Q 12-04-230JEE MainJEE Main 2018 (15 Apr, Shift 2)Medium
A current of $1\ \text{A}$ is flowing on the sides of an equilateral triangle of side $4.5\times10^{-2}\ \text{m}$. The magnetic field at the centre of the triangle will be:
Answer: (A) $4\times10^{-5}\ \text{Wb/m}^2$
The centre is at perpendicular distance $r = \dfrac{a}{2\sqrt3}$ from each side, and each side subtends $60^\circ$ on either side of the perpendicular:
$$B_\text{side} = \frac{\mu_0I}{4\pi r}(\sin60^\circ + \sin60^\circ) = \frac{\mu_0I\sqrt3}{4\pi r}$$
All three sides give fields in the same direction:
$$B = 3\times\frac{\mu_0I\sqrt3}{4\pi}\cdot\frac{2\sqrt3}{a} = \frac{18\times10^{-7}\times1}{4.5\times10^{-2}} = 4\times10^{-5}\ \text{Wb/m}^2$$
Solution by Sreeraj P, M.Sc Physics