Electromagnetic Induction and Alternating Current 2 Question 4

5. A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction B. At the position MNQ the speed of the ring is v and the potential difference developed across the ring is

(1996,2M)

(a) zero

(b) BvπR2/2 and M is at higher potential

(c) πBRv and Q is at higher potential

(d) 2RBv and Q is at higher potential

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Answer:

Correct Answer: 5. (d)

Solution:

  1. Induced motional emf in MNQ is equivalent to the motional emf in an imaginary wire MQ i.e.

eMNQ=eMQ=Bvl=Bv(2R)[l=MQ=2R]

Therefore, the potential difference developed across the ring is 2RBv with Q at higher potential.



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