First slide
Motional EMF
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 the figure. 

The potential difference developed across the ring when its speed ν, is 
 

Easy
Solution

Motional emf induced in the semicircular ring PQR is equivalent to the motional emf induced in the imaginary conductor PR. 

 i.e., εPQR=εPR=Bvl=Bv(2r)                                   (l= PR = 2r)

Therefore, potential difference developed across the ring is 2rBv with R is at highe rpotential. 

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