A long straight wire of a circular cross-section with radius '$$a$$' carries a steady current $$I$$. The current $$I$$ is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance $$r$$ from the centre of the wire is given by

To find the magnetic field $$B$$ at a distance $$r$$ from the center of the wire, we use Ampere's Circuital Law:

$$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} $$

Case 1: Inside the wire ($$r < a$$)

Since the current $$I$$ is uniformly distributed across the circular cross-section, the current enclosed ($$I_{enc}$$) by an Amperian loop of radius $$r$$ is proportional to the area:

$$ I_{enc} = I \left( \frac{\pi r^2}{\pi a^2} \right) = I \frac{r^2}{a^2} $$

Applying Ampere's Law:

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

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

Therefore, inside the wire, the magnetic field increases linearly with distance: {$$B \propto r$$}.

Case 2: Outside the wire

For a point outside the wire, the Amperian loop encloses the entire current $$I$$:

$$ I_{enc} = I $$

Applying Ampere's Law:

$$ B (2\pi r) = \mu_0 I $$

$$ B = \frac{\mu_0 I}{2\pi r} $$

Therefore, outside the wire, the magnetic field decreases inversely with distance: {$$B \propto \frac{1}{r}$$}