Is electromotive force always equal to potential difference?

[Ahmed1029]

In the case motional emf, there is a static magnetic field and a rectulgular loop that goes into the field region, then current is produced. There is no electric field, but there is an emf. However, Griffiths states that emf is equal to the potential difference between the source endpoints. But here, the potential difference is zero since E field is zero, while the emf is not. What am I getting wrong?

 

[anuttarasammyak]

Electric field is given by
$$E=-\nabla \phi-\frac{\partial A}{\partial t}$$
EMF comes from the second term which corresponds to time varying magnetic field.

 

[Ahmed1029]

Electric field is given by
$$E=-\nabla \phi-\frac{\partial A}{\partial t}$$
EMF comes from the second term which corresponds to time varying magnetic field.

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?

 

[anuttarasammyak]

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?

Let me know an example setting of "motional emf where the magnetic field is static” in your mind for investigation.

 

[Ahmed1029]

Let me know an example setting of "motional emf where the magnetic field is static” in your mind for investigation.

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?

 

[anuttarasammyak]

A rectangular conducting loop moving into a region of a static uniform magnetic field, cuasing flux to increase in time and producing an emf,

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.

 

[Ahmed1029]

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.

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?

 

[anuttarasammyak]

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

 

[hutchphd]

No discussion of this is complete without a nod to Prof Walter Lewin.

is the potential difference always zero between any two points on the ring/rectangle?

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 .

 

[vanhees71]

The fundamental Maxwell equations are the differential ones. Faraday's law is (in SI units)
$$\vec{\nabla} \times \vec{E} = -\partial_t \vec{B}.$$
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