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AB and CD are long straight conductor, distance d apart, carrying a current I. The magnetic field at the midpoint of BC is

  1. fraction numerator negative mu subscript 0 end subscript I over denominator 2 pi d end fraction stack k with hat on top    
  2. fraction numerator negative mu subscript 0 end subscript I over denominator pi d end fraction stack k with hat on top    
  3. fraction numerator negative mu subscript 0 end subscript I over denominator 4 pi d end fraction stack k with hat on top    
  4. fraction numerator negative mu subscript 0 end subscript I over denominator 8 pi d end fraction stack k with hat on top    

The correct answer is: fraction numerator negative mu subscript 0 end subscript I over denominator pi d end fraction stack k with hat on top


    The field at the midpoint of BC due to AB is open parentheses negative fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator d divided by 2 end fraction stack k with hat on top close parentheses and the same is due to CD. Therefore the total field is open square brackets negative open parentheses fraction numerator mu subscript 0 end subscript i over denominator pi d end fraction close parentheses stack k with hat on top close square brackets

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