# Hall effect over a conducting ring

• Trisztan
In summary: It goes into more detail about the Hall effect. In summary, the ring is made of conducting material. I was originally asked to find the potential difference between ##a## and ##b##. I did so using the Hall effect (and assuming it would work as per normal in this situation). This got me ##\Delta V = vBl##. However, I am now lost on how to find the "voltage around the ring". If I had to guess, I would say its 0 everywhere around the ring except at either ##a## and ##b## (depending on which you take to be the point where ##V=0##). Any help would be much appreciated.
Trisztan
Homework Statement
What is the voltage around a conducting ring of diameter ##l##, moving through a uniform magnetic field of magnitude ##B##, at speed ##v##?
Relevant Equations
Definition of potential difference:
$$\Delta V = -\int_i^f {\mathbf{E}\cdot \mathrm{d}\mathbf{s}}$$
Force magnitude equivalence under the Hall effect:
$$qE = qvB$$
This is the diagram provided in the question:

The ring is made of conducting material. I was originally asked to find the potential difference between ##a## and ##b##. I did so using the Hall effect (and assuming it would work as per normal in this situation). This got me ##\Delta V = vBl##.

However, I am now lost on how to find the "voltage around the ring". If I had to guess, I would say its 0 everywhere around the ring except at either ##a## and ##b## (depending on which you take to be the point where ##V=0##).

Any help would be much appreciated.

You need to do a line integral $$\int_a^b \mathbf{E}\cdot d\mathbf{s}=\int_a^b (\mathbf{v}\times \mathbf{B})\cdot d\mathbf{s}$$ where ##d\mathbf{s}## is a line element along the circular path.

@kuruman Sorry, but I don't understand how that will help me. Wouldn't that just give me ##-\Delta V## between ##a## and ##b##? I already have that, don't I?

It will give you the integral around the semicircle which is what (I think) the question is asking you to find. You wrote that you didn't know how to do it and I explained that to you. Maybe someone else has another answer that is more to your liking. Stick around.

BTW what you call the Hall effect is more often called "motional emf". Here is a brief description of the Hall effect.

berkeman
@kuruman No need to get so defensive man, your answer was perfectly to my liking. I simply didn't understand exactly what you were saying.

But I see now why you want to take the line integral around the semi-circle; that should probably be it. Thank you.

And yeah the link you mentioned is what I was referring to.

## What is the Hall effect?

The Hall effect is a phenomenon in which an electric current flowing through a conductor produces a transverse voltage difference perpendicular to the direction of the current. This effect was discovered by Edwin Hall in 1879.

## What is a conducting ring?

A conducting ring is a circular loop of material, typically made of metal, that allows electric current to flow through it. It is often used in experiments to demonstrate the Hall effect.

## How does the Hall effect work over a conducting ring?

When an electric current is passed through a conducting ring, a magnetic field is created around the ring. This magnetic field interacts with the moving charges in the current, causing them to experience a force perpendicular to both the direction of the current and the magnetic field. This force results in a voltage difference across the ring, which is known as the Hall voltage.

## What is the significance of the Hall effect over a conducting ring?

The Hall effect over a conducting ring is an important phenomenon in physics and is used in many practical applications, such as measuring the strength of magnetic fields and determining the type of charge carriers in a material.

## What factors affect the Hall effect over a conducting ring?

The Hall effect over a conducting ring is influenced by several factors, including the strength of the magnetic field, the amount of current flowing through the ring, and the type of material used for the ring. Additionally, the dimensions and shape of the ring can also affect the magnitude of the Hall voltage.

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