Physics-
General
Easy

Question

A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity omega. The induced e.m.f. between the two ends is

  1. fraction numerator 1 over denominator 2 end fraction B omega l to the power of 2 end exponent    
  2. fraction numerator 3 over denominator 4 end fraction B omega l to the power of 2 end exponent    
  3. B omega l to the power of 2 end exponent    
  4. 2 B omega l to the power of 2 end exponent    

The correct answer is: fraction numerator 1 over denominator 2 end fraction B omega l to the power of 2 end exponent


    If in time t. the rod turns by an angle q, the area generated by the rotation of rod will be equals fraction numerator 1 over denominator 2 end fraction l cross times l theta equals fraction numerator 1 over denominator 2 end fraction l to the power of 2 end exponent theta

    So the flux linked with the area generated by the rotation of rod
    phi equals B   open parentheses fraction numerator 1 over denominator 2 end fraction l to the power of 2 end exponent theta close parentheses cos invisible function application 0 equals fraction numerator 1 over denominator 2 end fraction B l to the power of 2 end exponent theta equals fraction numerator 1 over denominator 2 end fraction B l to the power of 2 end exponent omega t
    And so e equals fraction numerator d phi over denominator d t end fraction equals fraction numerator d over denominator d t end fraction open parentheses fraction numerator 1 over denominator 2 end fraction B l to the power of 2 end exponent omega t close parentheses equals fraction numerator 1 over denominator 2 end fraction B l to the power of 2 end exponent omega

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