A long, cylindrical wire having radius b carries a uniformly...

Using Ampere's circuital law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$

For a point inside the wire at a distance $a$ from the axis, consider a circular Amperian loop of radius $a$.

Since the current is uniformly distributed over the cross-section of radius $b$, the current density is:

$J = \dfrac{i}{\pi b^2}$

The current enclosed by the Amperian loop is:

$I_{enclosed} = J \times \pi a^2 = \dfrac{i}{\pi b^2} \times \pi a^2 = \dfrac{i a^2}{b^2}$

Applying Ampere's law:

$B(2\pi a) = \mu_0$\left( \dfrac{i a^2}{b^2} \right)<br><br>B = \dfrac{\mu_0 i a}{2\pi b^2}