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In summary, the potential difference between any two points on a ring or rectangle moving into a static magnetic field is not always zero.f

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- #2

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[tex]E=-\nabla \phi-\frac{\partial A}{\partial t}[/tex]

EMF comes from the second term which corresponds to time varying magnetic field.

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So in case of motional emf where the magnetic field is static, is the potential difference between any two points zero when current is going?

[tex]E=-\nabla \phi-\frac{\partial A}{\partial t}[/tex]

EMF comes from the second term which corresponds to time varying magnetic field.

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Let me know an example setting of "motional emf where the magnetic field is static” in your mind for investigation.So in case of motional emf where the magnetic field is static, is the potential difference between any two points zero when current is going?

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A rectangular conducting loop moving into a region of a static uniform magnetic field, cuasing flux to increase in time and producing an emf, called motional emf. There isn't any role for an electric field in the plot, but there is a current going, so can I just say the potential difference between any two points on the loop is zero, and the magnetic force is what causes the emf, which can't be equated to a potential difference?Let me know an example setting of "motional emf where the magnetic field is static” in your mind for investigation.

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Fig.17-6 of https://www.feynmanlectures.caltech.edu/II_17.html seems to be similar though ring not rectangular. Feynman analyses it by Flux law. Magnetic flux changes through the loop changes by motion of the loop. Other cases he explains in that chapter are also very interesting.A rectangular conducting loop moving into a region of a static uniform magnetic field, cuasing flux to increase in time and producing an emf,

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I understand the flux law but it doesn't answer my question, that is, is the potential difference always zero between any two points on the ring/rectangle?Fig.17-6 of https://www.feynmanlectures.caltech.edu/II_17.html seems to be similar though ring not rectangular. Feynman analyses it by Flux law. Magnetic flux changes through the loop changes by motion of the loop. Other cases he explains in that chapter are also very interesting.

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It is zero. E comes from ##\phi## and A (as posted #2). In this configuration ##\phi## has no contribution to E.

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Depends upon your definition. I would say the emf depends upon your path between the the points and your voltmeter will often not measure zero .is the potential difference always zero between any two points on the ring/rectangle?

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$$\vec{\nabla} \times \vec{E} = -\partial_t \vec{B}.$$

You can always use Stokes's theorem for an arbitrary surface ##A## with boundary ##\partial_A## (and the surface and its boundary can be time dependent)

$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$

Now when you want to take the time-derivative out of the surface integral you must take into account the change of the surface with time (if the surface is moving like in your example where you integrate along a wire loop moving into the magnetic field). The calculation then leads to

$$\mathcal{E} = \int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B}) = -\frac{\mathrm{d}}{\mathrm{d} t} \int_{A} \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$

Here, ##\vec{v}(t,\vec{r})## is the velocity field along the surface's boundary. The derivation of this complete version of Faraday's Law including moving areas/boundaries can be found in Wikipedia:

https://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof

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