A long straight wire carries a current of \( 20 \) A. What is the magnitude of the magnetic field at a point \( 40 \) cm from the wire?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 10^{-5} \) T
The field of a long straight wire (from Ampere’s circuital law) falls off as 1/r:
\( \displaystyle B = \frac{\mu_{0}I}{2\pi r} \)
Converting the distance to metres:
\( r = 40\ \mathrm{cm} = 0.40\ \mathrm{m} \)
\( \displaystyle B = \frac{(4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}) \times (20\ \mathrm{A})}{2\pi \times 0.40\ \mathrm{m}} \)
\( \displaystyle B = \frac{2\times10^{-7} \times 20}{0.40}\ \mathrm{T} = 1\times10^{-5}\ \mathrm{T} \)
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 – …
