A closed circuit in the form of a regular hexagon of side a carries a…

The correct answer is Option (4) → \( \displaystyle\frac{\sqrt{3}\mu_0 I}{\pi a} \)

Each side is a finite straight conductor. Its field at a point at perpendicular distance r is \( B = \frac{\mu_0 I}{4\pi r}(\sin\alpha_1 + \sin\alpha_2) \), where \( \alpha_1, \alpha_2 \) are the angles the two ends make with the perpendicular. All six sides give fields in the same direction at the centre, so the fields add.

For a regular hexagon each side subtends \( 60^\circ \) at the centre, so \( \alpha_1 = \alpha_2 = 30^\circ \), and the centre is at the distance of the apothem:

    \( \displaystyle r = \frac{a/2}{\tan30^\circ} = \frac{\sqrt{3}\,a}{2} \)

    \( \displaystyle B_{\mathrm{side}} = \frac{\mu_0 I}{4\pi r}(\sin30^\circ + \sin30^\circ) = \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{2\sqrt{3}\,\pi a} \)

    \( \displaystyle B = 6\times\frac{\mu_0 I}{2\sqrt{3}\,\pi a} = \frac{\sqrt{3}\,\mu_0 I}{\pi a} \)