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

A semicircular conducting ring acb of radius R moves with constant speed v in a plane perpendicular to the uniform magnetic field B as shown in the figure. Identify the correct statement.


A
VaVc=BRv
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
VbVc=BRv
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
VaVb=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C VaVb=0

To calculate motional emf, we take the effective length between the two points if the conductor is curved.

For points a and b, ab=2R is the effective length which is moving without cutting the field lines.

So, VaVb=0

Also, for points a and c, effective length is oc=R.

Hence, ε=BRv

From right-hand rule, c is at the higher potential.

Therefore, VcVo=VcVa=BRv

Similarly, VcVb=BRv

Hence, option (C) is correct.


flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Algebraic Identities
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon