A long thick straight wire of radius R is carrying a steady current I, which is uniformly distributed across its cross section. The magnetic field at a distance r (r < R) from the axis of the wire varies as:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( B \propto r \)
Apply Ampere’s circuital law to a circle of radius r (r < R) inside the wire. Because the current is spread uniformly, the circle encloses only the fraction of the current that flows through its area.
\( \displaystyle \oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\mathrm{enc}} \)
\( \displaystyle I_{\mathrm{enc}} = I\,\frac{\pi r^{2}}{\pi R^{2}} = \frac{Ir^{2}}{R^{2}} \)
\( \displaystyle B(2\pi r) = \mu_0\,\frac{Ir^{2}}{R^{2}} \)
\( \displaystyle B = \frac{\mu_0 I}{2\pi R^{2}}\,r \)
\( B \propto r \)
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 – “… Now the current enclosed Ie is not I, but is less than this value. 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,
\( \displaystyle \begin{array}{l} B(2\pi r) = \mu_0 \frac{I r^2}{a^2} \\ B = \left(\frac{\mu_0 I}{2\pi a^2}\right) r \qquad [4.15(b)] \\ B \propto r \qquad (r \lt a) \end{array} \)”