Length element in Ampere's law

[Nway]

Homework Statement

Problem below

Relevant Equations

Problem below

Why is the length element (vector ds) perpendicular to the current?
I though length element should be parallel to current.

Would anybody be kind enough to help.

 

[TSny]

The title of your thread mentions the Biot-Savart law. But the solution in the textbook is using Ampere's law. See the section of the solution called "Categorize". Also, see the caption of Figure 29.13.

 

[Nway]

The title of your thread mentions the Biot-Savart law. But the solution in the textbook is using Ampere's law. See the section of the solution called "Categorize". Also, see the caption of Figure 29.13.

Edited.

 

[Nway]

The title of your thread mentions the Biot-Savart law. But the solution in the textbook is using Ampere's law. See the section of the solution called "Categorize". Also, see the caption of Figure 29.13.

I though length element should be parallel to current?

 

[TSny]

I just though length element should also be in direction of current.

Review the mathematical statement of Ampere's law. Be sure that you can state the meaning of all of the symbols that occur in the mathematical statement of the law. That should help you see why the vector $\vec{ds}$ does not have to be in the direction of the current.

 

[Nway]

Review the mathematical statement of Ampere's law. Be sure that you can state the meaning of all of the symbols that occur in the mathematical statement of the law. That should help you see why the vector $\vec{ds}$ does not have to be in the direction of the current.

Sir my textbook says that vector $\vec{ds}$ is just a small length element. Nothing to do with direction of current?

 

[SammyS]

Sir my textbook says that vector $\vec{ds}$ is just a small length element. Nothing to do with direction of current?

Please give your textbook's statement of Ampere's Law .

 

[Nway]

Please give your textbook's statement of Ampere's Law .

Ok I'll find it.

 

[Nway]

Please give your textbook's statement of Ampere's Law .

Where $\vec{ds}$ is equivalent notation to $\vec{dl}$

 

[TSny]

Your textbook uses $\vec{dl}$ for a length element. The textbook should provide some explanation of the meaning of the integral on the left side of Ampere's law. Hopefully, the book also has some figures that help illustrate the meaning of the integral. This video might help.

 

[kuruman]

Why is the length element (vector ds) perpendicular to the current?

Because the plane of the circle in which this element lies is perpendicular to the current. Any vector lying in that plane has, by definition, no component parallel to the current.

 

[Nway]

Because the plane of the circle in which this element lies is perpendicular to the current. Any vector lying in that plane has, by definition, no component parallel to the current.

Sorry sir. What do you mean the plane of current? I don't understand because when there is a current in a wire like here:

the vector $\vec{ds}$ is parallel to current. I know that the vector perpendicular to the current will have no component parallel to current. However, why is that vector not parallel to the current like in the image above?

 

[nasu]

The direction of the vector $d\vec{s}$ depends on the path of integration. It is not related to the current, in general. In your problem the integral is along a circular loop. The plane of this loop is perpendicular to the current. There was no mention of the "plane of the current". @kuruman wrote "the plane of the circle". The circle is this loop along which the integral is done. You can see this in the figure in the OP. The blue circle with dashed line. This is the integration path and $d\vec{s}$ is tangent to this path at any point.

 

[SammyS]

Sorry sir. What do you mean the plane of current? I don't understand because when there is a current in a wire like here:
View attachment 320993
the vector $\vec{ds}$ is parallel to current. I know that the vector perpendicular to the current will have no component parallel to current. However, why is that vector not parallel to the current like in the image above?

Look at what Ampere's law says. You are not integrating along the current carrying wire. The wire passes through an open surface. The line integral is along a (closed) path which is along the boundary of the open surface.

This is NOT the Biot–Savart law which has you integrate along the current carrying wire.

 

[Nway]

The direction of the vector $d\vec{s}$ depends on the path of integration. It is not related to the current, in general. In your problem the integral is along a circular loop. The plane of this loop is perpendicular to the current. There was no mention of the "plane of the current". @kuruman wrote "the plane of the circle". The circle is this loop along which the integral is done. You can see this in the figure in the OP. The blue circle with dashed line. This is the integration path and $d\vec{s}$ is tangent to this path at any point.

Thank you sir. I see now maybe.

So if $d\vec{s}$ for a straight wire will be in direction of current because I am integrating along wire, right?

 

[Nway]

Look at what Ampere's law says. You are not integrating along the current carrying wire. The wire passes through an open surface. The line integral is along a (closed) path which is along the boundary of the open surface.

This is NOT the Biot–Savart law which has you integrate along the current carrying wire.

Sir thank you. I assume that $d\vec{s}$ has same meaning for Ampere's law as with Biot–Savart law? I am now correct to say that $d\vec{s}$ gives direction of path at each differential point along a path.

However I feel confused about that $d\vec{s}$ would be tangent to circular path. Why must this be true?

 

[kuruman]

Sorry sir. What do you mean the plane of current? I don't understand because when there is a current in a wire like here:
View attachment 320993
the vector $\vec{ds}$ is parallel to current. I know that the vector perpendicular to the current will have no component parallel to current. However, why is that vector not parallel to the current like in the image above?

That's a different $d\vec s$ from the one you asked about in the original post. Look at your picture below that you yourself circled in yellow. It shows a circle defined by a blue dotted line and current $I$ with an arrow in the direction of the current. The current is perpendicular to the circle and to all vectors in the plane of that circle including the $d\vec s$ shown therein.

