PX157 - C4b - magnetic field of an infinite straight wire

step 1: symmetry arguments
- $| \vec{B} |$ is constant at fixed radius, $R$ , independent of $ϕ$ and $z$
- the biot-savart law : $\vec{B} ⊥ I d \vec{l} ⟹ B_{z} = 0$
- solenoidal condition on cylindrical surface: $B_{R} = 0$
- magnetic field is azimuthal: $\vec{B} = B_{ϕ} (R) \hat{ϕ}$
step 2: path
- a path, $C$ , that is circular and centred on the wire, $d \vec{l} = d l \hat{ϕ} :$

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

step 3: enclosed current
- $I_{e n c l .} = I$
step 4: summarize $\begin{array}{r} B_{ϕ} (R) 2 π R = μ_{0} I \\ ∴ \vec{B} = \frac{μ_{0} I}{2 π R} \hat{ϕ} \end{array}$