A closed circuit in the form of a regular hexagon of side a carries a current I. What is the magnetic field at the centre of the hexagon?

$\frac{\sqrt{3}μ_0I}{πa}$

The correct answer is Option (4) → $\frac{\sqrt{3}μ_0I}{πa}$

The correct expression for the magnetic field at the center of the hexagon is -

$B=\sqrt{3}\frac{μ_0I}{πa}$

from right angled ΔAMO,

$\tan\phi=\frac{AM}{OM}$

$⇒\tan 30°=\frac{a/2}{r}$

$⇒r=\frac{a}{2\tan 30}=\frac{\sqrt{3}a}{2}$

Magnetic field along AB,

$B_{AB}=\frac{μ_0}{4π}×\frac{I}{r}(\sin θ+\sin θ)$

$B_{AB}=\frac{μ_02I}{4πr}\sin 30°$

$=\frac{μ_0I\sqrt{3}}{πa}$