logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
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
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...