Q 12-04-078JEE MainJEE Main 2024 (6 Apr, Shift 1)Easy
A current element $\Delta\vec l = \Delta x\,\hat i$ is placed at the origin and carries a large current $I = 10\ \text{A}$. The magnetic field on the $y$-axis at a distance of $0.5\ \text{m}$ from the element, for $\Delta x$ of $1\ \text{cm}$ length, is
Answer: (A) $4\times10^{-8}\ \text{T}$
By the Biot–Savart law, with the element perpendicular to the line joining it to the point:
$$dB = \frac{\mu_0}{4\pi}\frac{I\,\Delta x}{r^2} = 10^{-7}\times\frac{10\times0.01}{(0.5)^2} = 4\times10^{-8}\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics