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A small square loop of wire of side i is placed inside a large square loop of slide $L( L >>1)$. If the loops are coplanar and there centers coinside. the mutual induction of the system is directly proportional to :

\[\begin {array} {1 1} (a)\;\frac{L}{l} & \quad (b)\;\frac{l}{L} \\ (c)\;\frac{L^2}{l} & \quad (d)\;\frac{l^2}{L} \end {array}\]
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$ (d)\;\frac{l^2}{L} $
answered Nov 7, 2013 by pady_1

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