Q 12-04-129JEE MainJEE Main 2023 (6 Apr, Shift 1)Medium
Two identical circular wires of radius $20\ \text{cm}$ and carrying current $\sqrt2$ A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is ______ $\times10^{-8}$ T. (Take $\pi=3.14$)
Numerical value type. Enter your answer.
Answer: 628
Each loop gives $B=\dfrac{\mu_0I}{2r}=\dfrac{4\pi\times10^{-7}\times\sqrt2}{0.4}=\sqrt2\pi\times10^{-6}\ \text{T}$ at the centre, and the two fields are perpendicular.
$$B_{net}=\sqrt2\times\sqrt2\pi\times10^{-6}=2\pi\times10^{-6}=6.28\times10^{-6}\ \text{T}=628\times10^{-8}\ \text{T}$$
Solution by Sreeraj P, M.Sc Physics