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

Question

Aicharya · Accepted Answer

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.

$(B_{1}) = \frac{k i}{a} (\sin (60°) - \sin 30°) = \frac{k i}{2 a} (\sqrt{3} - 1)$

$B_{n e t} = 12 \times B_{1} = \frac{6 k i}{a} (\sqrt{3} - 1) a n d (k = \frac{μ_{0}}{4 π})$

A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure.…

Solution