A current element of length \( d\vec{l} = 2\hat{\imath} \) mm placed at the origin carries a current \( I = 2 \) A. The magnetic field on the y axis at a distance of \( 2 \) cm from origin is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1\,\mu\mathrm{T} \)
The field of a small current element is given by the Biot-Savart law, \( dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\sin\theta}{r^{2}} \), where \( \theta \) is the angle between \( d\vec{l} \) and the line joining the element to the point.
The element is along the x-axis and the point is on the y-axis, so \( \theta = 90^\circ \).
\( dl = 2\,\mathrm{mm} = 2\times10^{-3}\,\mathrm{m} \)
\( I = 2\,\mathrm{A} \)
\( r = 2\,\mathrm{cm} = 2\times10^{-2}\,\mathrm{m} \)
\( \displaystyle dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\sin\theta}{r^{2}} = 10^{-7}\times\frac{2\times(2\times10^{-3})\times1}{(2\times10^{-2})^{2}} \)
\( \displaystyle dB = 10^{-7}\times\frac{4\times10^{-3}}{4\times10^{-4}} = 10^{-6}\,\mathrm{T} = 1\,\mu\mathrm{T} \)
Its direction is along \( d\vec{l}\times\hat{r} = \hat{\imath}\times\hat{\jmath} = \hat{k} \), the \( +z \) direction.
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 – …
