If a copper rod carries a direct current, the magnetic field due to it will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → both inside and outside of the rod
A steady current produces a magnetic field in the whole space around it, and Ampere’s circuital law gives its value. For a rod of radius R carrying current I uniformly over its cross-section, take a circular loop of radius r about the axis.
Outside the rod (\( r \gt R \)): the loop encloses the whole current I.
\( \displaystyle B\,(2\pi r) = \mu_0 I \Rightarrow B = \frac{\mu_0 I}{2\pi r} \)
Inside the rod (\( r \lt R \)): the loop encloses only the fraction \( r^{2}/R^{2} \) of the current.
\( \displaystyle B\,(2\pi r) = \mu_0 I\,\frac{r^{2}}{R^{2}} \Rightarrow B = \frac{\mu_0 I r}{2\pi R^{2}} \)
Both values are non-zero, so the field exists inside as well as outside the rod.
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 – “… Figure (4.14) shows a plot of the magnitude of B with distance r from the centre of the wire.”
