Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
0 votes

A thin semicircular ring of radius 'a' is falling with its plane vertical to a horizontal magnetic field induction $\overrightarrow {B}$ .At the given position PQR the speed of ring is 'v' , the potential difference developed across the ring.

$(a)\;\text{zero} \\(b)\;\text{P will be with higher potential 2aBv} \\ (c)\;\text{R will be with higher potential 2aBv}\\ (d)\;\text{R will be with higher potential $\pi Bv $} $

Can you answer this question?

1 Answer

0 votes
Since the falling object is a ring the projection of length of the ring along right angle is equal to $PR=2a$
$emf= Blv$
$\qquad= B 2aV$
and induced current in anti clock wise direction.
So,R will be with higher potential 2aBv
Hence c is the correct answer.
answered Mar 26, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App