1,845
edits
Changes
Jump to navigation
Jump to search
Line 3:
Line 3: − + − + + + + + + + + + + + + + + − Gauss' Law for Magnetism:+ − + − + − + + + + + + + + + + + + + − {| class="wikitable" style="margin: 1em auto 1em auto" |<math>\boldsymbol{\nabla \times E} + \frac{\partial \boldsymbol{B}}{\partial t}= 0</math>|align="right" width="200"| (3) |} − Ampere's Law:+ − − {| class="wikitable" style="margin: 1em auto 1em auto" |<math>\boldsymbol{\nabla \times B} - \mu_0\epsilon_0\frac{\partial \boldsymbol{E}}{\partial t}= 0 </math>|align="right" width="200"| (4)|} − − − {| class="wikitable" style="margin: 1em auto 1em auto" − |<math>\vec{\nabla}\times\vec{D}=\frac{\rho_{ext}}{\epsilon_0}</math> − |align="right" width="200"| (1) − |}
no edit summary
These are the Maxwell's Equations we will be using to solve for regions "I" and "II" in our approximation of the Michelson interferometer.
These are the Maxwell's Equations we will be using to solve for regions "I" and "II" in our approximation of the Michelson interferometer.
Gauss' Law:
{|align=center
{|class="wikitable" style="margin: 1em auto 1em auto" |<math>\boldsymbol{\nabla \cdot E} = 0 </math>|align="right" width="200"|(1)|}
|Gauss' Law:
|Gauss' Law for Magnetism:
|-
|<math>\boldsymbol{\nabla \cdot E} = 0 </math>
|<math>\boldsymbol{\nabla \cdot B} = 0</math>
|-
|height="20"| ||
|-
|Faradays's Law:
|Ampere's Law:
|-
|width="400"|<math>\boldsymbol{\nabla \times E} + \frac{\partial \boldsymbol{B}}{\partial t}= 0</math>
|width="400"|<math>\boldsymbol{\nabla \times B} - \mu_0\epsilon_0\frac{\partial \boldsymbol{E}}{\partial t}= 0 </math>
|}
== In the Presence of Charges and Dielectric Media ==
{| class="wikitable" style="margin: 1em auto 1em auto" |<math>\boldsymbol{\nabla \cdot B} = 0</math>|align="right" width="200"| (2)|}
{|align=center
|Gauss' Law:
Faradays's Law:
|Gauss' Law for Magnetism:
|-
|<math>\boldsymbol{\nabla \cdot D} = \rho </math>
|<math>\boldsymbol{\nabla \cdot B} = 0</math>
|-
|height="20"| ||
|-
|Faradays's Law:
|Ampere's Law:
|-
|width="400"|<math>\boldsymbol{\nabla \times E} + \frac{\partial \boldsymbol{B}}{\partial t}= 0</math>
|width="400"|<math>\boldsymbol{\nabla \times H} - \frac{\partial \boldsymbol{D}}{\partial t}= \boldsymbol{j} </math>
|}
Where <math>\boldsymbol{D} = \epsilon_0 \boldsymbol{E}</math> and <math>\boldsymbol{B} = \mu_0 \boldsymbol{H}</math>.