What is the magnetic field on the y-axis at a distance of 0.7 m in the following diagram. Take current element \( \Delta x\,\hat{\imath} \) with current of 10 A and Δx = 1 cm.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2\times10^{-8}\,\hat{k}\,\mathrm{T} \)
The field of a small current element follows the Biot–Savart law, \( d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{l}\times\vec{r}}{r^{3}} \). Here the element is at the origin along x and the point is on the y-axis.
\( d\vec{l} = \Delta x\,\hat{\imath} = 10^{-2}\,\hat{\imath}\,\mathrm{m},\quad \vec{r} = 0.7\,\hat{\jmath}\,\mathrm{m},\quad I = 10\,\mathrm{A} \)
Since \( d\vec{l} \) is perpendicular to \( \vec{r} \), the magnitude is
\( \displaystyle dB = \frac{\mu_0}{4\pi}\,\frac{I\,\Delta x}{r^{2}} = 10^{-7}\times\frac{10\times10^{-2}}{(0.7)^{2}} \)
\( dB = \frac{10^{-8}}{0.49} \approx 2\times10^{-8}\,\mathrm{T} \)
Its direction is that of \( \hat{\imath}\times\hat{\jmath} = +\hat{k} \), so \( d\vec{B} \approx 2\times10^{-8}\,\hat{k}\,\mathrm{T} \).
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Magnetic Field Due to a Current – …
