How is Faraday's Law true if we cannot define a potential energy corresponding to a non conservative field?

Faradays Law states that, $$EMF = -\frac{d\phi}{dt}$$ where the EMF is the due to the induced E-field. But, this induced E-field is non conservative. We cannot define a potential energy corresponding to a non conservative field. How then, can we define potential and hence EMF?

• What do you think $\phi$ is in your equation? Commented Aug 13, 2018 at 6:55
• The magnetic flux? Commented Aug 13, 2018 at 16:10
• Potential is a change in work (per unit charge). \int E \cdot dl (left side of Faraday's law) is the work done, so this is your potential. Commented Apr 18, 2022 at 14:55

The EMF is simply the electric field integrated around the loop:

$$\operatorname{EMF}=-\oint\vec E\cdot d\vec l$$

The electric field can be written in two parts, in terms of the scalar potential $V$ and the vector potential $\vec A$.

$$\vec E=-\nabla V-{\partial\vec A\over\partial t}$$

If you plug this into the integral above, it's easy to show that the first part vanishes. That is, the EMF has units of voltage, but it has nothing to do with the conservative scalar potential $V$.

Faraday's law states $$\varepsilon=-\frac{d\Phi}{dt}$$ where $\Phi$ is called $\textbf{magnetic flux}$ (is not a potential). It is defined by a flux integral $$\Phi=\int_{\Sigma} \,\textbf{B} \bullet\textbf{da}$$ Using the definition of electromotive force we obtain the "integral form" of Faraday's law $$\oint_{\gamma}\, \textbf{E}\bullet\textbf{dl}=\int_{\Sigma} \,\frac{\partial}{\partial t}\textbf{B} \bullet\textbf{da}$$ And using Stokes' Theorem we obtain the "differential form" $$\nabla\times\textbf{E}=-\frac{\partial\textbf{B}}{\partial t}$$

• Maybe worth mentioned that $\varepsilon$ is not a vector field? Commented Aug 13, 2018 at 5:26
• i meant that the induced electric field, is non conservative, how can we define a potential or EMF onding to a non conservative field? Commented Aug 13, 2018 at 16:09
• I'm aware that $\phi$ is the magnetic flux, not a potential. Commented Aug 13, 2018 at 16:10
• Since $\nabla\cdot\textbf{B}=0$ we can always write $\textbf{B}=\nabla\times\textbf{A}$, where $\textbf{A}$ is called "the vector potential". Putting this in Faraday's law we can write $\textbf{E}=-\nabla V-\frac{\partial\textbf{A}}{\partial t}$ as you can check. Also you can check chapter 10 of Griffiths for more details. Commented Aug 14, 2018 at 6:24

The $EM F$ is a potential difference, therefore, it remains the same for any gauge.

$$EM F=V_1-V_2=(V_1+k)-(V_2+k)$$

for any $k \in \mathbb{R}$.

• This doesn't answer the question.
– Chris
Commented Aug 13, 2018 at 12:31
• Doesn't the question ask why the we can define a $EM F$ when we can't define a potential? Commented Aug 13, 2018 at 12:37
• The question has nothing to do with gauges. A non-conservative field can't have a potential written for it, regardless of gauge.
– Chris
Commented Aug 13, 2018 at 14:39