Question

The magnetic field at the origin due to a current element i d→l placed at a position →r is
(a) μ0i4πd→l ×→r r3
(b) -μ0i4π→r ×d→l r3
(c) μ0i4π→r ×d→l r3
(d) -μ0i4πd→l ×→r r3

Answer 1

The magnetic field at the origin due to a current element i d→l placed at a position →r is $\frac{\mu_{0} t}{4 \pi} \frac{\overrightarrow{d \imath} \times \vec{r}}{r^{3}}$ and $\frac{\mu_{0} i}{4 \pi} \frac{\overrightarrow{d \imath} \times \vec{r}}{r^{2}}$

Explanation:

The magnetic field at the origin is given by the current element $i \overrightarrow{d l}$  located at $\vec{r}$  

$d \vec{B}=\frac{\mu_{0} i}{4 \pi} \frac{\overrightarrow{d l} \times \vec{r}}{r^{3}}$

The cross-product property depends on,

Cross product of Vector A and B is equal to cross product of vector B and vector A  

It is  

$\begin{aligned}&\vec{A} \times \vec{B}=-\vec{B} \times \vec{A}\\&d \vec{B}=-\frac{\mu_{0} i}{4 \pi} \frac{\overrightarrow{d l} \times \vec{r}}{r^{3}}\end{aligned}$

Therefore the correct answer is  

$\begin{aligned}&\text { (a) } \frac{\mu_{0} i}{4 \pi} \frac{\overline{d \imath} \times \vec{r}}{r^{3}}\\&\text { (b) }-\frac{\mu_{0} i}{4 \pi} \frac{\overline{d \imath} \times \vec{r}}{r^{3}}\end{aligned}$

Answer 2

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Option A and B is correct