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Question: Answered & Verified by Expert
A thin semicircular conducting ring PQR 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:
PhysicsElectromagnetic InductionNEETNEET 2014
Options:
  • A Zero
  • B Bvπr2/2 and P is at higher potential
  • C πrBV and R is at higher potential
  • D 2rBv and R is at higher potential
Solution:
2359 Upvotes Verified Answer
The correct answer is: 2rBv and R is at higher potential

Here we have to calculate the emf of the conducting ring when it is falling

So, we have to calculate with the following formulae


emf=VBleq=VB2R

where R is at higher potential and P is at lower potential.

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