A) zero
B) \[Bv\pi {{r}^{2}}/2\]. and P is at higher potential
C) \[\pi rBv\] and R is at higher potential
D) 2rBv and R is at higher potential
Correct Answer: D
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.You need to login to perform this action.
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