An infinitely long cylindrical wire of radius $a$ carries a steady current $I$ uniformly distributed across its cross-section. Determine the magnetic field $B$ at a distance $r$ from the axis for: $(a) r > a$,$(b) r = a$,$(c) r < a$,and $(d)$ at the axis $(r = 0)$.

(N/A) Using Ampere's circuital law: $\oint B \cdot dl = \mu_0 I_{\text{enclosed}}$.
$(a)$ For $r > a$: The enclosed current is $I$. Thus,$B(2\pi r) = \mu_0 I$,which gives $B = \frac{\mu_0 I}{2\pi r}$.
$(b)$ For $r = a$: The magnetic field is $B = \frac{\mu_0 I}{2\pi a}$.
$(c)$ For $r < a$: The current density $J = \frac{I}{\pi a^2}$. The enclosed current $I_{\text{enclosed}} = J(\pi r^2) = I \frac{r^2}{a^2}$. Applying Ampere's law: $B(2\pi r) = \mu_0 I \frac{r^2}{a^2}$,which gives $B = \frac{\mu_0 I r}{2\pi a^2}$.
$(d)$ At the axis $(r = 0)$: Substituting $r = 0$ into the expression for $r < a$,we get $B = 0$.