The generic directed length element $d\vec s$ is given to any path. It could be in the same direction as the current or perpendicular to the current. It could have no relation to current when you are writing the work done by a force along a path as $W=\int \vec{F}\cdot d\vec s.$

 

[Nway]

That's a different $d\vec s$ from the one you asked about in the original post. Look at your picture below that you yourself circled in yellow. It shows a circle defined by a blue dotted line and current $I$ with an arrow in the direction of the current. The current is perpendicular to the circle and to all vectors in the plane of that circle including the $d\vec s$ shown therein.

The generic directed length element $d\vec s$ is given to any path. It could be in the same direction as the current or perpendicular to the current. It could have no relation to current if you are writing the work done by a force as $W=\int \vec{F}\cdot d\vec s.$View attachment 320995

Thanks sir. So why dose $d\vec{s}$ have to be perpendicular to the current (in the plane of the circle)? Maybe a better restatement of my confusion: Is there a generalized definition for $d\vec{s}$ that is is Bio-Savart and Ampere's law?

The only definition for $d\vec{s}$ that I'm aware of is from the work equation. Where $d\vec{s}$ is the differential displacement due to a force $F$. However, I don't see that definition solve my E&M confusion.

 

[nasu]

It does not have to be in any way. It is up to you to choose the integration path according to the problem you try to solve. $d\vec{s}$ is the path element for any path, no matter what is the problem to be solved. You pick the integration path to solve the problem, usualy to solve it in the easiest way. You need to understand how the laws used (here Biot-Savart and Ampere) work in order to understand how do you choose the integration path. Once you do that, you don't have to make any choice about $d\vec{s}$. It is tangent to the chosen path.

 

[Nway]

It does not have to be in any way. It is up to you to choose the integration path according to the problem you try to solve. $d\vec{s}$ is the path element for any path, no matter what is the problem to be solved. You pick the integration path to solve the problem, usualy to solve it in the easiest way. You need to understand how the laws used (here Biot-Savart and Ampere) work in order to understand how do you choose the integration path. Once you do that, you don't have to make any choice about $d\vec{s}$. It is tangent to the chosen path.

Thank you very much sir. I still don't understand why $d\vec{s}$ is tangent to path. It is similar to how the velocity vector is tangent to a circular path in uniform circular motion?

 

[kuruman]

Thank you very much sir. I still don't understand why $d\vec{s}$ is tangent to path. It is similar to how the velocity vector is tangent to a circular path in uniform circular motion?

Yes, it is similar to that.

 

[Nway]

Yes, it is similar to that.

Sir thank you. What are the differences?

 

[nasu]

It is similar to how the velocity vector is tangent to a circular path in uniform circular motion?

The velocity is tangent to the path for any motion. Do you know the definition of the velocity vector?

 

[Nway]

The velocity is tangent to the path for any motion. Do you know the definition of the velocity vector?

Yes I do sir. Velocity vector is time integral of position vector.

I don't see how velocity vector (and therefore position vector) is tangent for straight line motion? I guess they are parallel for this special case.

 

[nasu]

No, the velocity is not the integral of position. And the position vector is not tangent to the trajectory. Your "therefore" does not make sense. Velocity and position vector don't have the same direction, in general. You need to review the basic notions about motion.

 

[Nway]

No, the velocity is not the integral of position. And the position vector is not tangent to the trajectory. Your "therefore" does not make sense. Velocity and position vector don't have the same direction, in general. You need to review the basic notions about motion.

Yep sorry sir I got it the wrong way round.

 

[nasu]

The velocity is $\vec{v}=\frac{d\vec{s}}{dt}$ so the two vectors ( $\vec{v}$ and $d\vec{s}$) must be along the same direction.

 

[Amraha]

I am not sure that rather Ampere's law relates to Bio-Savart Law or not, but both of these are helpful in finding magnetic field distributions. Moreover, Ampere's law takes symmetry into account!

 

[nasu]

Ampere' law per se does not take any symmetry into account. You may use symmetry to simplify the solution but the law itself does not require or imply any symmetry.

 

[SammyS]

Ampere' law per se does not take any symmetry into account. You may use symmetry to simplify the solution but the law itself does not require or imply any symmetry.

Yes, it is true that Ampere's law does not demand anything in the way of symmetry.

However, to use it in determining the magnetic field at a particular location generally requires the use of symmetry arguments.

 

[Nway]

It does not have to be in any way. It is up to you to choose the integration path according to the problem you try to solve. $d\vec{s}$ is the path element for any path, no matter what is the problem to be solved. You pick the integration path to solve the problem, usualy to solve it in the easiest way. You need to understand how the laws used (here Biot-Savart and Ampere) work in order to understand how do you choose the integration path. Once you do that, you don't have to make any choice about $d\vec{s}$. It is tangent to the chosen path.

Can $\vec{ds}$ also be though of as the displacement of a positive test using the right hand rule?

 

[nasu]

Can $\vec{ds}$ also be though of as the displacement of a positive test using the right hand rule?

I don't know what you mean by this.

 

[SammyS]

Can $\vec{ds}$ also be though of as the displacement of a positive test using the right hand rule?

No, $\displaystyle \vec{ds}$ cannot be thought of as the displacement of a positive test charge, using the right hand rule?

In using Ampere's law, you are integrating the magnetic field along a closed path.