Thread: conductor and magnet problem special relativity

1. In the conductor magnet problem that start a basis for special relativity, when the conductor is moving while the magnet is stationary, it is considered that the emf is equal to q B x V. but when the conductor is moving towards the pole of the magnet both B and V are in the same direction making the result equals to zero. This is not the case when the magnet is moving towards the conductor. Can somebody explain please.

3. Whether an electric field or a magnetic field is observed is dependent on the observers motion, relative to the charge. So if I am stationary to a charge I will observe an electric field. As I start to move slowly relative to the charge I will observe slightly less electric field and a bit of a magnetic field. As I move faster relative to the charge I will observe less electric field and more magnetic field. It's quite amazing.... The only difference between electric and magnetic fields is the relative motion of the observer!

4. Could be. original consideration of Einstein in his paper. If you refer to this http://www.sigmapisigma.org/radiatio...ics_fall05.pdf

The diagram shows when the observer is at rest with magnet, the emf force = q B x V
Since B and V are in the same direction B x V has to be zero.

5. No "could be" about it. It's the way it is....

6. When a conductor move towards the magnet, many literature say that the force on a charge as q VxB. But since the conductor is moving towards the magnet and the magnetic field is in the same direction V x B should be zero. This is so obvious but many literature,when explaining Einstein's conductor and magnet problem uses V x B?

Would like to know some insight. Thanks

7. Originally Posted by Harsha Kumar
When a conductor move towards the magnet, many literature say that the force on a charge as q VxB. But since the conductor is moving towards the magnet and the magnetic field is in the same direction V x B should be zero. This is so obvious but many literature,when explaining Einstein's conductor and magnet problem uses V x B?

Would like to know some insight. Thanks

9. Originally Posted by Harsha Kumar
I tried to imagine what your actual question might be since your original makes no sense. But I guess we have to start from what you actually wrote.

Here's where there's a problem:

Originally Posted by HK
since the conductor is moving towards the magnet and the magnetic field is in the same direction...
Why are you assuming that the magnetic field must be in the same direction? You are working hard to invent a paradox where none exists.

The Lorentz force law tells us that the magnetic force is indeed qv X B. If v and B are fully collinear, then there will be no force on the test charge. But if v and B have any orthogonal component, then there will be a force.

So, carefully plot v and B; don't just make some random (and logically self-inconsistent) assumption. Then compute the vector cross-product v X B. Multiply by q. That's the Lorentz force.

It's that simple.

10. tk421: Thank you for answering.

I tried to draw a diagram here, but could not.

My question is when a magnet(assume a bar magnet) and conductor (a closed loop) moves towards (or away from) each other in such a way that a magnetic pole is getting closer/away to/from the loop. Then the flux inside the loop changes creating a emf in the conductor loop. From Maxwell Faraday equation this can be calculated. However from the magnet frame, I have seen many are using the q V x B to find the force on the charges on the loop. But since the direction of magnetic field is away from it's pole and the conductor loop is moving in the same line, B x V should be zero.

Where have I mistaken?

For example if you see the first figure of the following web page you will see what bothers me.

Thanks

11. Originally Posted by Harsha Kumar

I tried to draw a diagram here, but could not.
Unfortunately, the forum software seems not to permit the attachment of pictures.

My question is when a magnet(assume a bar magnet) and conductor (a closed loop) moves towards (or away from) each other in such a way that a magnetic pole is getting closer/away to/from the loop. Then the flux inside the loop changes creating a emf in the conductor loop. From Maxwell Faraday equation this can be calculated. However from the magnet frame, I have seen many are using the q V x B to find the force on the charges on the loop. But since the direction of magnetic field is away from it's pole and the conductor loop is moving in the same line, B x V should be zero.

Where have I mistaken?

For example if you see the first figure of the following web page you will see what bothers me.

Thanks
Ah, yes, I think I now understand where your problem lies (but you will be the final judge of that!). You appear to be assuming that the magnet produces field lines that are perfectly straight throughout all space. If that were the case (it isn't), then B and v would be perfectly collinear, and no Lorentz force would act.

However, the field lines actually form closed loops (think about the classical pictures of field maps formed with iron filings). The curvature that is necessarily implied by that closure assures the existence of components of B that are orthogonal to v, thus producing a nonzero v X B.

I hope that helps!

12. Thanks tk421

"The curvature that is necessarily implied by that closure assures the existence of components of B that are orthogonal to v...." is the part that i had not noticed before. I learned it from you. Thanks again. However, is there any proof available for that?

Actually, I am devicing a theory that is different to relativity. Already 99% completed the book. However this part was keeping me from proceeding. Truly thankful to you for the lesson. (you may refer EMS Theory - Home, Pl still being developed)

13. Originally Posted by Harsha Kumar
Thanks tk421

"The curvature that is necessarily implied by that closure assures the existence of components of B that are orthogonal to v...." is the part that i had not noticed before. I learned it from you. Thanks again. However, is there any proof available for that?
Ultimately it comes down to the fact that the divergence of B is zero ( in words, there are no magnetic monopoles). A bar magnet thus produces a dipole field; field lines emanating from the north pole eventually find their way to the south pole. There's no geometric way to produce the implied closed paths without curving somewhere.

14. Thank you.

Can you recommend some literature where I can find the proof for above? (proof for orthogonal magnetic field component equaling B)

15. Originally Posted by Harsha Kumar
Thank you.

Can you recommend some literature where I can find the proof for above? (proof for orthogonal magnetic field component equaling B)
The outline of the proof I offered is not to demonstrate the "orthogonal magnetic field componenet equaling B" (sic). Rather, it is to demonstrate that there must exist a component of B that points at some angle not along the vector v. I'm not sure if that's what you meant, but it's what I meant.

16. MODERATOR ACTION: I merged the two threads on the same topic in different forums into this one thread.

17. Originally Posted by Harsha Kumar
But since the direction of magnetic field is away from it's pole and the conductor loop is moving in the same line, B x V should be zero. Where have I mistaken?
Look at fig 1 in the paper, you can see that does NOT have a constant direction and it curves AWAY from the magnet pole such that it is NOT parallel with . Do you see that?

