Question

The magnetic field at the origin due to a current element $id\overrightarrow{l}$ placed at a position $\overrightarrow{r}$ is
(a) $\frac{{\mathrm{\mu }}_{0}i}{4\pi }\frac{d\overrightarrow{l}\times \overrightarrow{r}}{r^3}$
(b) $-\frac{{\mathrm{\mu }}_{0}i}{4\pi }\frac{\overrightarrow{r}\times d\overrightarrow{l}}{r^3}$
(c) $\frac{{\mathrm{\mu }}_{0}i}{4\pi }\frac{\overrightarrow{r}\times d\overrightarrow{l}}{r^3}$
(d) $-\frac{{\mathrm{\mu }}_{0}i}{4\pi }\frac{d\overrightarrow{l}\times \overrightarrow{r}}{r^3}$

Answer 1

(a) $\frac{{\mathrm{\mu }}_{0}i}{4\pi }\frac{d\overrightarrow{l}\times \overrightarrow{r}}{r^3}$
(b) $-\frac{{\mathrm{\mu }}_{0}i}{4\pi }\frac{\overrightarrow{r}\times d\overrightarrow{l}}{r^3}$

The magnetic field at the origin due to current element $id\overrightarrow{l}$ placed at a position $\overrightarrow{r}$ is given by
$d\overrightarrow{B}=\frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}i\frac{\overrightarrow{r}\times d\overrightarrow{l}}{r^3}$
According to the cross product property,
$$\begin{gathered} \overrightarrow{A}\times \overrightarrow{B}=-\overrightarrow{B}\times \overrightarrow{A} \\ \Rightarrow d\overrightarrow{B}=-\frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}i\frac{\overrightarrow{r}\times \overrightarrow{dl}}{r^3} \end{gathered}$$