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Question:

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?

2rBvandRis at higher potential

Bvπr²/2andPis at higher potential

Zero

πrBvandRis at higher potential

Solution:

emf=VBl=Bv(2r)where R is at higher potential and P is at lower potential