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A very simple question. It is about Maxwell's 3rd equation of electromagnetics also known as Faraday's law. $$\vec\nabla\times\vec E = -\frac{\partial \vec B}{\partial t}$$ In all books and resources I've exhausted always takes the constant of proportionality for the equation as '1' like above equation. Can anyone please tell me which units are used for electric and magnetic field to make the proportionality constant of the equation '1'? Because, I think it would be a great coincidence if conventional units(V-m or Tesla) are used and we still get constant of proportionality '1'.

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    $\begingroup$ Please include the equation you are referring to. Don't make us guess. To enter equations, use MathJax Here’s a MathJax tutorial $\endgroup$
    – garyp
    Commented Sep 23, 2016 at 13:06
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Bort
    Commented Sep 23, 2016 at 13:07
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    $\begingroup$ V cross B has the same units as E. So E/meters=B/seconds. This is in SI with electric field being newtons per coulomb (or volts per meter) $\endgroup$
    – user12029
    Commented Sep 23, 2016 at 13:07
  • $\begingroup$ Of course, other choices of units are possible. See the comparisons made among SI, Gaussian, and Heaviside Lorentz units at Wikipedia. $\endgroup$ Commented Sep 23, 2016 at 13:42
  • $\begingroup$ @Bort , In the wikipedia page the 4th equation ends up having $$ \frac{1}{c^2} $$ before $$ -\frac{\partial \vec E}{\partial t} $$ in SI system! $\endgroup$ Commented Sep 23, 2016 at 13:56

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I suppose you are referring to the equation

$$\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}. $$

In fact this equation is true (i.e. you get a proportional constant of 1) if you use the following SI-units:

  • $[\nabla] = \frac{1}{\text{m}}$
  • $[\mathbf{E}] = \frac{\text{V}}{\text{m}}$
  • $[\mathbf{B}] = \text{T} = \frac{\text{Vs}}{\text{m}^2}$
  • $[t] = \text{s}$

Inserting these units yields:

Left hand side: $\frac{1}{\text{m}} \cdot \frac{\text{V}}{\text{m}} = \frac{V}{\text{m}^2}$

Right hand side: $\text{T} \cdot \frac{1}{\text{s}} = \frac{Vs}{\text{m}^2} \cdot \frac{1}{\text{s}} = \frac{V}{\text{m}^2}$

As you can see, we get the same units on both sides of the equation. In other words, we get a proportional constant of "1" if we use standard SI-units.

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    $\begingroup$ @schlumma You showed that the constant is dimensionless. Do you also mean that even the value of this dimensionless constant is '1'? $\endgroup$ Commented Sep 23, 2016 at 13:43
  • $\begingroup$ Yes, that's true. However the derivation of this property is not so easy, it can be done by using Gauge transformations. Maybe the following is helpful for you: In all other Maxwell equations (in SI units) you have proportional constants unequal to 1. But they're chosen in a way, that you have the "1" in Faraday's law. If you switch to another unit system (e.g. cgs), you'll get proportional constants in all Maxwell equations. $\endgroup$
    – schlunma
    Commented Sep 23, 2016 at 13:59
  • $\begingroup$ Thanks @schlumma and everyone. I will jump over Gauge transformation now! $\endgroup$ Commented Sep 23, 2016 at 14:03

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