Line 11:
Line 11:
<math>\boldsymbol{\nabla \times E} + \frac{\partial \boldsymbol{B}}{\partial t}= 0</math>
<math>\boldsymbol{\nabla \times E} + \frac{\partial \boldsymbol{B}}{\partial t}= 0</math>
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+
Ampere's Law:
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+
<math>\boldsymbol{\nabla \times B} - \mu_0\epsilon_0\frac{\partial \boldsymbol{E}}{\partial t}= 0 </math>