A symmetric star shaped conducting wire loop is carrying a steady...

The given points (1, 2, 3, 4, 5, 6) makes 360^o angle at ‘O’. Hence angle made by vertices 1 and 2 with ‘O’ is 60^o.

Direction of magnetic field at ‘O’ due to each segment is same. Since it is symmetric star shape, magnitude will also be same.

Magnetic field due to section BC.

$\left({B}_{1}\right) = \dfrac{ki}{a}\left(\sin \left(60°\right) - \sin 30°\right) = \dfrac{ki}{2a}\left(\sqrt{3} - 1\right)$

${B}_{net} = 12 \times {B}_{1} = \dfrac{6ki}{a}\left(\sqrt{3} - 1\right)and\left(k = \dfrac{{\mu}_{0}}{4\pi}\right)$