An element \( d\vec{l} = 0.5\,\hat{\jmath}\,\mathrm{cm} \) is placed at the origin and carries a large current \( I = 10\,\mathrm{A} \). The magnetic field at point (50 cm, 0) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2\times10^{-8}\,(-\hat{k})\,\mathrm{T} \)
The field of a small current element is given by the Biot-Savart law:
\( \displaystyle d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{l}\times\hat{r}}{r^{2}} \)
The element is along \( \hat{\jmath} \) and the point lies on the \( x \)-axis, so \( \hat{r} = \hat{\imath} \) and \( r = 50\,\mathrm{cm} = 0.5\,\mathrm{m} \). Also \( dl = 0.5\,\mathrm{cm} = 5\times10^{-3}\,\mathrm{m} \), and \( d\vec{l} \) is perpendicular to \( \hat{r} \).
Magnitude:
\( \displaystyle dB = 10^{-7} \times \frac{10 \times 5\times10^{-3}}{(0.5)^{2}} = 2\times10^{-8}\,\mathrm{T} \)
Direction: \( \hat{\jmath}\times\hat{\imath} = -\hat{k} \), so
\( \displaystyle d\vec{B} = 2\times10^{-8}\,(-\hat{k})\,\mathrm{T} \)
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 Element, Biot-Savart Law – …
