Electromagnetic Force on a Charge inside a Volume
📂Electrodynamics Electromagnetic Force on a Charge inside a Volume Overview The electromagnetic force experienced by all charges in volume V \mathcal{V} V
F = ∮ S T ⋅ d a − ϵ 0 μ 0 d d t ∫ V S d τ
\mathbf{F} =\oint_{\mathcal{S}} \mathbf{T} \cdot d\mathbf{a} -\epsilon_{0}\mu_{0}\dfrac{d}{dt}\int_{\mathcal{V}} \mathbf{S} d\tau
F = ∮ S T ⋅ d a − ϵ 0 μ 0 d t d ∫ V S d τ
S \mathcal{S} S V \mathcal{V} V T \mathbf{T} T Maxwell stress tensor S \mathbf{S} S Poynting vector
Derivation By Lorentz force law , the force experienced by a charge is
F = q ( E + v × B )
\mathbf{F}=q(\mathbf{E} + \mathbf{v} \times \mathbf{B})
F = q ( E + v × B )
If we express the charge quantity in terms of charge density, it becomes q = ∫ ρ d τ q=\int \rho d\tau q = ∫ ρ d τ
F = ∫ V ρ ( E + v × B ) d τ
\mathbf{F}=\int_{\mathcal{V}} \rho (\mathbf{E} + \mathbf{v} \times \mathbf{B})d\tau
F = ∫ V ρ ( E + v × B ) d τ
Volume current density is therefore ρ v = J \rho\mathbf{v}=\mathbf{J} ρ v = J
F = ∫ V ( ρ E + J × B ) d τ
\mathbf{F}=\int_{\mathcal{V}} (\rho\mathbf{E} + \mathbf{J} \times \mathbf{B})d\tau
F = ∫ V ( ρ E + J × B ) d τ
Differentiating both sides to represent the force experienced by the charge per unit volume,
d F d τ = f = ρ E + J × B
\dfrac{d\mathbf{F}}{d \tau}=\mathbf{f}=\rho\mathbf{E} + \mathbf{J} \times \mathbf{B}
d τ d F = f = ρ E + J × B
Maxwell’s Equations
( i ) ρ = ϵ 0 ∇ ⋅ E \mathrm{(i)}\ \rho=\epsilon_{0} \nabla \cdot \mathbf{E} ( i ) ρ = ϵ 0 ∇ ⋅ E ( i v ) J = 1 μ 0 ∇ × B − ϵ 0 ∂ E ∂ t \mathrm{(iv)}\ \mathbf{J}=\dfrac{1}{\mu_{0}}\nabla \times \mathbf{B}-\epsilon_{0}\dfrac{\partial \mathbf{E}}{\partial t} ( iv ) J = μ 0 1 ∇ × B − ϵ 0 ∂ t ∂ E
Using Maxwell’s equations to express f \mathbf{f} f
f = ϵ 0 ( ∇ ⋅ E ) E + 1 μ 0 ( ∇ × B ) × B − ϵ 0 ∂ E ∂ t × B
\begin{equation}
\mathbf{f} = \epsilon_{0}(\nabla \cdot \mathbf{E})\mathbf{E} + \dfrac{1}{\mu_{0}}(\nabla \times \mathbf{B})\times \mathbf{B} - \epsilon_{0} \dfrac{\partial \mathbf{E}}{\partial t} \times \mathbf{B}
\end{equation}
f = ϵ 0 ( ∇ ⋅ E ) E + μ 0 1 ( ∇ × B ) × B − ϵ 0 ∂ t ∂ E × B
To adjust the last term on the right side of ( 1 ) (1) ( 1 )
∂ ∂ t ( E × B ) = ∂ E ∂ t × B + E × ∂ B ∂ t
\dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B}) = \dfrac{\partial \mathbf{E}}{\partial t} \times \mathbf{B} + \mathbf{E} \times \dfrac{\partial \mathbf{B}}{\partial t}
