Question

The magnetic field at the origin due to the current flowing in the wire as shown in figure below is

• A. $-\frac{{\mu }_{0}I}{8\pi a}(\hat{i}+\hat{k})$
• B. $\frac{{\mu }_{0}I}{2\pi a}(\hat{i}+\hat{k})$
• C. $\frac{{\mu }_{0}I}{8\pi a}(-\hat{i}+\hat{k})$
• D. $-\frac{{\mu }_{0}I}{4\pi a\sqrt{2}}(\hat{i}-\hat{k})$

Answer 1

The correct option is C $\frac{{\mu }_{0}I}{8\pi a}(-\hat{i}+\hat{k})$



Due to $\left(2\right)$ and $\left(3\right)$ magnetic field at $O$ is zero, since $O$ lies on the line of flow of the current $I$.
The field is due to (1)
The vector joining $\left(1\right)$ to $O$ is in the x-z plane and is $\overrightarrow{r}=-a\hat{i}-a\hat{k}$
$r=\sqrt{2}a$.
Current in (1) is along $\hat{j}$

So, $\overrightarrow{B}=\frac{{\mu }_{0}I}{4\pi r}\hat{x}$ where $\hat{x}=\frac{\hat{j}\times \overrightarrow{r}}{|\overrightarrow{r}|}$
$\hat{x}=\frac{\hat{j}\times a(-\hat{i}-\hat{k})}{r}=\frac{a}{r}(\hat{k}-\hat{i})$

$\overrightarrow{B}=\frac{{\mu }_{0}Ia}{4\pi r^2}(\hat{k}-\hat{i})$
$\overrightarrow{B}=\frac{{\mu }_{0}I}{8\pi a}(\hat{k}-\hat{i})$