Testwiki:Reference desk/Archives/Science/2022 December 10

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December 10

Maxwell's third and fourth equations

Using Formulation in SI units convention and ignoring the conventional current term of Ampere-Maxwell equation (focusing solely on the displacement current term), we have: $\begin{array}{cc}{\oint }_{\partial \mathrm{\Sigma }}& 𝐁\cdot \mathrm{d}\mathit{\ell }=={\mu }_0{\epsilon }_0\frac{\mathrm{d}}{\mathrm{d}t}\underset{\mathrm{\Sigma }}{\iint }𝐄\cdot \mathrm{d}𝐒\\ \end{array}$

So my question is: why is the Maxwell-Faraday equation in the following form: ${\oint }_{\partial \mathrm{\Sigma }}𝐄\cdot \mathrm{d}\mathit{\ell }=-\frac{\mathrm{d}}{\mathrm{d}t}\underset{\mathrm{\Sigma }}{\iint }𝐁\cdot \mathrm{d}𝐒$ and not in the following form: ${\oint }_{\partial \mathrm{\Sigma }}𝐄\cdot \mathrm{d}\mathit{\ell }=-{\mu }_0{\epsilon }_0\frac{\mathrm{d}}{\mathrm{d}t}\underset{\mathrm{\Sigma }}{\iint }𝐁\cdot \mathrm{d}𝐒$ ?

Many thanks, 173.209.130.10 (talk) 21:29, 10 December 2022 (UTC)

The units wouldn't match. The SI unit of the electric field is $[E]=1\frac{N}{As}$, of the magnetic field $[B]=1\frac{N}{Am}$. With that the unit of the left hand side is $\frac{N}{As}\mathrm{\cdot }\mathrm{m}=\frac{Nm}{As}$, of the right hand side it is $\frac{1}{s}\frac{N}{Am}\mathrm{\cdot }{\mathrm{m}}^{\mathrm{2}}=\frac{Nm}{As}$, i.e. the same, as it must be. The factor ${\epsilon }_0{\mu }_0=1/c^2$ with unit $\frac{s^2}{m^2}$, and if you insert that as you propose, you mess up the units of the right hand side. --Wrongfilter (talk) 21:47, 10 December 2022 (UTC)

Using natural units in which $c=1,$ the identities take on a more symmetric appearance.  --Lambiam 05:13, 11 December 2022 (UTC)