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A uniform copper wire of length “L” is bent into a circular coil of two turns and a current “i” is passed through it. The coil now behaves like a magnetic dipole of moment

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i L² /4 π

i L² /2 π

i L² /16 π

i L² /8 π

answer is B.

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Detailed Solution

L = 2 × 2πr
$\begin{array}{l} r = \frac{L}{4 π} \\ M = niA \\ = 2 xix π {(\frac{L}{4 π})}^{2} \\ = \frac{i L^{2}}{8 π} \end{array}$

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