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A small square loop of wire of side l is placed inside a large square loop of side L(L>>l) . If the loops are coplanar and their centres coincide, the mutual induction of the system is directly proportional to :

Options

  1. AL l
  2. Bl L
  3. CL^2 l
  4. Dl^2 L

Correct answer

D. l^2 L

Step-by-step solution

Considering the larger loop to be made up of four rods each of length L , the field at the centre, i.e., at a distance ( L 2 ) from each rod, will be B=4 ₀ 4 l d [ + ] i.e., B=4 ₀ 4 I ( L 2 ) 2 45 i.e., B₁= ₀ 4 8 2 L I So, the flux with smaller loop ₂=B₁ S₂= ₀ 4 8 2 L l^2 I and hence, M= ₂ I =2 2 ₀ l^2 L or M l^2 L

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