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Electric Charges And Fields

Question
CBSEENPH12039871
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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

  • zero

  • Bvπr squared divided by 2 space and space straight P is at higher potential
  • πrBv and R is at higher potential
  • 2 rBv space and space straight R spaceis at higher potential

Solution

D.

2 rBv space and space straight R spaceis at higher potential

For motional emf,e = Bv x (2r)
                         = 2rBv
R will be at higher potential.
The electrons of the wire will move towards end P due to due to electric force and at end R the excess positive charge will be left.

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point? 

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