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. [figure] The potential difference developed across the ring when its speed is v, is
- zero
- Bv r22 and P is at higher potential
- rBv and R is at higher potential
- 2rBv and R is at higher potential
Answer
(4) 2rBv and R is at higher potential
Solution
The emf induced in a curved conductor moving in a uniform field equals that in the straight line joining its ends.
Effective length = PR = 2r (perpendicular to both v and B)
= B (2r) v = 2rBv
Direction: force on a positive charge F = q( v B); with v downward and B into the page, v B points towards R.
Positive charges collect at R, so R is at higher potential.
Answer: (4)