A cylindrical conductor carries a steady electric current. Which of...

For a steady current $I$ to flow through a cylindrical conductor, an electric field $\vec{E}$ must exist inside it to drive the charge carriers, given by $\vec{J} = \sigma \vec{E}$.

Since the current density $\vec{J}$ is non-zero, the electric field $\vec{E}$ is non-zero everywhere inside the conductor, including its axis.

By the boundary condition for electric fields, the tangential component of the electric field must be continuous across the surface of the conductor. Therefore, an electric field also exists in the vicinity outside the conductor.

Using Ampere's circuital law, the magnetic field at a distance $r$ from the axis inside the conductor is given by $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$.

At the axis, $r = 0$, which means the enclosed current is zero. Thus, the magnetic field at the axis is zero.

In the vicinity outside the conductor, the magnetic field is $B = \dfrac{\mu_0 I}{2 \pi r}$, which is non-zero.

Therefore, only the magnetic field at the conductor's axis is zero.