Q 12-04-138JEE MainJEE Main 2023 (1 Feb, Shift 2)Easy
As shown in the figure, a long straight conductor with semicircular arc of radius $\dfrac{\pi}{10}$ m is carrying current $I=3$ A. The magnitude of the magnetic field at the center $O$ of the arc is (The permeability of the vacuum $=4\pi\times10^{-7}\ \text{N A}^{-2}$)
Answer: (D) $3\ \mu\text{T}$
The straight parts lie along the line through $O$ and give no field. The semicircle gives
$$B=\frac{\mu_0I}{4r}=\frac{4\pi\times10^{-7}\times3}{4\times\pi/10}=3\times10^{-6}\ \text{T}=3\ \mu\text{T}$$
Solution by Sreeraj P, M.Sc Physics