theta

\Delta \phi =-63\times 10^{-4}wb

(b) Now, calculating the induced emf of the coil is given as ;

e=\frac{-\Delta \phi }{\Delta t}

e=\frac{63\times 10^{-4}}{\Delta t}V

Assuming coil is removed within a millisecond

e=\frac{63\times 10^{-4}}{10^{-3}}V=6.3V

Posted by

Safeer PP

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years
anam-khan

CBSE Class 12 accountancy board exam analysis 2024 out; paper rated easy, balanced

Latest Question