QJEEAdv17Ph69b3: JEE Advanced 2017 Physics - Class 12 JEEAdv17Ph69b3 Solution: A symmetric star shaped conducting wire loop is carrying a…

The star shape is composed of 12 wires. Thus, the total magnetic field at centre is 12 times of magnetic field due to one wire.

Let us first calculate the magnetic field due to element AB. Perpendicular distance of centre O from element AB, OP = a

Angle subtended by centre at element are $$\theta$$₁ = 30$${}^{\circ }$$, $$\theta$$₂ = 60$${}^{\circ }$$ as shown in figure.

$$\therefore$$ $$\overrightarrow{B}=\frac{{\mu }_{0}I}{4\pi a}(\cos {30}^{\circ }-\cos {60}^{\circ })\odot$$

$$=\frac{{\mu }_{0}I}{4\pi a}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)\odot =\frac{{\mu }_{0}I}{8\pi a}(\sqrt{3}-1)\odot$$

Since the direction of magnetic field due to each element side is out of the paper

$$\therefore$$ Net magnetic field at O $$=12\times \frac{{\mu }_{0}I}{8\pi a}(\sqrt{3}-1)$$

$$=\frac{{\mu }_{0}I}{4\pi a}6(\sqrt{3}-1)$$