Physics-
General
Easy

Question

Figure shows the cross-sectional view of the hollow cylindrical conductor with inner radius 'R' and outer radius '2R', cylinder carrying uniformly distributed current along it's axis. The magnetic induction at point 'P' at a distance fraction numerator 3 R over denominator 2 end fractionfrom the axis of the cylinder will be

  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 circled times    
  3. fraction numerator square root of 2 mu subscript 0 end subscript i over denominator pi a end fraction    
  4. fraction numerator 5 mu subscript 0 end subscript i over denominator 36 pi R 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 circled times


    According to question resistance of wire ADC is twice that of wire ABC. Hence current flows through ADC is half that of ABC i.e. 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 Þ 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 centre O due to wire AB and BC (part 1 and (b) 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 4 5 to the power of o end exponent over denominator a divided by 2 end fraction circled times equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 square root of 2 i subscript 1 end subscript over denominator a end fraction circled times
    and magnetic field at centre O due to wires AD and DC (i.e. part 3 and (d) B subscript 3 end subscript equals B subscript 4 end subscript 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¤
    Also i1 = 2i2. So (B1 = B(b) > (B3 = B(d)
    Hence net magnetic field at centre O
    B subscript n e t end subscript equals left parenthesis B subscript 1 end subscript plus B subscript 2 end subscript right parenthesis 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 left parenthesis 2 minus 1 right parenthesis circled times equals fraction numerator square root of 2 mu subscript 0 end subscript i over denominator 3 pi a end fraction circled times

    rightwards double arrow B equals fraction numerator mu subscript 0 end subscript i over denominator 2 pi open parentheses fraction numerator 3 R over denominator 2 end fraction close parentheses end fraction cross times open curly brackets fraction numerator open parentheses fraction numerator 3 R over denominator 2 end fraction close parentheses to the power of 2 end exponent minus R to the power of 2 end exponent over denominator left parenthesis 2 R to the power of 2 end exponent right parenthesis minus R to the power of 2 end exponent end fraction close curly brackets equals fraction numerator 5. mu subscript o end subscript i over denominator 36 pi r end fraction.

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