Electromagnetic Induction - Result Question 4

4. 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 :

[2014]

(a) Zero

(b) Bvπr2/2 and P is at higher potnetial

(c) πrBv and R is at higher potnetial

(d) 2rBv and R is at higher potential

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Answer:

Correct Answer: 4. (d)

Solution:

  1. (d) Rate of decreasing of area of semicircular

ring =dAdt=(2r)V

From Faraday’s law of electromagnetic induction

e=dθdt=BdAdt=B(2rV)

As induced current in ring produces magnetic field in upward direction hence R is at higher potential.



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