How Does Magnetic Flux Density B Relate to E and H in Electromagnetic Waves?

[faber]

Hi,
We were told to show that the magnetic flux density B obeys a homogenous wave equation. This case applies to electromagnetic waves in a homogenous, linear, uncharged conductor.
Now I know that the wave equation for magnetic flux density is as follows.

[ tex ] \nabla^2-\epsilon\mju \frac {\deltaB} {\deltaT}=0 [ \ tex ]

However I am a little confused on what the solution of the wave equation will be for B. I have the solution for E and H and know they are both orthogonal how is B related to these?

 

[OldTee]

There're two auxiliary equations that relate E and H to D and B via material properties. They have it in Wikipedia http://en.wikipedia.org/wiki/Maxwell's_equations [/URL]

 

[Graeme Robinson]

The electromagnetic wave equation, also known as the Maxwell's equations, describes the behavior of electromagnetic waves in a medium. It is composed of two parts, the electric field equation and the magnetic field equation. The electric field equation is given by:

[ tex ] \nabla^2E-\mu\epsilon \frac{\deltaE}{\delta t} = 0 [ \tex ]

And the magnetic field equation is given by:

[ tex ] \nabla^2B-\mu\epsilon \frac{\deltaB}{\delta t} = 0 [ \tex ]

These equations are derived from the fundamental laws of electromagnetism, namely Gauss's law, Faraday's law, and Ampere's law. In a homogeneous, linear, and uncharged conductor, the permittivity and permeability are constant, thus the equations simplify to:

[ tex ] \nabla^2E-\mu\epsilon \frac{\deltaE}{\delta t} = 0 [ \tex ]

[ tex ] \nabla^2B-\mu\epsilon \frac{\deltaB}{\delta t} = 0 [ \tex ]

Now, to show that the magnetic flux density B obeys a homogeneous wave equation, we can substitute the value of B from the second equation into the first equation and simplify:

[ tex ] \nabla^2\left(\frac{1}{\mu}\nabla^2B\right)-\epsilon\mu \frac{\delta^2B}{\delta t^2} = 0 [ \tex ]

Using the vector identity, [ tex ] \nabla\times(\nabla\times\mathbf{A}) = \nabla(\nabla\cdot\mathbf{A})-\nabla^2\mathbf{A} [ \tex ]

We can rewrite the above equation as:

[ tex ] \nabla^2\left(\frac{1}{\mu}\nabla\times(\nabla\times\mathbf{B})\right)-\epsilon\mu \frac{\delta^2B}{\delta t^2} = 0 [ \tex ]

Since we know that [ tex ] \nabla\times\mathbf{E} = -\frac{\delta\mathbf{B}}{\delta t} [