A wire element of length 5 mm is placed along the x-axis at the origin and it carries a current of 5.0 A. The magnetic field at a point on y-axis 0.5 m away from the element is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 10 nT
The field of a small current element is given by the Biot–Savart law. The element lies along the x-axis and the point is on the y-axis, so the angle between the element and the line joining it to the point is 90°.
\( \displaystyle dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^{2}} \)
\( \displaystyle dB = 10^{-7} \times \frac{5.0 \times 5\times10^{-3} \times \sin 90^\circ}{(0.5)^{2}} \)
\( \displaystyle dB = 10^{-7} \times \frac{2.5\times10^{-2}}{0.25} \)
\( \displaystyle dB = 1\times10^{-8}\,\mathrm{T} = 10\,\mathrm{nT} \)
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 – …
