Q 12-04-105JEE MainJEE Main 2023 (29 Jan, Shift 1)Medium
The magnitude of magnetic induction at mid-point $O$ due to current arrangement as shown in figure will be
Answer: (D) $\dfrac{\mu_0I}{\pi a}$
$O$ lies on the line of the horizontal wires AB and DE, so they give no field at $O$.
Each vertical wire (BC and ET) is semi-infinite with its end level with $O$, at perpendicular distance $a/2$:
$$B_1=\frac{\mu_0I}{4\pi(a/2)}=\frac{\mu_0I}{2\pi a}$$
Current flows down BC (to the left of $O$) and up ET (to the right of $O$); by the right-hand rule both fields at $O$ point out of the page and add:
$$B=2\times\frac{\mu_0I}{2\pi a}=\frac{\mu_0I}{\pi a}$$
Solution by Sreeraj P, M.Sc Physics