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Question

A current passing through a coil of self inductance of 2mH changes at the rate of 20m A s to the power of negative 1 end exponent. The emf induced in the coil is

  1. 10 blank capital mu v  
  2. 40 blank mu V  
  3. 10 blank m V  
  4. 40 blank m V  

The correct answer is: 40 blank m V


    By Faraday’s second law, induced emf
    e equals negative fraction numerator N d ϕ over denominator d t end fraction blank w h i c h blank g i v e s blank e equals negative L fraction numerator d I over denominator d t end fraction
    therefore blank open vertical bar e open vertical bar equals close close 2 cross times 10 to the power of negative 3 end exponent cross times 20 cross times 10 to the power of negative 3 end exponent V equals 40 mu V

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