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A conducting straight wire PQ of length ...

A conducting straight wire `PQ` of length `l` is fixed along as diameter of a non conducting ring as shown in the figure. The ring is given a pure rolling motion as shown. Field `B` is horizontal in direction and perpendicular to the plane of ring. The magnitude of induced EMF in the wire PQ at the position shown in figure will be

A

`Bvl`

B

`2Bvl`

C

`3Bvl/2`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
A

`omega v/R=v/(l//2)=(2v)/l`
`e=(Bomegal^2)/w=(B((2v)/l)l^2)/2=Bvl`
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