A long straight wire of circular cross section with radius 'a' carries steady current 'I' which is uniformly distributed across the cross-section. The magnetic field in the region \( r \lt a \) and \( r \gt a \) is represented by:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4)
Ampere's circuital law is applied on a circle of radius \( r \) coaxial with the wire: \( B\,(2\pi r) = \mu_0 I_{\mathrm{enclosed}} \).
Inside the wire (\( r \lt a \)) the circle encloses only part of the current. Since the current is spread uniformly over the cross-section,
\( \displaystyle I_{\mathrm{enclosed}} = I\,\frac{\pi r^{2}}{\pi a^{2}} \;\Rightarrow\; B = \frac{\mu_0 I r}{2\pi a^{2}} \propto r \)
So the field is zero on the axis and rises linearly to its largest value at the surface \( r = a \).
Outside the wire (\( r \gt a \)) the whole current is enclosed:
\( \displaystyle B = \frac{\mu_0 I}{2\pi r} \propto \frac{1}{r} \)
So the field falls as \( 1/r \), a curve that keeps decreasing but never reaches zero. Only the graph of Option (4) has a straight rise up to \( r = a \) followed by such a curve.
Note: A similar question was also asked in CUET (UG) 2023 on 05-06-2023 (Slot 2).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Ampere’s Circuital Law – …
