Physics-
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Easy

Question

In the given figure net magnetic field at O will be

  1. fraction numerator 2 mu subscript 0 end subscript i over denominator 3 pi a end fraction square root of 4 minus pi to the power of 2 end exponent end root    
  2. fraction numerator mu subscript 0 end subscript i over denominator 3 pi a end fraction square root of 4 plus pi to the power of 2 end exponent end root    
  3. fraction numerator 2 mu subscript 0 end subscript i over denominator 3 pi a to the power of 2 end exponent end fraction square root of 4 plus pi to the power of 2 end exponent end root    
  4. fraction numerator 2 mu subscript 0 end subscript i over denominator 3 pi a end fraction square root of left parenthesis 4 minus pi to the power of 2 end exponent right parenthesis end root    

The correct answer is: fraction numerator mu subscript 0 end subscript i over denominator 3 pi a end fraction square root of 4 plus pi to the power of 2 end exponent end root



    Magnetic field at 0 due to
    Part (a) : B subscript 1 end subscript equals 0
    Part (b) : B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator pi i over denominator left parenthesis a divided by 2 right parenthesis end fraction circled times (along –Z-axis)
    Part (c) : B subscript 3 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator left parenthesis a divided by 2 right parenthesis end fraction open parentheses downwards arrow close parentheses (along – Y-axis)
    Part (d) : B subscript 4 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator pi i over denominator left parenthesis 3 a divided by 2 right parenthesis end fraction¤(along +Z-axis)
    Part (e): B subscript 5 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator i over denominator left parenthesis 3 a divided by 2 right parenthesis end fraction open parentheses downwards arrow close parentheses (along – Y-axis)
    B subscript 2 end subscript minus B subscript 4 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator pi i over denominator a end fraction open parentheses 2 minus fraction numerator 2 over denominator 3 end fraction close parentheses equals fraction numerator mu subscript 0 end subscript i over denominator 3 a end fraction circled times (along – Z-axis)
    B subscript 3 end subscript plus B subscript 5 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 1 over denominator a end fraction open parentheses 2 plus fraction numerator 2 over denominator 3 end fraction close parentheses equals fraction numerator 8 mu subscript 0 end subscript i over denominator 12 pi a end fraction open parentheses downwards arrow close parentheses (along – Y-axis)
    Hence net magnetic field
    B subscript n e t end subscript equals square root of left parenthesis B subscript 2 end subscript minus B subscript 4 end subscript right parenthesis to the power of 2 end exponent plus left parenthesis B subscript 3 end subscript plus B subscript 5 end subscript right parenthesis to the power of 2 end exponent end root equals fraction numerator mu subscript 0 end subscript i over denominator 3 pi a end fraction square root of pi to the power of 2 end exponent plus 4 end root

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