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The magnetic field at the centre of an equilateral triangular loop of side $2L$ carrying a current $i$ is

$\begin {array} {1 1} (a)\;\large\frac{9\mu_0i}{4 \pi L} & \quad (b)\;\large\frac{\mu_0i}{ \pi L} \\ (c)\;\large\frac{3\mu_0i}{4 \pi L} & \quad (d)\;\large\frac{\mu_0i}{2 \pi L} \end {array}$


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$B =3 \bigg[\large\frac{\mu_0i}{4 \pi r}$$ ( \sin60^{\circ} + \sin60^{\circ}) \bigg]$
$r = \large\frac{L}{\sqrt3}$
Ans : (a)
answered Feb 22, 2014 by thanvigandhi_1

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