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Question

Find magnetic field at O

  1. fraction numerator 5 mu subscript 0 end subscript i theta over denominator 24 pi r end fraction    
  2. fraction numerator mu subscript 0 end subscript i theta over denominator 24 pi r end fraction    
  3. fraction numerator 11 mu subscript 0 end subscript i theta over denominator 24 pi r end fraction    
  4. Zero    

The correct answer is: fraction numerator 5 mu subscript 0 end subscript i theta over denominator 24 pi r end fraction


    B subscript 1 end subscript equals B subscript 3 end subscript equals B subscript 5 end subscript equals 0
    B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator theta i over denominator 3 r end fraction circled times comma   B subscript 4 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator theta i over denominator 2 r end fraction
    and B subscript 6 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator theta i over denominator r end fraction
    thereforeNet magnetic field at O,
    B subscript n e t end subscript equals B subscript 2 end subscript minus B subscript 4 end subscript plus B subscript 6 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator theta i over denominator r end fraction open parentheses fraction numerator 1 over denominator 3 end fraction minus fraction numerator 1 over denominator 2 end fraction plus 1 close parentheses equals fraction numerator 5 mu subscript 0 end subscript theta i over denominator 24 pi r end fraction

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