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Faradays law in different units?

  1. Aug 29, 2011 #1
    This is maddening and i cannot find a concise explaination anywhere despite the simplicity of this question.

    I keep coming across faradays law expressed as
    [tex] \text{curl}\left(\mathbf{E}\right) = -\frac{1}{C^2} \frac{\partial\mathbf{B}}{\partial t}
    Im only used to working in SI units so all i can do is guess that this is expressed in gaussian units?
    [tex] \text{curl}\left(\mathbf{E}\right) = -\frac{1}{C} \frac{\partial\mathbf{B}}{\partial t}

    Can anyone explain these 2 different expressions of faradays law?
  2. jcsd
  3. Aug 29, 2011 #2


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    The latter uses Gaussian, or Lorentz-Heaviside units. I have to admit to never seeing the former in a general expression for Faraday's law before. Do you have a reference for where you "came across" this form?
  4. Aug 30, 2011 #3
    hmm, the 1/c^2 expression i've only come across once in some lecture slides for uni electromagnetism module. If you've never come it expressed in that form then it could simply be a misprint/error. As the uni course is in recess for summer break here until october it's dificulty in simply asking the lecturer who wrote them. Thanks for clearing that up Hootenanny, i'll just take it as a misprint. :)
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