∂ t ∂ ( E × B ) = ∂ t ∂ E × B + E × ∂ t ∂ B
Faraday’s Law
∂ B ∂ t = − ∇ × E \frac{\partial \mathbf{B}}{\partial t}=-\nabla \times \mathbf{E} ∂ t ∂ B = − ∇ × E
Reorganizing the above formula using Faraday’s Law,
∂ E ∂ t × B = ∂ ∂ t ( E × B ) + E × ( ∇ × E )
\dfrac{\partial \mathbf{E}}{\partial t} \times \mathbf{B} = \dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B}) + \mathbf{E} \times (\nabla \times \mathbf{E} )
∂ t ∂ E × B = ∂ t ∂ ( E × B ) + E × ( ∇ × E )
Substituting it into ( 1 ) (1) ( 1 )
f = ϵ 0 ( ∇ ⋅ E ) E + 1 μ 0 ( ∇ × B ) × B − ϵ 0 [ ∂ ∂ t ( E × B ) + E × ( ∇ × E ) ] = ϵ 0 [ ( ∇ ⋅ E ) E − E × ( ∇ × E ) ] − 1 μ 0 [ B × ( ∇ × B ) ] − ϵ 0 ∂ ∂ t ( E × B )
\begin{align}
\mathbf{f} &= \epsilon_{0}(\nabla \cdot \mathbf{E})\mathbf{E} + \dfrac{1}{\mu_{0}}(\nabla \times \mathbf{B})\times \mathbf{B} - \epsilon_{0} \left[ \dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B}) + \mathbf{E} \times (\nabla \times \mathbf{E} ) \right] \nonumber
\\ &= \epsilon_{0} \bigg[ (\nabla \cdot \mathbf{E})\mathbf{E} - \mathbf{E} \times (\nabla \times \mathbf{E}) \bigg] - \dfrac{1}{\mu_{0}} \bigg[ \mathbf{B} \times (\nabla \times \mathbf{B} ) \bigg] -\epsilon_{0}\dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B} )
\end{align}
f = ϵ 0 ( ∇ ⋅ E ) E + μ 0 1 ( ∇ × B ) × B − ϵ 0 [ ∂ t ∂ ( E × B ) + E × ( ∇ × E ) ] = ϵ 0 [ ( ∇ ⋅ E ) E − E × ( ∇ × E ) ] − μ 0 1 [ B × ( ∇ × B ) ] − ϵ 0 ∂ t ∂ ( E × B )
To make the formula symmetric, in other words, to make it look nicer, add ( ∇ ⋅ B ) B (\nabla \cdot \mathbf{B})\mathbf{B} ( ∇ ⋅ B ) B ( 2 ) (2) ( 2 ) ∇ ⋅ B = 0 \nabla \cdot \mathbf{B}=0 ∇ ⋅ B = 0
f = ϵ 0 [ ( ∇ ⋅ E ) E − E × ( ∇ × E ) ] + 1 μ 0 [ ( ∇ ⋅ B ) B − B × ( ∇ × B ) ] − ϵ 0 ∂ ∂ t ( E × B )
\begin{equation}
\begin{aligned}
\mathbf{f}
&= \epsilon_{0} \bigg[ (\nabla \cdot \mathbf{E})\mathbf{E} - \mathbf{E} \times (\nabla \times \mathbf{E}) \bigg] + \dfrac{1}{\mu_{0}} \bigg[(\nabla \cdot \mathbf{B})\mathbf{B} - \mathbf{B} \times (\nabla \times \mathbf{B} ) \bigg] \\
&\quad -\epsilon_{0}\dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B})
\end{aligned}
\end{equation}
f = ϵ 0 [ ( ∇ ⋅ E ) E − E × ( ∇ × E ) ] + μ 0 1 [ ( ∇ ⋅ B ) B − B × ( ∇ × B ) ] − ϵ 0 ∂ t ∂ ( E × B )
Multiplication Rule
∇ ( A ⋅ B ) = A × ( ∇ × B ) + B × ( ∇ × A ) + ( A ⋅ ∇ ) B + ( B ⋅ ∇ ) A
