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A small square loop of wire of side 'I' is placed inside a large square loop of side L (L > I). If the loops are coplanar and their centres coincide, the mutual inducation of the system is directly proportional to |
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Answer» Solution :Considering the large 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 xx (mu_(0))/(4 PI)(I)/(d)[sin alpha+sin BETA]` i.e., `B=4 xx (mu_(0))/(4phi)(I)/((L//2)) xx 2sin45` i.e., `B_(1)=(mu_(0))/(4PI)(8sqrt2)/(L)I`  So the flux linked with smaller loop `Phi_(2)=B_(1)S_(2)=(mu_(0))/(4pi)(8sqrt2 I)/(LO) l^(2)` and hence, `M=(phi_(2))/(I) =2sqrt2 (mu_(0))/(pi)(l^(2))/(L) implies M prop (l^(2))/(L)` |
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