A cylindrical, straight long wire having radius 'a', is carrying current 'I' uniformly distributed inside the wire. As per Ampere's circuital law, the magnetic field inside the wire at any distance \( r \), \( (r \lt a) \) is given by:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( B = \mu_0 I r/2\pi a^{2} \)
By symmetry the field lines are circles about the axis of the wire. Take a circular Amperian loop of radius \( r \lt a \). Because the current is uniformly distributed, the loop encloses only the fraction \( \pi r^{2}/\pi a^{2} \) of the total current. Ampere's circuital law gives
\( \displaystyle \oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\mathrm{enc}} \)
\( \displaystyle I_{\mathrm{enc}} = I\,\frac{\pi r^{2}}{\pi a^{2}} = \frac{I r^{2}}{a^{2}} \)
\( \displaystyle B\,(2\pi r) = \mu_0\,\frac{I r^{2}}{a^{2}} \)
\( \displaystyle B = \frac{\mu_0 I r}{2\pi a^{2}} \)
Inside the wire \( B \propto r \): it is zero on the axis and rises to \( \mu_0 I/2\pi a \) at the surface.
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 – “… Example 4.7 Figure 4.13 shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a.”
Ampere’s Circuital Law – “… Since the current distribution is uniform, the current enclosed is,
\( \displaystyle I_e = I\left(\frac{\pi r^2}{\pi a^2}\right) = \frac{I r^2}{a^2} \)
Using Ampere’s law,
\( \displaystyle B(2\pi r) = \frac{\mu_0 I r^2}{a^2} \)
\( \displaystyle B = \left(\frac{\mu_0 I}{2\pi a^2}\right) r \qquad [4.15(b)] \)”