Three rings, each with equal radius ' $r$ ' are placed mutually perpendicular to each other and each having…
Three rings, each with equal radius ' $r$ ' are placed mutually perpendicular to each other and each having centre at the origin of coordinate system. ' I ' is current passing through each ring. The magnetic field value at the common centre is
zero
$(\sqrt{3}-1) \frac{\mu_0 \mathrm{I}}{2 \pi r}$
$\sqrt{3} \frac{\mu_0 \mathrm{I}}{2 r}$
$\sqrt{2} \frac{\mu_0 I}{2 r}$
Solution
The magnetic field at the centre of the ring is
$\begin{aligned}
& \mathrm{B}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{i}}{\mathrm{r}}(2 \pi)=\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}} \\
& \therefore \quad \overrightarrow{B_1}=\frac{\mu_0 I}{2 r} \hat{i}, \overrightarrow{B_2}=\frac{\mu_0 I}{2 r} \hat{j}, \overrightarrow{B_3}=\frac{\mu_0 I}{2 R} \hat{k} \\
& \therefore \overrightarrow{\mathrm{~B}_0}=\overrightarrow{\mathrm{B}_1}+\overrightarrow{\mathrm{B}_2}+\overrightarrow{\mathrm{B}_3}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
& \therefore \quad B_0=\frac{\mu_0 I}{24} \sqrt{1^2+1^2+1^2}=\frac{\sqrt{3} \mu_0 I}{2 \mathrm{r}}
\end{aligned}$
