Physics-
General
Easy

Question

A small square loop wire of side l is placed inside a large squre loop of side L open parentheses L greater than greater than l close parentheses. If the loops are coplanar and their centres coincide, the mutual induction of the system is directly proportional to

  1. fraction numerator L over denominator l end fraction  
  2. fraction numerator l over denominator L end fraction  
  3. fraction numerator L to the power of 2 end exponent over denominator l end fraction  
  4. fraction numerator l to the power of 2 end exponent over denominator L end fraction  

The correct answer is: fraction numerator l to the power of 2 end exponent over denominator L end fraction


    Considering the larger loop to be made up of four rods each of length L, the field at the centre,
    i. e. comma at a distance open parentheses fraction numerator L over denominator 2 end fraction close parenthesesfrom each rod, will be

    B equals 4 cross times fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator l over denominator d end fraction open square brackets sin invisible function application alpha plus sin invisible function application beta close square brackets
    i e comma blank B equals 4 cross times fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator I over denominator open parentheses fraction numerator L over denominator 2 end fraction close parentheses end fraction cross times 2 sin invisible function application 45 degree
    i e comma blank B subscript 1 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 8 square root of 2 over denominator L end fraction I
    So, the flux with smaller loop
    ϕ equals B subscript 1 end subscript S subscript 2 end subscript equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction fraction numerator 8 square root of 2 over denominator L end fraction l to the power of 2 end exponent I
    and hence, M equals fraction numerator ϕ subscript 2 end subscript over denominator I end fraction equals 2 square root of 2 fraction numerator mu subscript 0 end subscript over denominator pi end fraction fraction numerator l to the power of 2 end exponent over denominator L end fraction
    or M proportional to fraction numerator l to the power of 2 end exponent over denominator L end fraction

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