A steady current $i$ flows in a small square loop of wire of side $L$ in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let $\overrightarrow {{\mu _1}} $ and $\overrightarrow {{\mu _2}} $ respectively denote the magnetic moments due to the current loop before and after folding. Then
IIT 1993, Diffcult
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(c) Initial magnetic moment $=$ $\mu_1 = iL^2$
After folding the loop, $M =$ magnetic moment due to each part $ = i\,\left( {\frac{L}{2}} \right) \times L = \frac{{i{L^2}}}{2} = \frac{{{\mu _1}}}{2}$
$==>$ ${\mu _2} = M\sqrt 2 = \frac{{{\mu _1}}}{2} \times \sqrt 2 = \frac{{{\mu _1}}}{{\sqrt 2 }}$
After folding the loop, $M =$ magnetic moment due to each part $ = i\,\left( {\frac{L}{2}} \right) \times L = \frac{{i{L^2}}}{2} = \frac{{{\mu _1}}}{2}$
$==>$ ${\mu _2} = M\sqrt 2 = \frac{{{\mu _1}}}{2} \times \sqrt 2 = \frac{{{\mu _1}}}{{\sqrt 2 }}$
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