Question

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 $4a$ .

• A. $\frac{\mu _{0} I}{4 \pi a} \, 3 \, \left[\sqrt{3} - 1\right]$
• B. $\frac{\mu _{0} I}{4 \pi a}6\left[\sqrt{3} - 1\right]$
• C. $\frac{\mu _{0} I}{4 \pi a}6 \, \left[\sqrt{3} + 1\right]$
• D. $\frac{\mu _{0} I}{4 \pi a}3\left[2 - \sqrt{3}\right]$

Answer 1

Answer: (B)

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.
$\left(B_{1}\right)=\frac{k i}{a}\left(\sin \left(60^{\circ}\right)-\sin 30^{\circ}\right)=\frac{k i}{2 a}(\sqrt{3}-1)$
$B_{n e t}=12 \times B_{1}=\frac{6 k i}{a}(\sqrt{3}-1)$ and $\left(k=\frac{\mu_{0}}{4 \pi}\right)$