Simplified Maxwell's Equation Proof

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
Von Neumann
Messages
101
Reaction score
4
Problem:

I'm trying to crudely prove the following:

[itex]\frac{{\partial}B}{{\partial}x}[/itex]=-[itex]\mu_{o}[/itex][itex]\epsilon_{o}[/itex][itex]\frac{{\partial}E}{{\partial}t}[/itex]

Solution (so far):

I can get the derivation, but the minus sign eludes me somehow...

Integrating over a thing rectangular loop of length [itex]l[/itex] and width [itex]dx[/itex], start with the following,

[itex]\oint{B{\cdot}dl}[/itex]=[itex]\mu_{o}\epsilon_{o}[/itex][itex]\frac{\partial{\Phi_{E}}}{\partial{t}}[/itex]

Then,

[itex]\oint{B{\cdot}dl}[/itex]=[itex](B+dE)l-Bl=dEl[/itex]

Also,

[itex]\Phi_{E}=EA=E(dx)(l)[/itex]

∴[itex]\frac{\partial{\Phi_{E}}}{\partial{t}}=[/itex][itex]\frac{\partial{E}}{\partial{t}}dxl[/itex]

Equating the equations above,

[itex]dEl=[/itex][itex]\mu_{o}\epsilon_{o}\frac{\partial{E}}{\partial{t}}dxl[/itex]

[itex]dE=[/itex][itex]\mu_{o}\epsilon_{o}\frac{\partial{E}}{\partial{t}}dx[/itex]

[itex]\frac{\partial{E}}{\partial{x}}=[/itex][itex]\mu_{o}\epsilon_{o}\frac{\partial{E}}{\partial{t}}[/itex]

Any advice is greatly appreciated.
 
Physics news on Phys.org