- #1

Ahmed1029

- 109

- 40

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Ahmed1029
- Start date

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

- #1

Ahmed1029

- 109

- 40

Physics news on Phys.org

- #2

anuttarasammyak

Gold Member

- 2,561

- 1,368

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

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

- #3

Ahmed1029

- 109

- 40

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 said:

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

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

- #4

anuttarasammyak

Gold Member

- 2,561

- 1,368

Let me know an example setting of "motional emf where the magnetic field is static” in your mind for investigation.Ahmed1029 said: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?

- #5

Ahmed1029

- 109

- 40

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

- #6

anuttarasammyak

Gold Member

- 2,561

- 1,368

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

- #7

Ahmed1029

- 109

- 40

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 said: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.

- #8

anuttarasammyak

Gold Member

- 2,561

- 1,368

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

Last edited:

- #9

hutchphd

Science Advisor

Homework Helper

- 6,716

- 5,779

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 .Ahmed1029 said:is the potential difference always zero between any two points on the ring/rectangle?

- #10

- 24,488

- 15,026

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

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

Electromotive force (EMF) is the energy per unit charge that is supplied by a source, such as a battery, to move charge through a circuit. Potential difference, on the other hand, is the difference in electrical potential between two points in a circuit. In simpler terms, EMF is the driving force that causes current to flow, while potential difference is the measure of the energy used by the current as it flows through the circuit.

No, electromotive force is not always equal to potential difference. EMF is the maximum potential difference that can be produced by a source, while potential difference can vary depending on the resistance and other factors in a circuit. In ideal conditions, EMF and potential difference may be equal, but in real-world circuits, they can differ.

Electromotive force and potential difference are related by Ohm's law, which states that the potential difference across a conductor is equal to the current flowing through it multiplied by its resistance. In other words, EMF is equal to the potential difference when there is no resistance in the circuit.

Yes, potential difference can be negative. This occurs when the direction of current flow is opposite to the direction of the electric field. In this case, the potential energy of the charge decreases as it moves through the circuit, resulting in a negative potential difference.

The equality of EMF and potential difference can be affected by factors such as the resistance of the circuit, the type of source providing the EMF, and the type of load connected to the circuit. Other factors such as temperature, material properties, and circuit configuration can also play a role in determining the relationship between EMF and potential difference.

- Replies
- 58

- Views
- 1K

- Replies
- 10

- Views
- 2K

- Replies
- 25

- Views
- 1K

- Replies
- 10

- Views
- 716

- Replies
- 17

- Views
- 2K

- Replies
- 7

- Views
- 1K

- Replies
- 5

- Views
- 2K

- Replies
- 20

- Views
- 2K

- Replies
- 4

- Views
- 1K

- Replies
- 6

- Views
- 981

Share: