NEET Solved Paper 2014 Question 32
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 figure. The potential difference developed across the ring when its speed is u, is [AIPMT 2014]
Options:
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
Show Answer
Answer:
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.
BETA
