| 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 u, is [NEET 2014] |
 |
A) zero
B) \[Bv\pi {{r}^{2}}/2\] and P is at higher potential
C) \[\pi rBv\] and R is at higher potential
D) \[2\,rBv\] and R is at higher potential
Correct Answer: C
Solution :
| For motional emf \[e=Bv\times (2r)=2rBv\] |
| R will be at higher potential, we can find it by using right hand rule. The electrons of wire will move towards end P due to electric force and at end R the excess positive charge will be left. |
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