Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity v=10 m/ s.

Q6.16 (b) Now assume that the straight wire carries a current of 50A and the loop is moved to the right with a constant velocity, V=10m/s. Calculate the induced emf in the loop at the instant when X=0.2m. Take a=0.1m and assume that the loop has a large resistance.

Answers (1)

best_answer

Given,

Current in the straight wire

I= 50A

Speed of the Loop which is moving in the right direction

 V=10m/s

Length of the square loop

a=0.1m

 distance from the wire to the left side of the square

X=0.2m

NOW,

Induced emf in the loop :

E=B_xav=\frac{\mu _0I}{2\pi x}av=\left ( \frac{4\pi*10^{-7}*50}{2\pi*0.2} \right )*0.1*10=5*10^{-5}V

Hence emf induced is 5*10^{-5}V.

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 Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers
anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4
anam-khan

When will CBSE Board exam 2025 date sheet be released for Class 10, 12?

Latest Question