Question

Three rings, each having equal radius $\mathrm{R}$, are placed mutually perpendicular to each other and each having its center at the origin of the coordinates system. If a current $\mathrm{I}$ is flowing through each ring, then the magnitude of the magnetic field at the common center is:

Answer 1

Step 1: Given parameters

Three rings, each having an equal radius $\mathrm{R}$.

Current $\mathrm{I}$ is flowing through each of them.

Step 2: Calculation

We know the formula for the magnetic field at the center of a circular loop of current:

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}}$

Where, $\mathrm{B}$ is the magnetic field, $\mathrm{I}$ is the current, $\mathrm{R}$ is the radius of the ring, and ${\mathrm{\mu }}_0$is the magnetic permeability.

When,

${\mathrm{B}}_{\mathrm{x}}$ is the magnetic field due to the ring in the y–z plane.

${\mathrm{B}}_{\mathrm{y}}$ is the magnetic field due to the ring in the x-z plane.

${\mathrm{B}}_{\mathrm{z}}$ is the magnetic field due to the ring in the x-y plane.

$$\begin{gathered} \overrightarrow{\mathrm{B}}=\overrightarrow{{\mathrm{B}}_{\mathrm{x}}}+\overrightarrow{{\mathrm{B}}_{\mathrm{y}}}\overrightarrow{+{\mathrm{B}}_{\mathrm{z}}} \\ =\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}}+\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}}+\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}} \\ \left|\overrightarrow{\mathrm{B}}\right|=\sqrt{3}\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}} \end{gathered}$$

Therefore, the magnitude of the magnetic field at the common center is $\left|\overrightarrow{\mathrm{B}}\right|=\sqrt{3}\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{R}}$.