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A conducting disc of radius `R` is rolling without sliding on a horizontal surface with a constant velocity `v`.A uniform magnetic field of strength `B` is applied normal to the plane of the disc.Find the `EMF` induced between (at this moment )
(a)`P&Q` , (b)`P&C` , (c )`Q&C`
(`C` is centre `P&Q` are opposite points on vertical diameter of the disc)
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Correct Answer - A::B::C
(a)`epsilon_(PQ)=1/2Bomega(2R)^(2) =1/2B(v/R)(2R)^(2)=2BvR`
(b)`epsilon_(PC)=1/2BomegaR^(2) =1/2B(v/R)R^(2)=(BvR)/2`
( c)`epsilon_(QC)=2Bv/R-(BvR)/2=3/2vBR`

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