A long straight wire carries a current of 35 A. The magnitude of the field B at a point 20 cm from the wire would be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 3.5\times10^{-5}\,\mathrm{T} \)
The field of a long straight current-carrying wire follows from Ampere’s circuital law, \( B = \frac{\mu_0 I}{2\pi r} \). The distance must be in metres: \( r = 20\,\mathrm{cm} = 0.20\,\mathrm{m} \).
\( \displaystyle B = \frac{4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}}\times35\,\mathrm{A}}{2\pi\times0.20\,\mathrm{m}} \)
\( B = 3.5\times10^{-5}\,\mathrm{T} \)
Shortcut: Use \( \frac{\mu_0}{2\pi} = 2\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \). Then \( B = 2\times10^{-7}\times\frac{35}{0.20} = 2\times10^{-7}\times175 = 3.5\times10^{-5}\,\mathrm{T} \).
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):
Exercises – …
