Home Physics Electromagnetic Induction Self Induction, Emf, Induced Current A conductor ABOCD moves along its bisector w…
Physics Electromagnetic Induction Self Induction, Emf, Induced Current MCQ (Single Correct)

A conductor ABOCD moves along its bisector with a velocity of 1 m/s through a perpendicular magnetic field of 1 wb/m 2 , as shown in fig. If all the four sides are of 1m length each, then the induced emf between points A and D is

A
0
B
1.41 volt
C
0.71 volt
D
None of the above

Share this question

For Instagram sharing, use “Apps” on mobile or copy the link.

Text Solution

Verified by Experts
The correct answer is:
B

There is no induced e.m.f. in the part and because they are moving along their length while e.m.f. induced between and i.e., between and can be calculate as follows
induced e.m.f. between and induced e.m.f.
betwween
and volt.

Prepare Smarter with CGP Edu

Get practice questions, solutions, and test series in one place.

Write a Review

Share your experience with this question and solution.

Commentary

Send your comment, doubt, correction, or feedback to admin.

Student Reviews

What students say about this solution

No reviews yet. Be the first to write a review.

Similar Questions

Explore conceptually related problems

CG
CGP Question Assistant Question Bank + AI Help
Hi! Type your question or upload one screenshot. First I will search related questions from CGP Edu Question Bank. If none match, type YES and I will solve it with AI.
Upload only one screenshot at a time. Flow: Question Bank first → If not matched, type YES for AI solution.