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For instance, the divergence of the magnetic field has units but it's equated to zero?

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  • $\begingroup$ It's a dot product; not a regular multiplication. There will be terms that add and subtract and those must equal zero for a closed surface. $\endgroup$ – JMac Jan 27 '17 at 12:59
  • $\begingroup$ How does that change anything? Even scalars have units. $\endgroup$ – Derby Moose Jan 27 '17 at 13:00
  • $\begingroup$ And dot product is only in the 2 divergence terms $\endgroup$ – Derby Moose Jan 27 '17 at 13:00
  • $\begingroup$ You don't get a single term from the equation. You get multiple terms with differing signs that can cancel out. You can have it happen with the continuity equations for fluids as well. It basically satisfies that the total going in is the total going out, or the total flux is 0. $\endgroup$ – JMac Jan 27 '17 at 13:03
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    $\begingroup$ Zero miter ,zero Kg,zero Newton .There are units after the zero $\endgroup$ – Lapmid Jan 27 '17 at 15:18
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How can Maxwell's equations be correct if the terms have different units on both sides of the equation?

Your premise is completely incorrect. Maxwell's equations always have the same units on both sides of the equation. There's two components two this:

  • Three of Maxwell's equations are traditionally phrased as equalities between dimensional quantities:

    • $\nabla \times \mathbf E = -\frac{\partial\mathbf B}{\partial t}$ has units of $\mathrm{V/m^2}$ on the left and $\mathrm{T/s}$ on the right, and both of those units coincide.
    • $\nabla \times \mathbf B = \mu_0\mathbf J + \mu_0\varepsilon_0 \frac{\partial\mathbf E}{\partial t}$ has units of $\mathrm{T/m}$ on the left and $\mathrm{N/Am^2}$ or $\mathrm{V\:s/m^3}$ on the right, and all of those units coincide.
    • $\nabla \cdot \mathbf E = \frac{1}{\varepsilon_0}\rho$ has units of $\mathrm{V/m^2}$ on the left and $\mathrm{C/F\:m^2}$ on the right, and both of those units coincide.
  • The fourth Maxwell equation, the magnetic divergence law $\nabla \cdot\mathbf B=0$, is perfectly consistent, because zero is zero regardless of what units it is applied to.* If you will you can see the right-hand-side of that equation as $0\:\mathrm{T/m}$, but really, you gain nothing by that.

    For more on the physical dimensionality of zero, see Is $0\,\mathrm m$ dimensionless? and Should zero be followed by units?, and the many linked questions on the right-hand sidebar from those two.

* unless you're talking about pathological cases like degrees Celsius or Fahrenheit, which are obviously excluded here.

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  • $\begingroup$ zero is zero regardless of what units it is applied to unless the unit is degree Celsius or some similarly shifted one. $\endgroup$ – Ruslan Jan 27 '17 at 14:59
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    $\begingroup$ @Ruslan That's a fairly pedantic nitpick, to be frank, but if you insist then I'll add it in. $\endgroup$ – Emilio Pisanty Jan 27 '17 at 16:06
  • $\begingroup$ Another exception is $o$ is not the same as $\vec{0}$, ie the number of dimensions can matter. $\endgroup$ – Sean E. Lake Jan 27 '17 at 16:14
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    $\begingroup$ @SeanLake That's yet another case that's irrelevant as far as I'm concerned. While in principle that's the case, there are absolutely no situations where using the same symbol for zeroes in all vector spaces will land you in trouble. $\endgroup$ – Emilio Pisanty Jan 27 '17 at 16:16
  • $\begingroup$ "There are absolutely no situations where using the same symbol for zeroes in all vector spaces will land you in trouble." I feel 100% sure this is false, but am insufficiently motivated to construct a counterexample. $\endgroup$ – WillO Jan 27 '17 at 17:17

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