For example if you see the first figure of the following web page you will see what bothers me.

Thanks
Look at fig 1 in the paper, you can see that does NOT have a constant direction and it curves AWAY from the magnet pole such that it is NOT parallel with , so . Do you see that?

18. The issue is that B is not constant, so an electric filed is generated by the dB/dt term by Faraday's law of induction as you move through it. As X0X said it is to do with the nature do the magnetic field.

19. OK. There is a flux cutting due to curvature of field lines. But how can it be equal to exactly B. B is the value that is used for the conductor frame, in Faraday maxwell equation. Edl = -d/dt int( Bds). B here is the average field strength cutting through a surface S (mostly in the direction of "magnet pole - center of conductor loop" line). How can it be the same B in the orthogonal direction as well.

20. Thanks

21. Originally Posted by Harsha Kumar
OK. There is a flux cutting due to curvature of field lines. But how can it be equal to exactly B. B is the value that is used for the conductor frame, in Faraday maxwell equation. Edl = -d/dt int( Bds). B here is the average field strength cutting through a surface S (mostly in the direction of "magnet pole - center of conductor loop" line). How can it be the same B in the orthogonal direction as well.
Nope, the vector which is used in the calculation of the Lorentz force , is tangent to the magnetic field force lines. Inside the magnet bar is aligned with the magnet bar so , outside the magnet bar is tangent to the magnetic field force lines that curve from the magnet bar, so . It is puzzling to see the famous rpenner stumbling on this question and piling up irrelevant answer on top of irrelevant answer to this simple question. It is clear that he doesn't understand the question.

22. It is puzzling to see the famous rpenner stumbling on

I don't see any answer from "rpenner"??

XoX, I think you have not got my question (may be the second question, hidden in the first one). I am asking how could that tangent field quantitatively be equal to "B", which is the B that is used in conductor frame. (Edl = -d/dt int(Bds))

23. Originally Posted by Harsha Kumar
It is puzzling to see the famous rpenner stumbling on

I don't see any answer from "rpenner"??
Click on the link. You are asking the same thing in the other forum but you aren't getting an answer.

XoX, I think you have not got my question (may be the second question, hidden in the first one).
I got the question, did you get the answer? Yes or no?

I am asking how could that tangent field quantitatively be equal to "B", which is the B that is used in conductor frame. (Edl = -d/dt int(Bds))
B is the magnetic induction produced by the permanent magnet.

24. Harsha, if B x v is zero the force will be zero. Are you you sure that you are comparing the same values of B? In one case B is at a point, in the other case is integrated over a surface. Why would you think that they are the same?

25. Jilan "if B x v is zero the force will be zero. Are you you sure that you are comparing the same values of B? In one case B is at a point, in the other case is integrated over a surface. Why would you think that they are the same?"

I am comparing the forces between the two frames, Conductor frame and magnet frame. They have to have equal forces.

Furthermore, my original question (which was written so badly so difficult to visualise), is, if the moving conductor (magnet frame) which is moving towards the pole of bar magnet at velocity V, and if the magnetic field is B (this is the magnetic field going across the loop, away from the magnet, parallel to surface vector of the surface enclosed (assume flat surface) by the loop), then people say the force is F = q V x B. If this is to be correct then the component of magnetic field parallel to surface (perpendicular to surface vector) at the rim (loop) should also be equal to B.

So at the points of the loop magnetic field should be such that it's components in the direction of the velocity of the loop and perpendicular to the velocity of loop both has to be equal to B.

I am wondering!!

26. to XOX : Yes I saw your answer. But my question is how do the components of curving magnetic field lines give same strength B in the perpendicular direction to the movement of conductor loop. Otherwise two forces in the magnet frame and the conductor frame cannot be equal.

27. Originally Posted by Harsha Kumar
In the conductor magnet problem that start a basis for special relativity, when the conductor is moving while the magnet is stationary, it is considered that the emf is equal to q B x V.
First let's be precise about this. You have the emf as a vector when in fact it's really a scalar defined as

where f is the sum of the driving force (e.g. a battery) and the electrostatic force. What you're referring to is motional emf which is related to Faraday's Law of Induction. The motional emf (Motional EMF) in this case is

where

Originally Posted by Harsha Kumar
.. but when the conductor is moving towards the pole of the magnet both B and V are in the same direction making the result equals to zero. This is not the case when the magnet is moving towards the conductor. Can somebody explain please.
That's too simplified since you seem to be thinking of a bar magnet and the field there is a bit complex and you haven't told us either the shape or orientation of the conductor. What it all ends up relating to is the cross product "x" between the velocity vector and the magnetic field vector. When they're parallel then the cross product vanishes, i.e. vxB = 0. When they're perpendicular the cross product doesn't vanish. See diagram here Motional Emf

28. Originally Posted by Harsha Kumar
to XOX : Yes I saw your answer. But my question is how do the components of curving magnetic field lines give same strength B in the perpendicular direction to the movement of conductor loop. Otherwise two forces in the magnet frame and the conductor frame cannot be equal.

I read all i get. But i think some I dont get. Like Rpenner. I am yet to familiarize with the forum.

I didn't see any post explaining how that "B" used in magnet frame and "B" used in conductor frame are quantitatively equal. I am discussing only the case where a bar magnet is moving into the conductor ring(loop). In the conductor frame Farady-Maxwell equation is used, in which "B" is perpendicular to the conductor loop plane. In the magnet frame, Lorentz force is used in which "B" is parallel to conductor loop plane.

These two "B" s have to be numerically equal for forces in magnet frame and conductor frame to be equal. I have not yet found any proof of making B parallel equal to B perpendicular.

30. Originally Posted by Harsha Kumar
I read all i get. But i think some I dont get. Like Rpenner. I am yet to familiarize with the forum.

