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NEET2014PhysicsElectromagnetic InductionActual

A thin semicircular conducting ring P Q R of radius r is falling with its plane vertical in a horizontal magnetic field B , as shown in figure. The potential difference developed across the ring when its speed is v , is:

Options

  1. AZero
  2. BB v π r 2 / 2 and P is at higher potential
  3. Cπ r B V and R is at higher potential
  4. D2 r B v and R is at higher potential

Correct answer

D. 2 r B v and R is at higher potential

Step-by-step solution

Here we have to calculate the emf of the conducting ring when it is falling So, we have to calculate with the following formulae e m f = V B l e q = V B 2 R where R is at higher potential and P is at lower potential.

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