PX157 - C4c - magnetic field on a long cylindrical conductor

assuming current is uniform in the wire, the current density, $J = \frac{I}{π a^{2}} = c o n s t a n t$
step 1: symmetry
- $\vec{B} = B_{ϕ} (R) \hat{ϕ}$
step 2: path

\oint_{C} \vec{B} \cdot d \vec{l} = 2 π R B_{ϕ}

step 3: enclosed current
- if $R > a : I_{e n c l .} = I$
- if $R < a : I_{e n c l .} = J π R^{2} = I {(\frac{R}{a})}^{2}$
step 4: summarize $\vec B = \begin{cases}$

\frac{\mu_{0}I}{2\pi R} \left(\frac{R}{a}\right)^{2} \hat\phi = \frac{\mu_{0}I}{2\pi a^{2}} \hat\phi;, & R<a
\end{cases}$$