I didn't see any post explaining how that "B" used in magnet frame and "B" used in conductor frame are quantitatively equal.
I see. based on your question it has become quite clear that you do not understand the Lorentz force. Let me try to make it quite clear to you:

the in the law is measured in the magnet frame. The whole exercise has NOTHING to do with "B used in the conductor frame". For a very simple reason, the emf is induced due to the MOTION of the conductor, at velocity with respect to....the magnet.

I am discussing only the case where a bar magnet is moving into the conductor ring(loop). In the conductor frame Farady-Maxwell equation is used, in which "B" is perpendicular to the conductor loop plane. In the magnet frame, Lorentz force is used in which "B" is parallel to conductor loop plane.
At this point I suggest that you try to understand what I posted above.

These two "B" s have to be numerically equal for forces in magnet frame and conductor frame to be equal. I have not yet found any proof of making B parallel equal to B perpendicular.
This is because you do not understand Lorentz law. I hope that after the above explanation, you will and you will stop asking questions that make no sense. There is no answer to questions that are based on misconceptions.

31. KJW, is there any way that we can draw diagrams in this forum? I don't think my writing has made it easy for most here to understand what I am asking.

I would make their lives easier if a diagram can be included.

32. Originally Posted by Harsha Kumar
I read all i get. But i think some I dont get. Like Rpenner. I am yet to familiarize with the forum.

I didn't see any post explaining how that "B" used in magnet frame and "B" used in conductor frame are quantitatively equal. I am discussing only the case where a bar magnet is moving into the conductor ring(loop). In the conductor frame Farady-Maxwell equation is used, in which "B" is perpendicular to the conductor loop plane. In the magnet frame, Lorentz force is used in which "B" is parallel to conductor loop plane.

These two "B" s have to be numerically equal for forces in magnet frame and conductor frame to be equal. I have not yet found any proof of making B parallel equal to B perpendicular.
Did you read and understand my post? If so then do you have any comments on it? Did it help you at all? If not then why?

33. Pysicist - Admiring your valuable effort to give some clue to clear my understanding, i may say that your answer is not exactly for my question. Let me clarify starting from the web page you refered to me - Motional Emf

Just after the eq (205)it says,

" where BT = Bcos is the component of the magnetic field which is perpendicular to the plane of the circuit.

The above sentence has to be changed to …. component of magnetic field which is perpendicular to the motion of the conductor
======================================
Now let us go to Equation (207)

What are the directions of v and B in eq (207). Surely perpendicular to each other! Therefore the magnetic field has to be perpendicular to the motion of the conductor.

In my question, in the magnet frame, when the conductor loop moves towards the bar magnet (As if a ring is moving to fit around a finger - plane of the conductor loop is perpendicular to its velocity), If we use B to find E0 , it has to be perpendicular to the direction of the velocity of the conductor. This is possible since the magnetic field lines do curve, and the component perpendicular to the velocity of the conductor should be quantitatively equal to “B”. Shall we call this Bcp. CP stands for conductor plane as this B is parallel to conductor plane, since conductor loop is moving perpendicular to its own plane.

However, in the conductor frame, conductor loop is not moving. What we get is change of magnetic flux occurring in side the loop. This change creates an electric force equal to qE, where E = vB. B here is now perpendicular to the close surface surrounded by the loop (we use Bds). If we assume the flat surface this B is parallel to the velocity of the conductor loop. Let us say this Bpcp, Pcp stands for perpendicular to conductor plane.

We however know that the value of the emf has to be same for both conductor and magnet frames. Which means Bcp shoul be equal to Bpcp.

My question from the beginning has been, how this Bcp = Bpcp. Can you show me a mathematical proof that it is so.

34. Physicist :a diagram similar to my question is available in fig 35 of Magnetic Induction - Magnetic induction. You would see Bpcp marked there, perpendicular to the conductor loop plane.

35. Originally Posted by Harsha Kumar
My question from the beginning has been, how this Bcp = Bpcp. Can you show me a mathematical proof that it is so.
Absolutely. However it will take me a minute or two to absorb what you've posted here and another few minutes to think about it and write a response. Right now I don't have the time but will get back to you later today.

Cheers!

36. Originally Posted by Harsha Kumar
Physicist :a diagram similar to my question is available in fig 35 of Magnetic Induction - Magnetic induction. You would see Bpcp marked there, perpendicular to the conductor loop plane.
Are you studying the EMF created by time varying fields or motional EMF, i.e. the EMF generated by the motion of conductors through magnetic fields?

37. Originally Posted by Harsha Kumar
Physicist :a diagram similar to my question is available in fig 35 of Magnetic Induction - Magnetic induction. You would see Bpcp marked there, perpendicular to the conductor loop plane.
I have explained this to you several times, let's try one last time,maybe you'll get it. Look at your fig.35. The velocity of the loop, is parallel with the magnetic induction ONLY IN THE CENTER of the loop. Everywhere else, and ARE NOT PARALLEL. Can you see that? Still no?

38. X0X, I dont thinks that's the issue. The question seems to be how to prove that the force is same in both frames of reference. It might be related to the fact that div B =0 and that the curl of a divergence is also zero. Perhaps? ......

39. Originally Posted by Jilan
X0X, I dont thinks that's the issue. The question seems to be how to prove that the force is same in both frames of reference.
Force cannot be the same in both frames of reference since force is frame variant in SR.

It might be related to the fact that div B =0 and that the curl of a divergence is also zero. Perhaps? ......
Nope.

40. OK Abbas there you have it. Force as well as energy is frame dependent. You cannot expect them to be the same. However as I suspect that we are talking about low velocities here I will continue to see seek an answer to your very interesting question.

41. Originally Posted by Jilan
OK Abbas there you have it. Force as well as energy is frame dependent. You cannot expect them to be the same. However as I suspect that we are talking about low velocities here I will continue to see seek an answer to your very interesting question.

There is no Abbas in this thread. There is a Kumar who has no clue about the Lorentz force and keeps asking the same question over and over.

42. Ah yes sorry, Kumar.

43. X0X, I dont thinks that's the issue. The question seems to be how to prove that the force is same in both frames of reference
I agree. This is so far the closest someone has understood the question. B is related as for both frames of reference the force is related to B. But magnet frame the B is in the direction perpendicular to relative motion and in the conductor frame it is (taken from Bds) parallel to motion

I am not worried about what SR says, but only what classical Maxwell/Faraday/Lorentz force says. If x0x (or anyone else) can mathematically show how the forces are same in both frames, that will be a great breakthrough.

44. Originally Posted by Harsha Kumar
I agree. This is so far the closest someone has understood the question. B is related as for both frames of reference the force is related to B. But magnet frame the B is in the direction perpendicular to relative motion and in the conductor frame it is (taken from Bds) parallel to motion
No, it is not parallel. But you are incapable of seeing that.

I am not worried about what SR says, but only what classical Maxwell/Faraday/Lorentz force says.
Good, because this is precisely what I explained to you several times. The conclusion is that you do not understand the very basics about the field generated by a magnet, nor do you understand the Lorentz force.

If x0x (or anyone else) can mathematically show how the forces are same in both frames, that will be a great breakthrough.
That would not be a breakthrough, it would be an imbecility because force is frame-variant.

45. X0x, classically it cannot be. So how do we prove it for the case v<<c?

46. Originally Posted by Harsha Kumar
I am not worried about what SR says, but only what classical Maxwell/Faraday/Lorentz force says. If x0x (or anyone else) can mathematically show how the forces are same in both frames, that will be a great breakthrough.
That's trivial. It's a well-known fact in Newtonian dynamics that force is an invariant quantity between inertial frames. If one frame is moving relative to another in the x direction then the transformation between frames is the Galilean transformation given by

x' = x - Vt
y' = y
z' = z

Velocity therefore transforms by differentiating the Galilean transformation

V_x' = V_x - V
V_y' = V_y
V_z' = V_z

Acceleration transforms by differentiating this to get

A_x' = A_x
A_y' = A_y
A_z' = A_z

That means that A' = A, i.e. acceleration is invariant. Since in Newtonian mechanics mass is invariant we multiply this to get

F = mA => mA' = mA ===> F' = [B]F i.e. force is invariant.

Originally Posted by Harsha Kumar
Pysicist - Admiring your valuable effort to give some clue to clear my understanding, i may say that your answer is not exactly for my question. Let me clarify starting from the web page you refered to me - Motional Emf
That's a mistake. I didn't suggest that you pay attention to that page. I merely referred you to the diagram in that page. The page I referred you to is here
Motional EMF

Originally Posted by Harsha Kumar
Just after the eq (205)it says,

" where BT = Bcos is the component of the magnetic field which is perpendicular to the plane of the circuit.
Dou you know why? First off it's only that portion of the field that exerts force on the charge. Also you can think of it as flux = B dot A where B is the magnetic field and A is the area vector which means it has the magnitude of the area and a direction normal to the surface. The dot product is he cosine between those two vectors.

Originally Posted by Harsha Kumar
What are the directions of v and B in eq (207).
You know what the direction of B is since it's the direction of the magnetic field. The direction of v is the direction that the old frame is moving in. That's how you get the electric field in the new frame.

Originally Posted by Harsha Kumar
In my question, in the magnet frame, when the conductor loop moves towards the bar magnet..
What is it with you and bar magnets? You need to first get the principles down before you complicate the problem by using a highly complex field which does not easily subject itself to analysis.

Originally Posted by Harsha Kumar
If we use B to find E0....
What does that mean? I.e. what does it mean to find B using E_0?

Sorry but your question is too confusing to make much sense out of. Please try rephrasing it starting with the essence of what exactly it is that you wish to know. It's still not clear to me yet. I thought I knew when you asked the question in post #1 but then it appears like you changed it after I posted the answer. Please clarify. Thanks.

47. x0x,
Good, because this is precisely what I explained to you several times. The conclusion is that you do not understand the very basics about the field generated by a magnet, nor do you understand the Lorentz force
If x0x can mathematically show how classical Maxwell laws and your Lorentz force law, prove the forces in conductor frame and magnet frame are equal, that would be more useful than making general statements.

(now pl. I am discussing the case where conductor loop moving towards the bar magnet, while the axis of bar magnet is parallel to the direction of the velocity of the conductor)

48. Originally Posted by Harsha Kumar
x0x,

If x0x can mathematically show how classical Maxwell laws and your Lorentz force law, prove the forces in conductor frame and magnet frame are equal, that would be more useful than making general statements.
That would not be useful since it is false. Force is frame-variant. Get that?

(now pl. I am discussing the case where conductor loop moving towards the bar magnet, while the axis of bar magnet is parallel to the direction of the velocity of the conductor)
But the lines of force produced by the magnet are not parallel to the magnet. Get that?

49. I am asking the question on a very specific case only. I see many of you kindly trying to explain me the general principles. That is not necessary.

I am thankful to all of you for attempting to understand my stupid-looking question and giving answers. I recommend that the question is not stupid.

Also I am not expecting anything from SR. This is only about application of classical Maxwell equations and Lorentz force law, and only in the very specific case I am interested in.

Let me rephrase the Question;

" A conductor (closed loop conductor) moves towards a bar magnet in the direction of the axis (axis that connects the two poles) of the bar magnet, concentrically.

The forces acting on the conductor are quantitatively the same when they are analysed from either conductor frame or magnet frame.

Show me mathematically the force on the conductor, 1. looking from the magnet frame 2. looking from conductor frame and show they are equal."

Thanks

Since the approaches to calculate forces in two frames are different and both approaches need "B", but in the conductor frame it is in the direction of the axis of the magnet and in the magnet loop it is in the direction perpendicular to the axis, the confusion exists. This is possible since the magnetic field lines in this particular case which we are discussing are curving. But what really bothers me is how at any given distance from the magnet we get same value for components of B in two perpendicular directions. = Please ignore this paragraph if it is not clear.

50. Originally Posted by Harsha Kumar
I am asking the question on a very specific case only. I see many of you kindly trying to explain me the general principles. That is not necessary. ...

Show me mathematically the force on the conductor, 1. looking from the magnet frame 2. looking from conductor frame and show they are equal."
In the days between your first and most recent posts, you could have looked at any number of references that explain this in great detail. I have already suggested one, by Nobelist E. M. Purcell. His explanation is detailed, clear, rigourously correct and accompanied by a rich collection of illuminating illustrations.

Posters who exert minimal effort to research answers to their own questions, but steadfastly repel answers offered by others stimulate speculations about motive that are unfavorable to the poster.

Purcell's book is widely available. I used the first edition; a second edition was subsequently produced. Ed Purcell knew what he was talking about. Better still, he could explain it to mere mortals.

51. Originally Posted by Harsha Kumar

Let me rephrase the Question;

" A conductor (closed loop conductor) moves towards a bar magnet in the direction of the axis (axis that connects the two poles) of the bar magnet, concentrically.

The forces acting on the conductor are quantitatively the same when they are analysed from either conductor frame or magnet frame.

Show me mathematically the force on the conductor, 1. looking from the magnet frame 2. looking from conductor frame and show they are equal."
This is a DIFFERENT question than the one you have been asking. tk421 is right, you have made no effort to understand the answers that we've given you. You have asked ad nauseaum why there is a Lorentz force, when, in you mind, . It was explained to you that the two vectors are NOT parallel. Do you get that?

Now, you are asking a different question. As mentioned by tk421 the answer can be found in Purcell and it can also be found in "The Feynman Lectures on Physics" vol II , chapter 13-6 (freely available on the internet). The short answer (that I gave you several times) is that Lorentz force is NOT frame invariant (force is not frame invariant), so your question above is meaningless. Feynman shows that the relation between the forces in the two frames in the case you are interested is:

The proof can be found at the location I told you. Please go away and study for a while.

52. Originally Posted by Harsha Kumar
Show me mathematically the force on the conductor.
That can't be done in closed form but has to be done numerically. Not all calculations like this can be done in closed form. In fact most can't. Especially not the one you selected. That's why I suggested that you choose a simpler magnetic field like the one I showed you in the diagram. That's also why I kept asking why you wanted to use a bar magnet. However you never answered that question so I was unable to help at that time.

53. 09-17-2014, 01:28 PM
Thank you.

Can you recommend some literature where I can find the proof for above? (proof for orthogonal magnetic field component equaling B)
tk421 : I asked this from you. You never referred any book/literature. Now it is mentioned about Purcel. Thanks. You understood my question and answered partly, very early, but did not give any proof for how that orthogonal component is quantitatively equaling B.

About your comment on me not researching. It is a misconception, I post this question in this forum only after having done an intensive research. This too is part of my research. I am skeptical there are people who really know and willing to share the answer for my question, or guide me to find the answer.

x0x :
"The Feynman Lectures on Physics" vol II , chapter 13-6
Thanks, i will see.

In SR those forces are not equal in two frames. That is why I repeatedly said that my question is for the solution in Maxwell/lorentz force law approach. (not the relativistic form)

Physicist :
That can't be done in closed form but has to be done numerically. Not all calculations like this can be done in closed form. In fact most can't. Especially not the one you selected.
Thanks. At least you are saying the question has an answer, only numerically, though. However, I am seeking an answer only for the case I selected. This is discussed heavily in liturature, but nowhere I see a straight forward answer. If you can refer to me some literature where this numerical solution is available, that would be helpful.

54. I tried to download the Feynman book but it was taking too long. I think I might understand the question now. The flux through the ring has a cosine term, but the electric field has a sine term. So I think the question might be how does the cosine become a sine? It's to do with the fact that the rate of change of area required to contain the same flux changes with with distance and has a tan term (sin/cos). This means that the cosines cancel and you a left with a sine term. Here are my workings. Sorry it's scruffy.
Image - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting

55. Originally Posted by Harsha Kumar
Thanks. At least you are saying the question has an answer, only numerically, though. However, I am seeking an answer only for the case I selected. This is discussed heavily in liturature, but nowhere I see a straight forward answer. If you can refer to me some literature where this numerical solution is available, that would be helpful.
FYI - It's unlikely that you could find it. It might be out there. Perhaps someone wrote a book and thought that might be a fun numerical problem to do. You never know. But finding it is next to impossible.

Now let's start from the beginning -

Dear Harsha,

(sigh!) You're asking me to help you but you're not doing your part in working to solve this problem. I've posted explanations and corrections that went ignored by you. When you do that it makes it quite literally impossible to help you. Why are you doing that?

Note: Please don't get the impression that I'm irritated with you or anything like that. I'm merely frustrated at not getting answers to the questions that I asked you, that's all. You seem like you want to find the answer to your problem but then you don't give me the information I need to help you solve it. Do you see my point? Why do you need an exact solution anyway?

For example; nowhere in this thread have you explained what it is that you want to know. It wasn't even until post #10 that you mentioned that the conductor was a conducting loop. All you did is to say that when the velocity is in one direction it's zero and in another direction its not zero and you asked why. I couldn't answer that until we got the problem itself correct first. I.e. you have the definition of the EMF wrong and the expected results wrong. For example; suppose you have a bar magnet whose axis is the z-axis of a Cartesian coordinate system and its geometrical center is centered at r = (0, 0, 0). It's oriented such that the magnets north pole points in the +x direction and the south pole points in the -z direction. Therefore in the xy-plane the direction of the magnetic field is in the -z direction everywhere in that plane.

Case #1:

Let me walk you through your first post to show you what I mean. You wrote
In the conductor magnet problem that start a basis for special relativity, when the conductor is moving while the magnet is stationary, it is considered that the emf is equal to q B x V. ...
First off you had it backwards, i.e. the order of v and B is backwards. It should read qvxB, not qBxv. Then in post #27 I explained to you that this is wrong too, i.e.
Originally Posted by Physicisti
You have the emf as a vector when in fact it's really a scalar defined as

where f is the sum of the driving force (e.g. a battery) and the electrostatic force. What you're referring to is motional emf which is related to Faraday's Law of Induction.
I don't see anywhere that you recognize that you have to integrate this, i.e. you have to integrate

around the loop at an instant in time. When the loop is moving towards the magnet the EMF will increase because the flux through the loop will be increasing with time. When you're moving away from the magnet the EMF will decrease because the flux through the loop will be decreasing with time.

Do you understand now?

Originally Posted by Harsha Kumar
...but when the conductor is moving towards the pole of the magnet both B and V are in the same direction making the result equals to zero. This is not the case when the magnet is moving towards the conductor. Can somebody explain please.
You're leaving out an important fact. I.e. the v is the velocity of a current element and that current element changes direction as you integrate around the current loop. The only reason why v x B would be zero would be if v was either parallel or anti-parallel to B and I don't see that happening anywhere in this problem.

To understand the physics better you'd be much better off using simpler configurations such as a magnetic field that is uniform pointing in the +z direction but confined to the region of space between -1m < x < +1m and -1m < x < +1m and -0.001 m < z > +0.001m and use a square loop. Then it's very easy to work with and can be done in closed form. It's even physically realizable.

56. Physicist, did you see my workings on post #54? Did that make sense to you? It does seem to show why the integral over an area can become an integral around a loop. I value your opinion.

57. Originally Posted by Harsha Kumar

x0x : Thanks, i will see.

In SR those forces are not equal in two frames. That is why I repeatedly said that my question is for the solution in Maxwell/lorentz force law approach. (not the relativistic form)
Actually, you don't know what you are asking because you do not understand the basics about the Lorentz force. Sometimes you are asking the classical case, sometimes (when you start complaining about the two forces not being equal in the two frames: conductor and magnet) you are asking about the relativistic case. Bottom line is that you do not understand either the Lorentz law, nor do you make an effort to understand relativity. The Lorentz law, as Feynman teaches you in the reference I gave you, is a relativistic effect. Same as the reference to Purcell tk421 gave you (actually, Purcell simply replays the Feynman answer),

58. Originally Posted by Jilan
Physicist, did you see my workings on post #54? Did that make sense to you? It does seem to show why the integral over an area can become an integral around a loop. I value your opinion.
Neither you nor Peter (the "physicist") understand what Kumar is asking. The point is that Kumar doesn't understand Lorentz force, so his questions lack any meaning. There is no answer to meaningless questions.

59. Originally Posted by x0x
Actually, you don't know what you are asking because you do not understand the basics about the Lorentz force. Sometimes you are asking the classical case, sometimes (when you start complaining about the two forces not being equal in the two frames: conductor and magnet) you are asking about the relativistic case. Bottom line is that you do not understand either the Lorentz law, nor do you make an effort to understand relativity. The Lorentz law, as Feynman teaches you in the reference I gave you, is a relativistic effect. Same as the reference to Purcell tk421 gave you (actually, Purcell simply replays the Feynman answer),
X0X, it was always pretty clear that this question had nothing to do with relativity at all. Your statement here makes it clear that you are struggling to know what the question is ( and you are not alone in this). Once that is established it should be a a piece of cake, but I think it is related to the cosine v sine issue.

60. Originally Posted by Jilan
X0X, it was always pretty clear that this question had nothing to do with relativity at all.
Actually, you are wrong. The question has to do with the Lorentz force. The Lorentz force IS a relativistic effect. You should try reading the Feynman/Purcell reference.

61. Originally Posted by Jilan
X0X, it was always pretty clear that this question had nothing to do with relativity at all.
Since the OP never mentioned transforming from one frame to another in the first post then you're correct regarding that post. However when you wrote the following
Whether an electric field or a magnetic field is observed is dependent on the observers motion, relative to the charge. So if I am stationary to a charge I will observe an electric field. As I start to move slowly relative to the charge I will observe slightly less electric field and a bit of a magnetic field. As I move faster relative to the charge I will observe less electric field and more magnetic field. It's quite amazing.... The only difference between electric and magnetic fields is the relative motion of the observer!
that changed. That's a subject for relativity and that's when the OP started to talk about relativity by referring to this paper

However that part of Feynman's text is not applicable to this problem since the actual problem is not about changing frames.

That part of Feynman's text explains how to use the phenomena of Lorentz contraction to derive an expression for the charge density in the frame which is moving relative to the first in which the charge densities are equal and there is a uniform current density. That charge density creates an electric field. The magnetic field is still there so in this new frame there is both an electric and a magnetic field. But again, that's not relative to this problem and suggests that x0x is somewhat ignorant on this subject and thread.

62. Originally Posted by Physicist

That part of Feynman's text explains how to use the phenomena of Lorentz contraction to derive an expression for the charge density in the frame which is moving relative to the first in which the charge densities are equal and there is a uniform current density. That charge density creates an electric field. The magnetic field is still there so in this new frame there is both an electric and a magnetic field. But again, that's not relative to this problem
Actually, Peter, the Feynman text (like the Purcell text) explains the relativistic nature of the Lorentz force. It would do wonders for your education if you learned that.

63. Harsha, were you able to see my diagram in post #54? It's still unclear to me if that is what your question relates to.

64. Originally Posted by Harsha Kumar
tk421 : I asked this from you. You never referred any book/literature.
You are quite right -- I apologise! I confused your thread with another in which I did refer the OP to Purcell.

About your comment on me not researching. It is a misconception, I post this question in this forum only after having done an intensive research.
That surprises me. I just did a quick Google search using terms like "magnetism as consequence of relativity" and a great many articles, lecture notes, etc. show up. Feynman and Purcell figure prominently in many of the answers. I think your Google-fu needs an upgrade.

And you do seem to be shifting your question around. That causes confusion and frustration.

By now, at least, I hope that you understand why qv X B does not produce a zero result. The v and B vectors are not always parallel, so the cross-product yields a nonzero result. I have no idea why you are obsessing about the integral of B in that context. The Lorentz force involves qV X B, not qV X (integral of B). Plus, you haven't defined what you mean by such an integral; what definition I can discern that you are implicitly using yields a scalar result, which makes no sense in the context of the cross product.

65. Originally Posted by tk421

And you do seem to be shifting your question around. That causes confusion and frustration.
He'll continue to do that, it is a form of trolling. What he's trying to say is that mainstream science is wrong, he "discovered" a hole. He's doing the same exact thing in parallel, in another forum.

By now, at least, I hope that you understand why qv X B does not produce a zero result. The v and B vectors are not always parallel, so the cross-product yields a nonzero result.
I think that he'll never accept that, he got this answer from both of us, yet he keeps on going as if the answer never happened.

I have no idea why you are obsessing about the integral of B in that context.
This comes from "physicist" and "Jilan" plodding about in this thread and adding to the confusion with their answers. Check out their posts above.

66. Originally Posted by x0x
He'll continue to do that, it is a form of trolling. What he's trying to say is that mainstream science is wrong, he "discovered" a hole. He's doing the same exact thing in parallel, in another forum.
That's my suspicion, too, but I'll play dumb for now and try -- as you and others have -- to provide answers that might benefit other readers, even if the OP might be another anti-relativity broken ceramicist.

67. Feel not too good to read last comments by x0x and tk421

Firstly, my question was understood by tk421 as back as 17th, post #11. But tk421 gave a part answer, which I admired and requested for the balance. The balance is the proof that such orthogonal curvature is numerically equal to B. For which you have not given an answer yet. My question remained same, but few others trying to understand it posed me questions and non related ansewers, I tried to response, expecting a positive answer. No body has yet given the answer, some are yet to understand the question.

Out of the group who are responding I feel at least tk421 and Jilan have understood the question. Physicist is spending lot of his time to come to a recourse.

Please answer my question. It is now explained enough and tk421, Jillan and Phycist know what the question is.

I don't understand the fear statment by x0x that
He'll continue to do that, it is a form of trolling. What he's trying to say is that mainstream science is wrong, he "discovered" a hole. He's doing the same exact thing in parallel, in another forum.

68. Jilan please give me some time. Thanks anyway for your effort.

69. Jilan, that is for the conductor frame, in which resultant B is parallel to moving direction (magnet axis, conductor axis). What I am seeking is the proof for the magnet frame in which the resultant B is orthogonal to moving axis , where rate of flux cutting (as in the Lorentz force law) is important.

Further, I am not too sure how far the linear approximation is valid in this case. As divergence of B is zero, curvature may not be allowing us to go for linear approximation.

70. Originally Posted by tk421
That's my suspicion, too, but I'll play dumb for now and try -- as you and others have -- to provide answers that might benefit other readers, even if the OP might be another anti-relativity broken ceramicist.
...and the trolling continues. We have it right.

71. Originally Posted by Harsha Kumar

I don't understand the fear statment by x0x that

I am not worried, I KNOW what you are doing, it is called trolling. I and tk421 have explained to you, repeatedly, that is NOT parallel to , look at the this picture and this picture and..... It is that simple.

72. Originally Posted by Harsha Kumar
Out of the group who are responding I feel at least tk421 and Jilan have understood the question. Physicist is spending lot of his time to come to a recourse.

Please answer my question. It is now explained enough and tk421, Jillan and Phycist know what the question is.
I can't help you until you answer the questions I asked you. You're making mistakes and I can't help you respond to my questions. I showed you mistakes that you need to correct before we move on. E.g. you incorrectly think that EMF is a vector i..e. the following is wrong
...it is considered that the emf is equal to q B x V...
qvrB has the wrong units to be an EMF. The correct units are volts. In this case it comes from LvB or distance x speed x magnetic field or

m x m/s x T

That's not what you have in your opening post which is charge x magnetic field x speed which does not have units of an EMF. Also you're confusing the velocity of the loop with the velocity of the charges on which the magnetic force acts on.

Remember that EMF is the integral of force per unit charge!

73. Originally Posted by Physicist

Remember that EMF is the integral of force per unit charge!
...meaning that the integrand is . Kumar has been claiming that . You need to pay attention, Peter, it is a very simple claim, very easy to rebut, you are all over the map with your "explanations".

74. Originally Posted by Harsha Kumar
Jilan, that is for the conductor frame, in which resultant B is parallel to moving direction (magnet axis, conductor axis). What I am seeking is the proof for the magnet frame in which the resultant B is orthogonal to moving axis , where rate of flux cutting (as in the Lorentz force law) is important.

Further, I am not too sure how far the linear approximation is valid in this case. As divergence of B is zero, curvature may not be allowing us to go for linear approximation.

Harsha, it is in the magnet frame. Is is the conductor that is moving. The calculations consider how much the flux through the loop changes in a small time interval dt. It is possible to consider the curved magnetic field as being linear over a very small distance. The B at each point of the loop won't be orthogonal to the motion, just the sine component hence the sine term in the last expression.

75. Jilan, If it is the conductor that is moving, then we should calculate the rate of flux cutting by the conductor(this is what tk421 also said, I buy that view), instead of rate of change of total flux inside the loop. Then the force on the charge would be q v x B sin@, according to your diagram (sorry I don't find how to write teeta here). But what is given in many places is q v x B. there is no multiple(or division) on B.

76. Harsha, that is what v x B means.... V multiplied by B multiplied by the sine of the angle between them.

77. Originally Posted by Harsha Kumar
Jilan, If it is the conductor that is moving, then we should calculate the rate of flux cutting by the conductor(this is what tk421 also said, I buy that view), instead of rate of change of total flux inside the loop. Then the force on the charge would be q v x B sin@, according to your diagram (sorry I don't find how to write teeta here). But what is given in many places is q v x B. there is no multiple(or division) on B.
is the vector product (google "crossproduct"), not your regular scalar product. Don't pretend that you did not know it, you made a big deal about it being zero because . You have been told many times that the two vectors are NOT parallel.

78. Physicist : Thank you. Yes there is a typo, what I mean is the force, not emf. Perhaps on that you may have lowered the significance of the question. If you assume that I (Harsha ) know 4 maxwell equations and Lorentz force law quite well, (even their proof and many applications) both you and I are safe.

In the magnet frame, force on the loop is given by many writers as q v x B. since, now we have to calculate the rate of flux crossing by the loop instead of total change of flux inside the loop, we have to consider orthogonal component of the magnetic field at the rim of the loop. Then, yes, v x B does not equal to zero. But then how is it still be the same B and not any other value (Bsin@, B/n etc). Unless the orthogonal component is equal to B, the force cannot be q v x B.

So my question is, if we consider the orthogonal component of magnetic field, please mathematically prove that orthogonal component of the magnetic field at the rim of the loop is also quantitatively equal to B.

79. X0X, B x v is a vector product for sure, resulting in another vector. I don't think I said otherwise..

80. Originally Posted by Jilan
X0X, B x v is a vector product for sure, resulting in another vector. I don't think I said otherwise..
I wasn't talking to you.

81. Originally Posted by Harsha Kumar
So my question is, if we consider the orthogonal component of magnetic field, please mathematically prove that orthogonal component of the magnetic field at the rim of the loop is also quantitatively equal to B.
It isn't, it's B sine theta. Where theta is the angle between the motion and the magnetic field.

82. Originally Posted by Jilan
It isn't, it's B sine theta. Where theta is the angle between the motion and the magnetic field.
The above is not quite right, what you need to integrate is :

So, it is a mixed product , a crossproduct followed by a dot product, making it where is the angle between and .

83. Ok, but I was keeping it is since simple and assuming the motion is perpendicular to the area of the loop.

84. Originally Posted by Jilan
Ok, but I was keeping it is since simple and assuming the motion is perpendicular to the area of the loop.
That would mean is perpendicular on . But this is not the expression under the integral, so your claim is irrelevant. Did you study vector algebra?

85. Originally Posted by Harsha Kumar
In the magnet frame, force on the loop is given by many writers as q v x B.
That's not true by any means. To see this notice that the magnetic field varies over the conductor in a non-uniform magnetic field like the one you're speaking of. But in that expression you gave is a point expression which means that the field is evaluated at a point. So what point in the magnetic field do you think is used to calculate the force on the conducting loop?

Back to what I said, i.e. that your statement here is not correct whatsoever. To see what I mean you won't be able to show me not even one "writer" who says such a thing. Please find one of them and post exactly where they said it and exactly what they said and then we'll go from there.

Suppose we have your bar magnet aligned parallel to the z-axis and located such that it's center is at the origin of a Cartesian coordinate system. Let the current loop lie in the xy-plane and move in the +y direction. Then the velocity vector points in the +y direction. In the xy-plane the magnetic field is normal to that plane and points in the -z direction. So the cross product vxB is never zero and always has the same direction, contrary to your claim in the first post.

86. Physicist: you said
Suppose we have your bar magnet aligned parallel to the z-axis and located such that it's center is at the origin of a Cartesian coordinate system. Let the current loop lie in the xy-plane and move in the +y direction.
You see, in my question current loop is moving in Z direction. If you refer to Jilan's diagram on post #54, you will find it.

You said
To see what I mean you won't be able to show me not even one "writer" who says such a thing. Please find one of them and post exactly where they said it and exactly what they said and then we'll go from there.
I have given one example on post #10 (as early as that), see here and read the post #10 and post #19
Where have I mistaken?

For example if you see the first figure of the following web page you will see what bothers me.

See the magnet frame of diagram one of the article.

87. JIlan, you r right. My worry is quantitative figure, because for both conductor frame and the magnet frame the quantitative values should be equal. Then we can't have |B| in one frame and |Bsin@| in other frame.

88. Originally Posted by Harsha Kumar
JIlan, you r right. My worry is quantitative figure, because for both conductor frame and the magnet frame the quantitative values should be equal. Then we can't have |B| in one frame and |Bsin@| in other frame.
You don't. It is B tan@cos@ in one frame and B sin@ in the other (which are of course identical). The cos@ term comes from taking the component of the flux perpendicular to the plane of the loop and the tan @ term arises due to the motion. The sin @ term is just from the definition of a cross product.

It might help to consider a loop around a very very long magnet where the flux enclosed is approximately constant and B is approximately parallel to the motion. If that loop moves along the magnet you would get approximately no electric field generated in the loop as the flux through the loop is not changing. (@ is approximately zero). It's only when the loop gets towards the end of the magnet and the field starts diverging that you will start to measure an electric field around the loop. (@ is no longer zero).

89. Originally Posted by Harsha Kumar
I have given one example on post #10 (as early as that), see here and read the post #10 and post #19
No. You wrote In the magnet frame, force on the loop is given by many writers as q v x B. and that's not true and its not what that link you pointed to says. qvxB is the force on a single charged particle. It's not the force on the loop itself. You have to integrate that over all the moving charges in the current loop to find the force on the loop.

In any case V is the dirction of motion of the charges and they have a component in the direction of the current (i.e. parallel to the conductor) and in the direction of the conductors motion. In any case its never parallel to B in the diagram Jill posted.

90. Originally Posted by Harsha Kumar
So my question is, if we consider the orthogonal component of magnetic field, please mathematically prove that orthogonal component of the magnetic field at the rim of the loop is also quantitatively equal to B.
The "orthogonal" component? Orthogonal to what???

91. Yes Jilan. Congratulations. With that linear approximation you have solved the problem. Thank you.

It is great.

Thank you all KJW, tk421, x0x, physicist, rpenner and all others also for trying to help me to find the answer. Trophy goes to Jilan, however.

Coming to this forum solved me one problem.

I have many more!!! Oh! don't worry. Next time I will try to be more descriptive and try to come with diagrams. Jilan has nice way of presenting diagrams. I will try to use that.

So thank you everybody in this forum who tried to find the answer for me. Thank you again Jilan.

And big thank you for who develop and maintain this forum and all others those who maintain forums like this.