A current element \( \overrightarrow{\Delta l} = \Delta x\,\hat{\imath} \) (where \( \Delta x \) = 1 cm) is placed at the origin and carries a current 10 A. The magnetic field on y-axis at a distance of 0.5 m :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4\times10^{-8}\,\mathrm{T} \)
The field of a small current element follows the Biot–Savart law:
\( \displaystyle d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,\overrightarrow{\Delta l}\times\hat{r}}{r^{2}} \)
The element lies along the x-axis and the point lies on the y-axis, so \( \overrightarrow{\Delta l} \) is perpendicular to \( \hat{r} \) (\( \sin\theta = 1 \)), and the direction of the field is that of \( \hat{\imath}\times\hat{\jmath} = \hat{k} \), along the +z axis.
\( \Delta l = 1\,\mathrm{cm} = 10^{-2}\,\mathrm{m},\quad I\,\Delta l = 10\,\mathrm{A}\times10^{-2}\,\mathrm{m} = 0.1\,\mathrm{A\,m} \)
\( \displaystyle B = \frac{\mu_0}{4\pi}\,\frac{I\,\Delta l}{r^{2}} = 10^{-7}\times\frac{0.1}{(0.5)^{2}} \)
\( \displaystyle B = 10^{-7}\times\frac{0.1}{0.25} = 4\times10^{-8}\,\mathrm{T} \)
Note: A similar question was also asked in CUET (UG) 2025 on 13-05-2025 (Slot 2).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Magnetic Field Due to a Current – …
