An element \( d\vec{l} = \Delta x\hat{\imath} \) is placed at the origin and carries a current \( I \) in it. The magnetic field at a point having position \( \vec{r} = r\hat{k} \) will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\mu_{0}}{4\pi}\frac{I\Delta x}{r^{2}}(-\hat{\jmath}) \)
By the Biot–Savart law, the field of a current element is along \( d\vec{l}\times\vec{r} \), perpendicular to both the element and the position vector.
\( \displaystyle d\vec{B} = \frac{\mu_{0}}{4\pi}\frac{I\,d\vec{l}\times\vec{r}}{r^{3}} \)
\( d\vec{l}\times\vec{r} = (\Delta x\,\hat{\imath})\times(r\,\hat{k}) = \Delta x\,r\,(\hat{\imath}\times\hat{k}) \)
In the cyclic order \( \hat{\imath}\rightarrow\hat{\jmath}\rightarrow\hat{k} \), \( \hat{k}\times\hat{\imath} = \hat{\jmath} \), so reversing the order gives \( \hat{\imath}\times\hat{k} = -\hat{\jmath} \).
\( \displaystyle d\vec{B} = \frac{\mu_{0}}{4\pi}\frac{I\Delta x\,r}{r^{3}}(-\hat{\jmath}) = \frac{\mu_{0}}{4\pi}\frac{I\Delta x}{r^{2}}(-\hat{\jmath}) \)