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 $4 a$ . The magnitude of the magnetic field at the centre of the loop is

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Solution

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
$B_{1} = \frac{μ_{0} 1}{4 π (0 E)} (cos 30^{\circ} - cos 60^{\circ}) = \frac{μ_{0} 1}{4 π a} (\frac{\sqrt{3}}{2} - \frac{1}{2}) = \frac{μ_{0} 1}{4 π a} (\sqrt{3} - 1)$
Field due to full star is B
$B = 12 B_{1} = \frac{3}{2} \frac{μ_{0} ˚ 1}{4 π a} (\sqrt{3} - 1)$

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