Question

A symmetric star-shaped conducting wire loop is carrying a steady state current I as shown the figure. The distance between the diametrically opposite vertices of the star is $4a$. The magnitude of the magnetic field at the centre of the loop is

Answer 1

Magnetic field due to star is upward direction and equal to 12 times the field due to one section of star side.

Consider a side AD.

Field due to AD is $B_1$ at 0

$B_1=\frac{{\mu }_{0}1}{4\pi \left(0E\right)}(\cos {30}^{\circ }-\cos {60}^{\circ })=\frac{{\mu }_{0}1}{4\pi a}(\frac{\sqrt{3}}{2}-\frac{1}{2})=\frac{{\mu }_{0}1}{4\pi a}(\sqrt{3}-1)$

Field due to full star is B

$B=12B_1=\frac{3}{2}\frac{{\mu }_0\mathring{1}}{4\pi a}(\sqrt{3}-1)$