Question

Three circular coils each having two turns and radius $R$ are placed mutually perpendicular to each other and each having its centre at the origin of the coordinate system. If current $i$ is flowing through each coil, then the magnitude of the magnetic field at the common centre is:

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

Answer 1

The correct option is A $\frac{\sqrt{3}{\mu }_{0}i}{R}$

The magnetic field due to a circular current carrying coil of $n$ turns at its centre is,

$B=\frac{{\mu }_{0}ni}{2R}$

Hence, due to each coil the magnitude of magnetic field at centre $O$ will be

$B=\frac{2{\mu }_{0}i}{2R}=\frac{{\mu }_{0}i}{R}(\because n=2)$

Now irrespective of the direction of current in coils, the direction of magnetic field will be mutually perpendicular due to each coil.

Using right-hand thumb rule and the assumed direction of current as shown in figure.
Magnetic field due to coil $1$, coil $2$ & coil $3$ will be:

$B_1=B\hat{k};B_2=B\hat{i};B_3=B\hat{j}$

Thus, the net magnetic field at centre $O$ will be

${\overrightarrow{B}}_{net}={\overrightarrow{B}}_1+{\overrightarrow{B}}_2+{\overrightarrow{B}}_3$

${\overrightarrow{B}}_{net}=B\hat{k}+B\hat{i}+B\hat{j}$

$|{\overrightarrow{B}}_{net}|=\sqrt{B^2+B^2+B^2}=\sqrt{3}B$

After putting the value of $B$, we get

$\therefore |{\overrightarrow{B}}_{net}|=\frac{\sqrt{3}{\mu }_{0}i}{R}$

Hence, option $\left(a\right)$ is correct.