A long straight wire of a circular cross-section of radius 'a' is carrying steady current I. The current I is uniformly distributed across this cross-section. What will be the magnetic field in the region \( r \lt a \) and \( r \gt a \)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( B \propto r\ (r \lt a) \) and \( B \propto r^{-1}\ (r \gt a) \)
Apply Ampere's circuital law to a circle of radius \( r \) centred on the axis of the wire: \( B\,(2\pi r) = \mu_0 I_{\text{enclosed}} \).
Inside the wire (\( r \lt a \)): the current is uniform, so the circle encloses the fraction \( r^{2}/a^{2} \) of it.
\( \displaystyle B\,(2\pi r) = \mu_0\,\frac{I r^{2}}{a^{2}} \;\Rightarrow\; B = \frac{\mu_0 I r}{2\pi a^{2}} \;\Rightarrow\; B \propto r \)
Outside the wire (\( r \gt a \)): the whole current \( I \) is enclosed.
\( \displaystyle B = \frac{\mu_0 I}{2\pi r} \;\Rightarrow\; B \propto r^{-1} \)
The field rises linearly from zero at the axis to its maximum at the surface and then falls as \( 1/r \).
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 – “… Solution (a) Consider the case r > a. … The result is the familiar expression for a long straight wire
\( \displaystyle B = \frac{\mu_0 I}{2\pi r} \qquad [4.15(a)] \)
\( \displaystyle B \propto \frac{1}{r} \quad (r \gt a) \)”
Ampere’s Circuital Law – “… Now the current enclosed Ie is not I, but is less than this value. … Using Ampere’s law,
\( \displaystyle B = \left(\frac{\mu_0 I}{2\pi a^2}\right) r \qquad [4.15(b)] \)
\( \displaystyle B \propto r \quad (r \lt a) \)”