\nabla(\mathbf{A} \cdot \mathbf{B}) = \mathbf{A} \times (\nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A})+(\mathbf{A} \cdot \nabla)\mathbf{B}+(\mathbf{B} \cdot \nabla) \mathbf{A}
∇ ( A ⋅ B ) = A × ( ∇ × B ) + B × ( ∇ × A ) + ( A ⋅ ∇ ) B + ( B ⋅ ∇ ) A
Also, by using the multiplication rule,
∇ ( A ⋅ A ) = ∇ ( A 2 ) = 2 ( A ⋅ ∇ ) A + 2 A × ( ∇ × A )
\nabla(\mathbf{A} \cdot \mathbf{A} ) =\nabla(A^2) = 2(\mathbf{A} \cdot \nabla) \mathbf{A} + 2\mathbf{A} \times (\nabla \times \mathbf{A} )
∇ ( A ⋅ A ) = ∇ ( A 2 ) = 2 ( A ⋅ ∇ ) A + 2 A × ( ∇ × A )
Thus,
{ E × ( ∇ × E ) = − ( E ⋅ ∇ ) E + 1 2 ∇ ( E 2 ) B × ( ∇ × B ) = − ( B ⋅ ∇ ) B + 1 2 ∇ ( B 2 )
\begin{cases} \mathbf{E} \times (\nabla \times \mathbf{E} ) = -(\mathbf{E}\cdot\nabla) \mathbf{E} + \dfrac{1}{2}\nabla(E^2)
\\ \mathbf{B} \times (\nabla \times \mathbf{B} ) = -(\mathbf{B}\cdot\nabla) \mathbf{B} +\dfrac{1}{2}\nabla(B^2) \end{cases}
⎩ ⎨ ⎧ E × ( ∇ × E ) = − ( E ⋅ ∇ ) E + 2 1 ∇ ( E 2 ) B × ( ∇ × B ) = − ( B ⋅ ∇ ) B + 2 1 ∇ ( B 2 ) ( 3 ) (3) ( 3 )
f = ϵ 0 [ ( ∇ ⋅ E ) E + ( E ⋅ ∇ ) E − 1 2 ∇ ( E 2 ) ] + 1 μ 0 [ ∇ ⋅ B ) B + ( B ⋅ ∇ ) B − 1 2 ∇ ( B 2 ) ] − ϵ 0 ∂ ∂ t ( E × B )
\begin{align*}
\mathbf{f} &= \epsilon_{0} \bigg[ (\nabla \cdot \mathbf{E})\mathbf{E} +(\mathbf{E}\cdot\nabla)\mathbf{E} - \dfrac{1}{2}\nabla(E^2) \bigg] + \dfrac{1}{\mu_{0}} \bigg[ \nabla \cdot \mathbf{B}) \mathbf{B} +(\mathbf{B}\cdot\nabla)\mathbf{B} -\dfrac{1}{2}\nabla(B^2) \bigg] -\epsilon_{0}\dfrac{\partial}{\partial t}(\mathbf{E} \times \mathbf{B} )
\end{align*}
f = ϵ 0 [ ( ∇ ⋅ E ) E + ( E ⋅ ∇ ) E − 2 1 ∇ ( E 2 ) ] + μ 0 1 [ ∇ ⋅ B ) B + ( B ⋅ ∇ ) B − 2 1 ∇ ( B 2 ) ] − ϵ 0 ∂ t ∂ ( E × B )
Because the equation is too complex, using the Maxwell stress tensor T \mathbf{T} T ∇ ⋅ T \nabla \cdot \mathbf{T} ∇ ⋅ T Poynting vector S \mathbf{S} S
f = ∇ ⋅ T − ϵ 0 μ 0 ∂ S ∂ t
\mathbf{f} = \nabla \cdot \mathbf{T} - \epsilon_{0}\mu_{0}\dfrac{\partial \mathbf{S}}{\partial t}
f = ∇ ⋅ T − ϵ 0 μ 0 ∂ t ∂ S
Integrating both sides over the volume and using the divergence theorem on the first term on the right-hand side results in
F = ∮ S T ⋅ d a − ϵ 0 μ 0 d d t ∫ V S d τ
\mathbf{F} =\oint_{\mathcal{S}} \mathbf{T} \cdot d\mathbf{a} -\epsilon_{0}\mu_{0}\dfrac{d}{dt}\int_{\mathcal{V}} \mathbf{S} d\tau
F = ∮ S T ⋅ d a − ϵ 0 μ 0 d t d ∫ V S d τ
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