A long straight wire carries a current of \( 25\ \mathrm{A} \). What is the magnetic field at a point \( 50\ \mathrm{cm} \) from the wire?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 10\ \mu\mathrm{T} \)
By Ampere’s circuital law, the field of a long straight wire at distance \( r \) is
\( \displaystyle B = \frac{\mu_{0}I}{2\pi r} \)
\( r = 50\ \mathrm{cm} = 0.50\ \mathrm{m} \)
\( \displaystyle B = \frac{(4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}})(25\ \mathrm{A})}{2\pi(0.50\ \mathrm{m})} = \frac{2\times10^{-7}\times25}{0.50}\ \mathrm{T} \)
\( B = 1.0\times10^{-5}\ \mathrm{T} = 10\ \mu\mathrm{T} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
