A long straight wire of circular cross-section (radius $$a$$) is carrying steady current I. The current I is uniformly distributed across this cross-section. The magnetic field is

Inside the wire for the region $$r < a$$, the current enclosed by an Amperian loop of radius $$r$$ depends on the cross-sectional area:

$$I_{\text{enclosed}} = I \cdot \frac{\pi r^2}{\pi a^2} = I \frac{r^2}{a^2}$$

Applying Ampere's Law for this internal region: $$\oint B \cdot dl = \mu_0 I_{\text{enclosed}}$$

$$B \cdot (2\pi r) = \mu_0 I \frac{r^2}{a^2}$$ $$\implies$$ $$B = \frac{\mu_0 I r}{2\pi a^2}$$

This shows that $$B \propto r$$ inside the wire.

Outside the wire for the region $$r > a$$, the Amperian loop of radius $$r$$ encloses the total current: $$I_{\text{enclosed}} = I$$

Applying Ampere's Law for this external region: $$\oint B \cdot dl = \mu_0 I_{\text{enclosed}}$$

$$B \cdot (2\pi r) = \mu_0 I$$ $$\implies$$ $$B = \frac{\mu_0 I}{2\pi r}$$

This shows that $$B \propto \frac{1}{r}$$ outside the wire.