Question

Three rings, each having equal radius $R$, are placed mutually perpendicular to each other and each having its centre at the origin of co-ordinate system. If current I is flowing through each ring then the magnitude of the magnetic field at the common center is

• A. $\sqrt{3}$ $\frac{{\mu }_{0}I}{2R}$
• B. zero
• C. $(\sqrt{2}-1)\frac{{\mu }_{0}I}{2R}$
• D. $(\sqrt{3}-\sqrt{2})\frac{{\mu }_{0}I}{2R}$

Answer 1

Answer: (A)

$\sqrt{3}$ $\frac{{\mu }_{0}I}{2R}$

We know that when current flows through the ring, it generates magnetic field at the center of ring and its direction is normal to the plane of ring depending upon the direction of current flow in the ring.

Hence,

The magnetic field due to ring in y-z plane is

$\overrightarrow{B}=\frac{{\mu }_{0}I}{2R}(\pm \hat{i})$

The magnetic field due to ring in x–z plane is

$\overrightarrow{B}=\frac{{\mu }_{0}I}{2R}(\pm \hat{j})$

The magnetic field due to ring in x–y plane is

$\overrightarrow{B}=\frac{{\mu }_{0}I}{2R}(\pm \hat{k})$

Adding all the three equations, we get

$\overrightarrow{B}=\frac{{\mu }_{0}I}{2R}(\pm \hat{i})+\frac{{\mu }_{0}I}{2R}(\pm \hat{j})+\frac{{\mu }_{0}I}{2R}(\pm \hat{k})$

$|\overrightarrow{B}|=\frac{{\mu }_{0}I}{2R}\sqrt{3}$