A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of magnitude of magnetic fields \( B_1 \) at a distance \( a/2 \) and \( B_2 \) at a distance \( 2a \) from the axis of the wire.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1 : 1 \)
Use Ampere's law with a circular loop of radius \( r \) around the axis.
Inside the wire (\( r \lt a \)): only the current through the loop counts, \( I' = I\,r^{2}/a^{2} \).
\( \displaystyle B\,(2\pi r) = \mu_0\,\frac{I r^{2}}{a^{2}} \;\Rightarrow\; B_1 = \frac{\mu_0 I r}{2\pi a^{2}} = \frac{\mu_0 I}{4\pi a} \quad (r = a/2) \)
Outside the wire (\( r \gt a \)): the whole current \( I \) is enclosed.
\( \displaystyle B_2 = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{2\pi (2a)} = \frac{\mu_0 I}{4\pi a} \)
\( \displaystyle \frac{B_1}{B_2} = 1 : 1 \)
Note: NTA picked this question from an NCERT (2026-27) example and added the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Ampere’s Circuital Law – …
