Q 12-04-124JEE MainJEE Main 2023 (10 Apr, Shift 2)Easy
A straight wire carrying a current of $14\ \text{A}$ is bent into a semicircular arc of radius $2.2\ \text{cm}$ as shown in the figure. The magnetic field produced by the current at the centre $O$ of the arc is ______ $\times10^{-4}\ \text{T}$.
Numerical value type. Enter your answer.
Answer: 2
The straight parts lie on the line through $O$ and give no field. The semicircle gives
$$B=\frac{\mu_0I}{4r}=\frac{4\pi\times10^{-7}\times14}{4\times0.022}=\frac{22}{7}\times\frac{14\times10^{-7}}{0.022}=2\times10^{-4}\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics