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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.
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Edited.TSny said: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 said: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.
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 said:I just though length element should also be in direction of current.
Sir my textbook says that vector ##\vec{ds}## is just a small length element. Nothing to do with direction of current?TSny said: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.
Please give your textbook's statement of Ampere's Law .Nway said:Sir my textbook says that vector ##\vec{ds}## is just a small length element. Nothing to do with direction of current?
Ok I'll find it.SammyS said:Please give your textbook's statement of Ampere's Law .
SammyS said:Please give your textbook's statement of Ampere's Law .
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 said:Why is the length element (vector ds) perpendicular 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:kuruman said: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.
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.Nway said: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?
Thank you sir. I see now maybe.nasu said: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.
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.SammyS said: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.
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.Nway said: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?
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?kuruman said: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
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?nasu said: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.
Yes, it is similar to that.Nway said: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?
Sir thank you. What are the differences?kuruman said:Yes, it is similar to that.
The velocity is tangent to the path for any motion. Do you know the definition of the velocity vector?Nway said:It is similar to how the velocity vector is tangent to a circular path in uniform circular motion?
Yes I do sir. Velocity vector is time integral of position vector.nasu said:The velocity is tangent to the path for any motion. Do you know the definition of the velocity vector?
Yep sorry sir I got it the wrong way round.nasu said: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.
Yes, it is true that Ampere's law does not demand anything in the way of symmetry.nasu said: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.
Can ##\vec{ds}## also be though of as the displacement of a positive test using the right hand rule?nasu said: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.
I don't know what you mean by this.Nway said: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?Nway said:Can ##\vec{ds}## also be though of as the displacement of a positive test using the right hand rule?