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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Induction

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 $} $

1 Answer

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
 

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