For a long wire carrying a current of 10 A, the magnitude of magnetic field at a point 4 cm away from it will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 5 \times 10^{-5}\,\mathrm{T} \)
By Ampere's circuital law (or the Biot-Savart law), the field of a long straight current-carrying wire at distance \( r \) is
\( \displaystyle B = \frac{\mu_0 I}{2\pi r} \)
\( I = 10\,\mathrm{A} \)
\( r = 4\,\mathrm{cm} = 0.04\,\mathrm{m} \)
\( \mu_0/2\pi = 2\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \)
\( \displaystyle B = \frac{2\times10^{-7} \times 10}{0.04} = 5\times10^{-5}\,\mathrm{T} \)
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 – …
