JEE Main 2023 — Magnetic Effects of Current Question with Solution

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

$B_C= \frac{{\mu }_{o}I}{4\pi r}\theta .......................\left(1\right)$

where, ${\mu }_0$ is the permeability of free space, $I$ is the current through the conductor, $r$ is the radius of the circular conductor and $\theta$ is the angle subtended by the arc at the centre.

From the given figure, it is clear that the arc of the conductor makes an angle $\pi$ at its centre.

Substitute the known values of the parameters into equation (1) to calculate the required magnetic field.

$\begin{array}{rcl}{B}_C& =& \frac{4\mathrm{\pi }\times {10}^{-7} \mathrm{N} {\mathrm{A}}^{-2}\times 3 \mathrm{A}}{4\times \pi \times {\displaystyle \frac{\mathrm{\pi }}{10} \mathrm{m}}}\times \pi \\ & =& 3\times {10}^{-6} \mathrm{T}\times \frac{{10}^6 \mathrm{\mu T}}{1 \mathrm{T}}\\ & =& 3 \mathrm{\mu T}\end{array}$