Question

Magnetic field at the center of a circular coil of radius R and carrying a current i is :

• A. $\frac{{\mu }_{0}i}{2R}$
• B. $\frac{i}{2c^2{\epsilon }_{0}R}$
• C. $\frac{{\mu }_{0}i}{2\pi R}$
• D. $\frac{i{c}^2}{2{\epsilon }_{0}R}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{{\mu }_{0}i}{2R}$
B $\frac{i}{2c^2{\epsilon }_{0}R}$

According to Boit - savert's law the magnetic field $d\overrightarrow{B}$ at the centre O of the coil due to current $Id\overrightarrow{I}$ element is given by

$d\overrightarrow{B}=\frac{{\mu }_0}{4\pi }I\left(\frac{d\overrightarrow{l}\ast \overrightarrow{r}}{r^3}\right)$

$\therefore dB=\frac{{\mu }_0}{4\pi }I\frac{dl}{r^2}$

for each current element, angle between $d\overrightarrow{l}$ and $d\overrightarrow{r}$ is ${90}^{\circ }$

$\therefore dB=\frac{{\mu }_0}{4\pi }\frac{Idl}{r^2}$

$\therefore B=\int dB=\int \frac{{\mu }_0}{4\pi }\frac{Idl}{r^2}$

$=\frac{{\mu }_0}{4\pi }\frac{I}{r^2}\int dl$

where $\int dl=$ total length of coil

$\therefore B=\frac{{\mu }_{0}i}{4\pi r^2}\ast 2\pi r=\frac{{\mu }_{0}i}{2r}$

as we know

$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$ where c= velocity of light

$\therefore B=\frac{i}{2c^2{\epsilon }_{0}R}$