Physics-
General
Easy

Question

Figure shows a square loop A B C D with edge length a. The resistance of the wire A B C is r and that of A D Cis 2 r. The value of magnetic field at the centre of the loop assuming uniform wire is

  1. fraction numerator square root of 2 mu subscript 0 end subscript i over denominator 3 pi a end fraction ⨀    
  2. fraction numerator square root of 2 mu subscript 0 end subscript i over denominator 3 pi a end fraction ⨂    
  3. fraction numerator square root of 2 mu subscript 0 end subscript i over denominator pi a end fraction ⨀    
  4. fraction numerator square root of 2 mu subscript 0 end subscript i over denominator pi a end fraction ⨂    

The correct answer is: fraction numerator square root of 2 mu subscript 0 end subscript i over denominator 3 pi a end fraction ⨂


    According to question, resistance of wire A D C is twice that of wire A B C. Hence current flowing through A D C is half that of A B C comma blank i. e. comma fraction numerator i subscript 2 end subscript over denominator i subscript 1 end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction. Also i subscript 1 end subscript plus i subscript 2 end subscript equals i
    rightwards double arrow i subscript 1 end subscript equals fraction numerator 2 i over denominator 3 end fraction and i subscript 2 end subscript equals fraction numerator i over denominator 3 end fraction
    Magnetic field at centreO due to wire A B and B C (part 1 and 2) B subscript 1 end subscript equals B subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 i subscript 1 end subscript sin invisible function application 45 degree over denominator a divided by 2 end fraction ⨂ equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 square root of 2 i subscript 2 end subscript over denominator a end fraction ⨂
    And magnetic field at centreO due to wires A D and D C
    Also i subscript 1 end subscript equals 2 i subscript 2 end subscript. So open parentheses B subscript 1 end subscript equals B subscript 2 end subscript close parentheses greater than open parentheses B subscript 3 end subscript equals B subscript 4 end subscript close parentheses
    Hence net magnetic field at centreO
    B subscript n e t end subscript equals open parentheses B subscript 1 end subscript plus B subscript 2 end subscript close parentheses minus left parenthesis B subscript 3 end subscript plus B subscript 4 end subscript right parenthesis
    equals 2 cross times fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 square root of 2 cross times open parentheses fraction numerator 2 over denominator 3 end fraction i close parentheses over denominator a end fraction minus fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 square root of 2 open parentheses fraction numerator i over denominator 3 end fraction close parentheses cross times 2 over denominator a end fraction
    equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 4 square root of 2 i over denominator 3 a end fraction open parentheses 2 minus 1 close parentheses circled times equals fraction numerator square root of 2 mu subscript 0 end subscript i over denominator 3 pi blank a end fraction circled times

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