A long, straight wire having radius R carries a current which is...

For a long straight wire of radius $R$ carrying a steady current $I$ uniformly distributed across its cross-section, the magnetic field at a distance $r$ from the axis is obtained using Ampere's circuital law.

Inside the wire ($r < R$):
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$
$B(2\pi r) = \mu_0 \left(\dfrac{I}{\pi R^2}\right) \pi r^2$
$B = \dfrac{\mu_0 I r}{2\pi R^2}$
The magnetic field is directly proportional to $r$. At the axis ($r = 0$), the magnetic field is zero, which is its minimum magnitude.

Outside the wire ($r \ge R$):
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I$
$B(2\pi r) = \mu_0 I$
$B = \dfrac{\mu_0 I}{2\pi r}$
The magnetic field is inversely proportional to $r$.

The magnetic field increases linearly from zero at the axis to a maximum value at the surface ($r = R$), where $B = \dfrac{\mu_0 I}{2\pi R}$, and then decreases outside the wire.

Thus, the magnitude of the magnetic field is maximum at the wire's surface and minimum at the wire's axis.