The third wave restates the right hand curl rule that a current loop produces a \(B\) field perpendicular to the loop; a loop in the anti-clockwise sense produces a \(B\) field toward the observer.
From the previous post "Magnetic Genie",
\(\nabla \times Y=-4\pi \nabla \rho _{ e }-\cfrac { 4\pi }{ c^2 }\cfrac{\partial \, j_{ e }}{\partial t}\)
If \(c\) is upwards, then \(ic\) is to the left. Since \(\nabla \times Y\) is in anti-clockwise, then \(\nabla \rho _{ e }\) and \(\cfrac{\partial \, j_{ e }}{\partial t}\) are both in the clockwise sense because of the negative sign. This directions are consistent with the right hand rule for finding magnetic field due to a current loop.
A rotating disc with a certain non uniform charge distribution will generate a EMW when rotated...
\(\nabla \times Y=-4\pi \nabla \rho _{ e }\)
The presence of \(\rho_e\) creates an \(E\) field,
\( \nabla .E=4\pi \rho _{ e }\)
A rotating \(Y\) generates a changing \(B\) field and a rotating \(E\) field,
\( -\nabla \times E=-i\cfrac { 1 }{ c } \cfrac { \partial B }{ \partial t } +Y\)
So,
\( -\nabla (\nabla .E)+\nabla ^{ 2 }E=-i\cfrac { 1 }{ c } \cfrac { \partial \nabla \times B }{ \partial t } +\nabla \times Y\)
\( -\nabla (4\pi \rho _{ e })+\nabla ^{ 2 }E=\cfrac { 1 }{ c } \cfrac { \partial (-i\nabla \times B) }{ \partial t } -4\pi \nabla \rho _{ e }\)
But a rotating \(B\) in turn generates a changing \(E\) field,
\( -i\nabla \times B=\cfrac { 1 }{ c } \cfrac { \partial E }{ \partial t } \)
And we have
\( \nabla ^{ 2 }E=\cfrac { 1 }{ c } \cfrac { \partial }{ \partial t } \left\{ \cfrac { 1 }{ c } \cfrac { \partial E }{ \partial t } \right\} \)
\( \nabla ^{ 2 }E=\cfrac { 1 }{ c^{ 2 } } \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } \)
An \(E\) field wave!
Now consider a flat coil spiral disc or a loop carrying currents...
\(\nabla \times Y=-4\pi \nabla \rho _{ e }-\cfrac { 4\pi }{ c^2 }\cfrac{\partial \, j_{ e }}{\partial t}\)
we will now set \(-4\pi \nabla \rho _{ e }=0\) and consider only the flow of \(j_e\)
\(\nabla \times Y=-\cfrac { 4\pi }{ c^{ 2 } } \cfrac { \partial \, j_{ e } }{ \partial t } \)
We know that \(j_e\) creates a changing \(E\) field and around \(j_e\) a circular \(B\) field,
\( -i\nabla \times B=\cfrac { 1 }{ c } \cfrac { \partial E }{ \partial t } +\cfrac { 4\pi }{ c } j_{ e }\)
Differentiating wrt \(t\),
\( \cfrac { \partial \, }{ \partial t } \left\{ -i\nabla \times B \right\} =\cfrac { 1 }{ c } \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } +\cfrac { 4\pi }{ c } \cfrac { \partial \, j_{ e } }{ \partial t } \) --- (*)
The vortex of \(Y\) creates a changing \(B\) field and an \(E\) field along it circular path,
\( -\nabla \times E=-i\cfrac { 1 }{ c } \cfrac { \partial B }{ \partial t } +Y\)
\( -\nabla (\nabla .E)+\nabla ^{ 2 }E=-i\cfrac { 1 }{ c } \cfrac { \partial \nabla \times B }{ \partial t } +\nabla \times Y\)
But \( \nabla .E=4\pi \rho _{ e }\)
\( -\nabla (4\pi \rho _{ e })+\nabla ^{ 2 }E=\cfrac { 1 }{ c } \cfrac { \partial (-i\nabla \times B) }{ \partial t } -\cfrac { 4\pi }{ c^{ 2 } } \cfrac { \partial \, j_{ e } }{ \partial t } \)
Substitute (*) into the above and since we set \( \nabla \rho _{ e }=0\),
\( \nabla ^{ 2 }E=\cfrac { 1 }{ c^2 } \cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } \)
And we have a wave.
In both cases we can independently create a EMW, which is expected from superposition.
