Physics-
General
Easy

Question

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

  1. Zero    
  2. B nu pi R to the power of 2 end exponent divided by 2 and M is at higher potential    
  3. pi R B Vand Q is at higher potential    
  4. 2RBV and Q is at higher potential    

The correct answer is: 2RBV and Q is at higher potential


    Rate of decrease of area of the semicircular ring negative fraction numerator d A over denominator d t end fraction equals left parenthesis 2 R right parenthesis   V
    According to Faraday's law of induction induced emf
    e equals negative fraction numerator d phi over denominator d t end fraction equals negative   B fraction numerator d A over denominator d t end fraction equals negative   B   left parenthesis 2 R V right parenthesis

    The induced current in the ring must generate magnetic field in the upward direction. Thus Q is at higher potential.

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