PX157 - Dx1 - displacement current

\int_{C} \vec{B} \cdot d \vec{l} = μ_{0} I = μ_{0} \iint \vec{J} \cdot d \vec{S}

surface $A_{2}$ goes between the capacitor plates, but is still bounded by the curve $C :$

\oint_{C} \vec{B} \cdot d \vec{l} = 0

both results cannot be true at the same time
there is an electric field between the plates, from PX157 - B5a - gauss's law:

ϕ_{E} = \iint_{A_{2}} \vec{E} d \vec{S} = \frac{Q}{ϵ_{0}}

a current is defined as $I = \frac{d Q}{d t} :$

I_{D} = ϵ_{0} \frac{d ϕ_{E}}{d t}

- compable with $ϵ = - \frac{d ϕ_{B}}{d t}$

I_{D} = ϵ_{0} \frac{d ϕ_{E}}{d t} = ϵ_{0} \frac{d}{d t} \iint_{S} \vec{E} \cdot d \vec{S}

adding $I_{D}$ to PX157 - C4a - ampere's law:

\oint_{C} \vec{B} \cdot d \vec{l} = μ_{0} I + μ_{0} I_{D} = μ_{0} I + μ_{0} ϵ_{0} \frac{d ϕ_{E}}{d t}

for surface $A_{1} :$

\oint \vec{B} \cdot d \vec{l} = μ_{0} I

for surface $A_{2} :$

\oint \vec{B} \cdot d \vec{l} = μ_{0} I_{D}

between the capacitor plates: magnetic field is induced in the azimuthal direction

how important is the displacement current?

consider a solenoid with an oscillating current:

I_{s} (t) = I_{0} \cos (ω t)

\vec{B} = μ_{0} n I_{s} (t) \hat{z}

solenoid has a cross-section, $π a^{2}$
faraday's law (static) over a circular path with radius, $R$ :

\begin{aligned} \oint \vec{E} \cdot d \vec{l} & = - \frac{d ϕ_{B}}{d t} \\ E_{ϕ} 2 π R & = - μ_{0} n \frac{d I_{S}}{d t} π a^{2} \\ E_{ϕ} (t) & = - \frac{1}{2} μ_{0} n \frac{a^{2}}{R} \frac{d I_{S}}{d t} \end{aligned}

PX157 - B4a - electric flux:

ϕ_{E} = \int_{0}^{L} d z \int_{R_{1}}^{R_{2}} d R E_{ϕ} = - \frac{1}{2} μ_{0} n a^{2} L \ln (\frac{R_{2}}{R_{1}}) \frac{d I_{S}}{d t}

displacement current:

I_{D} (t) = ϵ_{0} \frac{d ϕ_{E}}{d t} = - \frac{1}{2} μ_{0} ϵ_{0} n a^{2} L \ln (\frac{R_{2}}{R_{1}}) ω^{2} I_{S}

\frac{| I_{D} |}{| I_{S} |} = \frac{1}{2} n a^{2} L \ln (\frac{R_{2}}{R_{1}}) \cdot μ_{0} ϵ_{0} \cdot ω^{2} = {(\frac{λ ω}{c})}^{2}

$I_{D}$ is important if $λ$ is large, or $ω$ is large
eg: of large $λ :$ jupiter-io