Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s ^- 1 in the positive x -direction in an environment containing a magnetic field in the positive z- direction.

Q6.12  A square loop of side 12 cm  with its sides parallel to X and Y  axes is moved with a velocity of 8\: cm\: s^{-1} in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10^{-3}Tcm^{-1}  along the negative x-direction (that is it increases by  10^{-3}Tcm^{-1}as one moves in the negative x-direction), and it is decreasing in time at the rate of 10^{-3}Ts^{-1}. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50m\Omega.

Answers (1)

best_answer

Given:

Side of the square loop

l=12cm=0.12m

Area of the loop:

A=0.12*0.12m^2=144*10^{-4}m^2

The resistance of the loop is:

R=4.5m\Omega = 4.5*10^{-3}\Omega

The velocity of the loop in the positive x-direction

v=8cm/s=0.08m/s

The gradient of the magnetic field in the negative x-direction

\frac{dB}{dx}=10^{-3}T/cm=10^{-1}T/m

Rate of decrease of magnetic field intensity

\frac{dB}{dt}=10^{-3}T/s

Now, the emf is being induced by means of both changing magnetic field with time and changing with space. So let us find out emf induced by both changing of space and time, individually.

Induced emf due to field changing with time:

e_{withtime}=\frac{d\phi }{dt}=A\frac{dB}{dt}=144*10^{-4}*10^{-3}=1.44*10^{-5}Tm^2/s

Induced emf due to field changing with space:

e_{withspace}=\frac{d\phi }{dt}=\frac{d(BA)}{dt}=A\frac{dB}{dx}\frac{dx}{dt}=A\frac{dB}{dx}v

e_{withspace}=144*10^{4}*10^{-1}*0.08=11.52*10^{-5}Tm^2/s

Now, the total induced emf is:

e_{total}=e_{withtime}+e_{withspace}=1.44*10^{-5}+11.52*10^{-5}=12.96*10^{-5}V

Total induced current:

I=\frac{e}{R}=\frac{12.96*10^{-5}}{4.5*10^{-3}}=2.88*10^{-2}A

Since the flux is decreasing, the induced current will try to increase the flux through the loop along the positive z-direction.

Posted by

Pankaj Sanodiya

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 Board Exams 2025: Counselling for students, parents from February 1
anam-khan

CBSE asks schools to include APAAR IDs in school ID after NTA says ‘not mandatory’ for NEET
anam-khan

CBSE Board Exams 2025: What the ethics code bans from exam centres

Latest Question