Physics-
General
Easy

Question

A steady current is set up in a cubic network composed of wires of equal resistance and length d as shown in figure. What is the magnetic field at the centre P due to the cubic network ?

  1. fraction numerator 0 over denominator 4 pi end fraction times fraction numerator 2 I over denominator d end fraction    
  2. 0    
  3. fraction numerator 0 over denominator 4 pi end fraction times fraction numerator 3 I over denominator square root of 2 d end fraction    
  4. fraction numerator 0 over denominator 4 pi end fraction times fraction numerator 8 pi I over denominator d end fraction    

The correct answer is: fraction numerator 0 over denominator 4 pi end fraction times fraction numerator 3 I over denominator square root of 2 d end fraction


    B equals fraction numerator mu subscript 0 end subscript over denominator 2 end fraction fraction numerator I over denominator R end fraction open parentheses 2 pi R equals n 2 pi r rightwards double arrow r equals fraction numerator R over denominator n end fraction close parentheses
    B to the power of ´ end exponent equals fraction numerator mu subscript 0 end subscript over denominator 2 end fraction fraction numerator I n over denominator open parentheses fraction numerator R over denominator n end fraction close parentheses end fraction equals n to the power of 2 end exponent B

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