PX157 - C8 - solenoid

a helical winding of wire, usually a cylinder
assuming the solenoid is long, so the edge effects can be neglected
assuming the number of turns per unit length, $n$ , is large, $n >> 1$ , so axial symmetry can be assumed
assuming the wire is wound forward and backward so there is no net current along the axis
using cylindrical coordinates, $(R, ϕ, z)$ , with $z$ -direction along the axis
the solenoid has radius, $a$ , and the current flows in the positive $ϕ$ -direction
by symmetry, $| \vec{B} |$ is a function of $R$ alone

\vec{B} (R) = B_{R} (R) \hat{R} + B_{ϕ} (R) \hat{ϕ} + B_{z} (R) \hat{z}

\oint_{C} \vec{B} \cdot d \vec{l} = μ_{0} I_{e n c l .}

- $d \vec{l} = d l \hat{ϕ} ⟹ \vec{B} \cdot d \vec{l} = B_{ϕ} d l :$

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

- $I_{e n c l .} = 0$ , no current passing though the cross-section:

B_{ϕ} 2 π R = 0 ⟹ B_{ϕ} = 0

step 2: solenoidal condition

$\subset \supset \iint_{S} \vec{B} \cdot d \vec{S} = 0$ $\subset \supset \iint_{S} \vec{B} \cdot d \vec{S} = \iint_{l e f t} \vec{B} \cdot (- \hat{z}) d S + \iint_{r i g h t} \vec{B} \cdot (+ \hat{z}) d S + \iint_{c u r v e d} \vec{B} \cdot \hat{R} d S$
- left and right cancel due to symmetry as solution does not depend on z
$\iint_{c u r v e d} \vec{B} \cdot \hat{R} d S = 0 ⟹ \iint B_{R} d S = B_{R} 2 π R L ⟹ B_{R} = 0$
step 3:
- choosing a rectangular path in the plane of constant $ϕ$ outside the solenoid and applying ampere's law:

\begin{aligned} \oint_{C} \vec{B} \cdot d \vec{l} & = \int_{a}^{b} \vec{B} \cdot d \vec{l} + \int_{b}^{c} \vec{B} \cdot d \vec{l} + \int_{c}^{d} \vec{B} \cdot d \vec{l} + \int_{d}^{e} \vec{B} \cdot d \vec{l} \\ = \int_{0}^{L} \vec{B} \cdot \hat{z} d z + \int_{R_{1}}^{R_{2}} \vec{B} \cdot \hat{R} d R + \int_{L}^{0} \vec{B} \cdot \hat{z} d z + \int_{R_{2}}^{R_{1}} \vec{B} \cdot \hat{R} d R \\ = \int_{L}^{0} B_{z} (R_{1}) d z + \int_{R_{1}}^{R_{2}} B_{R} d R + \int_{0}^{L} B_{z} (R_{2}) d z + \int_{R_{2}}^{R_{1}} B_{R} d R \\ s i n c e B_{R} & = 0 : \\ \oint_{C} \vec{B} \cdot d \vec{l} & = B_{z} (R_{1}) L - B_{z} (R_{2}) L = μ_{0} I_{e n c l .} \end{aligned}

- outside the solenoid, $I_{e n c l .} = 0 :$

B_{z} (R_{1}) = B_{z} (R_{2})

- $B_{z}$ is constant outside
- taking $R_{2} \to \infty : B_{z} (R_{2} \to \infty) = 0$
- $B_{z} = 0$ outside

step 4:
- choosing a rectangular path in a plane of constant $ϕ$ , in and out of the solenoid and applying ampere's law:

\oint_{c} \vec{B} \cdot d \vec{l} = B_{z} (R_{1}) L - B_{z} (R_{2}) L

- since $R_{2} > a$ (outside) $: B_{z} (R_{2}) = 0 :$

\begin{aligned} \oint_{C} \vec{B} \cdot d \vec{l} = B_{z} (R_{1}) L & = μ_{0} N I = μ_{0} n L I \\ B_{z} (R_{1}) & = μ_{0} n I \end{aligned}

∴ \vec{B} = {\begin{cases} \vec{0} & o u t s i d e \\ μ_{0} n I \hat{z} & i n s i d e \end{cases}

for a solenoid that is not infinitely long, all field lines must pass though the cross-section, $π a^{2}$ , but outside the solenoid, the field lines can pass though anywhere. therefore, the field density outside is much smaller in comparison to that inside the solenoid. hence, the above result is sensible