A thick long straight wire of radius \( a \) is carrying steady current \( I \). The current \( I \) is uniformly distributed across the cross-section. The ratio of magnetic fields at a distance \( a/2 \) and \( 3a/2 \) from the central axis of the wire is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 3:4 \)
Apply Ampere's circuital law to a circular path of radius \( r \) about the axis. Inside the wire only the enclosed current by the path counts, and outside it the whole current counts.
Inside (\( r \lt a \)): the enclosed current is \( I\,r^{2}/a^{2} \), so
\( \displaystyle B_{\mathrm{in}} = \frac{\mu_0 I r}{2\pi a^{2}} \)
Outside (\( r \gt a \)):
\( \displaystyle B_{\mathrm{out}} = \frac{\mu_0 I}{2\pi r} \)
At \( r = a/2 \) (inside) and \( r = 3a/2 \) (outside):
\( \displaystyle B_1 = \frac{\mu_0 I}{2\pi a^{2}} \times \frac{a}{2} = \frac{\mu_0 I}{4\pi a} \)
\( \displaystyle B_2 = \frac{\mu_0 I}{2\pi\,(3a/2)} = \frac{\mu_0 I}{3\pi a} \)
\( \displaystyle \frac{B_1}{B_2} = \frac{1/4}{1/3} = \frac{3}{4} \)
So the ratio is \( 3:4 \).
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 – …
