A current flows in the wire bent to form a hexagon loop of side 'a'. If the current flowing is I. What is the magnetic field at centre

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$\frac{\sqrt{3} μ_{0} i}{π a}$

$\frac{μ_{0} i}{π a}$

$\frac{μ_{0} π i}{2 a}$

$\frac{μ_{0} π i}{a}$

answer is A.

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Detailed Solution

From the dia
$\tan 30^{0} = \frac{a / 2}{r}$
$r = \frac{\sqrt{3}}{2} a$
$∴$ Field induction at centre
$B = \frac{6 μ_{0}}{4 π} \frac{2 i}{\sqrt{3 a}} . \sin 30^{0}$
$B = \frac{\sqrt{3} μ}{π a} . i$

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