JEE Main 2023 — Magnetic Effects of Current Question with Solution

The formula to calculate the magnetic field $\left(B\right)$ at the centre of a current carrying circular loop is given by

$B= \frac{{\mu }_{0}i}{2R} ...\left(1\right)$

where, ${\mu }_0$ is the permeability of free space, $i$ is the current and $R$ is the radius of the loop.

$B_V, B_H$ are the magnetic field due to the vertical and the horizontal loop, which by magnitude are equal.

The resultant magnetic field $\left(B_{net}\right)$ at the centre is, then, given by

$$\begin{gathered} B_{\text{net }}=\sqrt{B^2+B^2}=\sqrt{2}B \\ =\frac{\sqrt{2}\times 4\pi \times {10}^{-7}\times \sqrt{2}}{2\times 0.2} \end{gathered}$$

$=628\times {10}^{-8} \mathrm{T}$