A long wire with a small current element of length \( 1\,\mathrm{cm} \) is placed at the origin and carries a current of \( 10\,\mathrm{A} \) along the x-axis. The magnitude of the magnetic field, due to the element, on the y-axis at a distance \( 0.5\,\mathrm{m} \) from it, would be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4 \times 10^{-8}\,\mathrm{T} \)
The field of a small current element is given by the Biot-Savart law, where \( \theta \) is the angle between the element and the line joining it to the point.
\( \displaystyle dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^{2}} \)
The element lies along the x-axis and the point is on the y-axis, so the two are at right angles and \( \sin\theta = 1 \).
\( I = 10\,\mathrm{A} \)
\( dl = 1\,\mathrm{cm} = 10^{-2}\,\mathrm{m} \)
\( r = 0.5\,\mathrm{m} \)
\( \displaystyle dB = \frac{10^{-7} \times 10 \times 10^{-2}}{(0.5)^{2}} = \frac{10^{-8}}{0.25} = 4\times10^{-8}\,\mathrm{T} \)
Note: A similar question was also asked in CUET (UG) 2023 on 06-06-2023 (Slot 1).
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 – …
