JEE Main 2023 — Magnetic Effects of Current Question with Solution

The formula to calculate the magnetic field at any axial point situated at a distance $x$ from the centre of a circular current carrying conductor of radius $r$ can be written as

$B=\frac{{\mu }_{0}I{r}^2}{2{\left(r^2+x^2\right)}^{3/2}} ...\left(1\right)$

Substitute $0$ for $x$ in equation (1) to calculate the magnetic field $\left(B_c\right)$ at the centre of the conductor.

$\begin{array}{rcl}{B}_c& =& \frac{{\mu }_{0}I{r}^2}{2{\left(r^2+0^2\right)}^{3/2}}\\ & =& \frac{{\mu }_{0}I{r}^2}{2r^3}\\ & =& \frac{{\mu }_{0}I}{2r} ...\left(2\right)\end{array}$

Substitute $r$ for $x$ in equation (1) to calculate the magnetic field $\left(B_r\right)$ at the given distance from the centre of the conductor.

$\begin{array}{rcl}{B}_r& =& \frac{{\mu }_{0}I{r}^2}{2{\left(r^2+r^2\right)}^{3/2}}\\ & =& \frac{{\mu }_{0}I{r}^2}{4\sqrt{2}{r}^3}\\ & =& \frac{{\mu }_{0}I}{4\sqrt{2}r} ...\left(3\right)\end{array}$

Divide equation (2) by equation (3) to calculate the required ratio of the magnetic fields.

$\begin{array}{rcl}\frac{B_c}{B_r}& =& \frac{\frac{{\mu }_{0}I}{2r}}{\frac{{\mu }_{0}I}{4\sqrt{2}r}}\\ & =& 2\sqrt{2}\\ & =& \sqrt{8}\end{array}$