CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Two conducting rings P and Q of radii r and 2r rotate uniformly in opposite directions with centre of mass velocities 2v and v respectively on a conducting surface S. There is a uniform magnetic field of magnitude B perpendicular to the plane of the rings. The potential difference between the highest points of the two rings is

A
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
8Bvr
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4Bvr
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
16Bvr
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 8Bvr
EMF between the top and bottom ends of the small ring is B×2v×2r=4Bvr

EMF between the top and the bottom ends of the larger ring is B×v×4r=4Bvr
So the potential difference between the two top ends is 4Bvr+4Bvr=8Bvr (due to opposite polarity)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Motional EMF
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon