# What is wrong with Ampere's force law?

1. Apr 24, 2010

### Dunnis

Code (Text):

+                   -     +          -
| I2= 1A            |     | I2= 1A   |        I2= 1A
|===================|     |==========|     + ==================infinite >> -
|                     |                    |
|r=1m                 |r=1m                |r=1m
|                     |                    |
|===================|     |==========|     + ==================infinite >> -
| I1= 1A            |     | I1= 1A   |        I1= 1A
|<------- 9m ------>|     |<-- 1m -->|
+                   -     +          -

Ampere's Force Law: F= 2* mu0/(4Pi*r) * I1*I2; is this the force felt by each one single wire, or the total force between the two? There is apparently something wrong with this equation, it gives the same result regardless of different wire lengths, therefore it seem useless, no? Also it will produce different result than what would you "normally" get by using the Lorentz force equation: F= I*L x B. Yes?

2. Apr 24, 2010

### Antiphon

It's the force experienced by each 1-meter section of the two infinitely long wires.

It will produce the same result as the force equation.

3. Apr 24, 2010

### Dunnis

Why "infinitely long wires"? Are you saying this equation can not actually be used in the real world with non-infinitely long wires?

F= 2* mu0/(4Pi*r) * I1*I2; where do you see any wire lengths in this equation? Do you not see it will produce the same result whether wires are actually 3cm, 5m or infinitely long?

No, it will not. Why did you say that? For start the Lorentz force does have the wire length as relevant parameter, F= I*L x B.

2* mu0/(4Pi*r)*I1*I2 = I1*L x B ?

So, you are saying these two equations are the same, and they both calculate the same thing - the force acting on one single wire?

4. Apr 24, 2010

### Antiphon

The equation you put up applies to infinitely long wires. You wires are short so the equation doesn't apply to your wires.

Forces always act on a pair of entities. Otherwise you would not conserve momentum and you could lift youself up by your own bootstraps.

Both equations should produce the force PER UNIT LENGTH of infinitely long wires.

5. Apr 24, 2010

### Born2bwire

Yep, the equations work out exactly the same once you note that the left-hand side is the force per unit length as Antiphon stated above. If it wasn't per unit length then logically the total force would have to be either infinite or zero since you have an infinite number of elements over which the force can act.

6. Apr 24, 2010

### Dunnis

What equation? You mean Ampere's force law works ONLY with infinite wires? But that's the whole problem, that's what I said and you are now saying it as if that makes some sense or can be used for something, while it actually makes it incorrect and useless.

What do you do with equation that only applies to infinite wires?

Should? Well, they don't, they both actually have units in Newtons.

Lorentz force, PER ONE METER:

F= I*L x B = 1A * 1m x 10^-7 N/(A*m) = 1*10^-7 Newtons

Ampere's force law, PER %#@$?: F= 2*mu0/(4Pi*r)*I1*I2 = 2* 10^-7 N/A^2 *1A*1A = 2*10^-7 Newtons Per unit length, yes - PER ONE METER, why do you need infinity for that? Do you think two wires that ACTUALLY are one meter in length would attract more, less, or with the same force as two infinite wires, per meter? And, how about per meter per second? Last edited: Apr 24, 2010 7. Apr 24, 2010 ### Dunnis No they do not work the same as I demonstrated above. Why did you say that? Do you know math? If yes then show me your calculation and your logic, there is no need to pull false assertions out of thin air. -- What equation, what left hand side is the 'force per one meter' and how am I supposed to see that? What are you talking about? Ampere's force law works ONLY with infinite wires and is therefore useless and can not be applied in the real world with normal wires - is this what you are trying to say? Last edited: Apr 24, 2010 8. Apr 24, 2010 ### Born2bwire The magnetic field from an infinite wire is $$\mathbf{B} = \frac{I\mu}{2\pi r}\hat{\phi}$$ Which we can easily see works out to be the same when we calculate the per unit length force. They should of course work out to be the same as the force law is nothing more than a combination of Ampere's Law and the Lorentz Force. We use Ampere's Law to get the magnetic field of the wire and then the Lorentz Force to find the resulting force. So Ampere's Force Law works for any general case. However, you are using the case that was worked out assuming the magnetostatic case of infinite wires. 9. Apr 24, 2010 ### Dunnis Previously: What equation, what left hand side is the 'force per one meter' and how am I supposed to see that? Both equations have units in Newtons, yes/no? Easily, eh? Then demonstrate your calculation and show us exactly where do you get this "unit length", commonly known as 'one meter', from? That equation too does not have any such variables, what in the world are you talking about? -- "Infinite wire", again!? There is no such thing as infinite wire, so please - where in THE REAL WORLD can I use that equation of yours and what purpose can it possibly have in real life and practical applications? Wrong. Not Ampere's law, but Biot-Savart law. Those equations of yours work ONLY with infinite wires as we both keep repeating, it is only that you need to realize infinity is UNDEFINED, while we need EXACT equation if we are to define the unit of ampere. When you take an instrument and measure how many amperes there are flowing in some conductor, what equations do you think those instruments are based on, the ones for INFINITE WIRES of undefined lengths, or the equations for NORMAL WIRES with certain and defined length? Biot-Savart law and Lorentz force: d2F = mu0/4pi*r^3 * i1*dl1 x (i2*dl2 x r) http://www.bipm.org/en/si/si_brochure/chapter1/1-2.html Wrong. You keep contradicting yourself. If it works ONLY for infinite wires, then it is not general and it does not work for any other cases, but ONLY for infinite wires, hence it does not work at all. I already told you all this couple of weeks ago, do you ever learn? Code (Text): + - + - | I2= 1A | | I2= 1A | I2= 1A |===================| |==========| + ==================infinite >> - | | | |r=1m |r=1m |r=1m | | | |===================| |==========| + ==================infinite >> - | I1= 1A | | I1= 1A | I1= 1A |<------- 9m ------>| |<-- 1m -->| + - + - So, what result do you get for these situation, the same one? Last edited: Apr 24, 2010 10. Apr 24, 2010 ### Born2bwire *Shrug* It's trivial. The force from the Lorentz Force for a given length L over infinite wires is $$F = \left| \int^L_0 dz I_1\hat{z} \times \mathbf{B} \right| = \frac{I_1I_2\mu L}{2\pi r}$$ So dividing both sides by the length$L$and we get back what is predicted by the equation given previously. The equation for the force between two infinite wires is the force per unit length, so its units are N/m, which we regain when we divide the force derived for a length L (in N) by the length L (m). Biot-Savart Law is a special case of Ampere's Law. Ampere's Law is the general law applicable to any situation while Biot-Savart assumes magnetostatics. As for the examples, of course the results will be different. Since you already know about the Biot-Savart law then you can derive Ampere's Force Law for magnetostatics and use them to find the forces in the first two examples while the last one can be handled in the case I show above. EDIT: I would also note that the Lorentz force you give in your original post is incorrect. The Lorentz force is defined for a point charge, thus the force due to a current is defined as the force from an infinitesimal current element. That is, $$\mathbf{F} = q\left(\mathbf{E} + \mathbf{v}\times\mathbf{B}\right) = Id\hat{\ell}\times\mathbf{B}$$ If there is no electric field. To get the total force one must integrate over the entire current source as I have done above. 11. Apr 24, 2010 ### Dale ### Staff: Mentor Hi Dunnis, look at the units of your equation. In SI units F is in N/m. That is where the length comes in. Go ahead and work it out for yourself in detail, you always need to keep track of the units. No, in the equation F= 2* mu0/(4Pi*r) * I1*I2 you can determine that F has units of N/m. Last edited: Apr 24, 2010 12. Apr 24, 2010 ### ZapperZ Staff Emeritus Or maybe you misunderstood the response that you got! Have you ever considered that as a possibility? You will note that there is an almost unanimous response here trying to correct you. Ampere's law works in general! However, it doesn't mean that you can easily get an analytical solution for all cases! It is easiest when the problem has high symmetry, such as a long infinite wire, a long infinite solenoid, etc, where edge effect and other issues do not come into play. For other cases, you may have to resort to either a numerical solution, use Biot-Savart, or use Green's function approach. Zz. 13. Apr 24, 2010 ### Dunnis Everyone said Ampere's force law (and later Ampere's law for B field) work ONLY with infinite wires, but is that supposed to mean "work in general" or "useless"? Let's find out, can they be applied in real world or not? Like this: Code (Text): CASE A: CASE B: + - + - | I2= 1A | | I2= 1A | |===================| |==========| | | |r=1m |r=1m | | |===================| |==========| | I1= 1A | | I1= 1A | |<------- 9m ------>| |<-- 1m -->| + - + - CASE A: $$B_1 = \frac{\mu_0*I_1}{2\pi r} = 2* \frac{10^-7_{N/A^2} * 1_A}{1_m} = 2 * 10^-7 \ N/A*m$$ CASE B: $$B_1 = \frac{\mu_0*I_1}{2\pi r} = 2* \frac{10^-7_{N/A^2} * 1_A}{1_m} = 2 * 10^-7 \ N/A*m$$ Are these correct results? Is that per unit length, per infinity, per what? Last edited: Apr 24, 2010 14. Apr 24, 2010 ### Dunnis I already did that, see above, and here is again: Lorentz force, PER ONE METER: F= I*L x B = 1A * 1m x 10^-7 N/(A*m) = 1*10^-7 Newtons Ampere's force law, PER %#@$?:

F= 2*mu0/(4Pi*r)*I1*I2 = 2* 10^-7 N/A^2 *1A*1A = 2*10^-7 Newtons

Can I see you "calculation" and dimensional analysis now, how did you come up with your conclusion?

15. Apr 24, 2010

### Born2bwire

Ampere's Force Law works for the general case. It is a marriage of Ampere's Law and the Lorentz Force, all of which are generally valid (though Ampere's Law is not sufficient by itself to describe an electrodynamic system in its entirety).

You, however, are using a special case of Ampere's Force Law that has been derived for the specific situation of two magnetostatic infinite wires.

You did it incorrectly. As Dalespam stated, work out the units yourself explicitly and you will see that they do not match up. More specifically, the first equation, your Lorentz Force, is not the force per meter (N/m), that is the force (N). The second one is the force per meter (N/m). Oh yeah, and as I mentioned before you used an incorrect equation for the magnetic field in the first equation too.

Last edited: Apr 24, 2010
16. Apr 24, 2010

### vela

Staff Emeritus
Actually, if you look back over the thread, you'll see only you said that. What the others said was the formula you wrote down, not Ampere's law, applies to the case of infinitely long wires.

Since I imagine you're just going to keep repeating yourself, saying you already did work out the units, I'll just ask, What happened to the 4πr?

17. Apr 24, 2010

### elect_eng

I have to warn you all that Dunnis is very stubborn, arrogant and rude. I've seen this in his previous posts. He clearly does not know basic EM theory, and probably does not have the full foundation to properly learn the subject well. He is often asking basic questions and then refusing to accept the answers.

There is nothing wrong with being ignorant and asking questions to educate oneself. Afer all, this is how we all learn. However, this must be done with humility and not with arrogance, rudeness and disbelief.

Dunnis doesn't seem to understand that many of us have studied EM theory and practice for decades, even after taking several undergrad and grad level courses on the subject. Some of us even use EM theory in everyday engineering and scientific work throughout our lives. Why does he come here asking questions as if we know the answer, and then turn around and tell us that everything we say is wrong? It's very strange and very annoying.

Until Dunnis learns the basic principles of self-education, he will make little progress. I for one will not waste any more of my time trying to talk to him, at least until I see evidence that he has changed his attitude. Also, if he doesn't change his attitude, he should be banned from these forums, in my opinion. He is not helping anyone here, and is not allowing anyone here to help him.

Last edited: Apr 24, 2010
18. Apr 24, 2010

### Born2bwire

True, but it is always a bit irresponsible to leave incorrect or misleading statements alone. It seems that people often pull up old posts here on the forums when they are looking for answers as indicated by the occasional necro thread coming up every so often. In addition, I find that Physics Forums has enough web presence to come up in response to Google searches. So while the immediate goal may never be obtained, we can at least provide a cogent explanation to the uninitiated about why and how he is wrong in his posts. Besides, most of this stuff is general enough that I only need to spend as much time in my answers as it takes to type out the post so little of my own time is lost.

19. Apr 24, 2010

### Staff: Mentor

In your analysis you neglected r. So mu0 has units of N/A², r has units of m, and I1 and I2 each have units of A. So you get

2*mu0/(4Pi*r)*I1*I2 -> N/A²/m*A*A= N/m

20. Apr 24, 2010

### elect_eng

I agree completely. This is a Catch-22 for sure. I was forced to respond to him when he interrupted another thread I was talking in. I normally don't like to call the moderators, but it was necessary to shut down that thread just to prevent his misinformation from propagating.

However, this guy will keep on going like the energizer bunny, so I just thought I'd warn you. You may be in for a rough ride.

21. Apr 24, 2010

### Dunnis

Oops, true. Thank you.

Lorentz force, PER ONE METER:

F= I*L x B = 1A * 1m x 10^-7 N/(A*m) = 1*10^-7 N

Ampere's force law, PER %#@\$?:

F= 2*mu0/(4Pi*r)*I1*I2 = 2* 10^-7 N/A^2 *1A*1A = 2*10^-7 N/m

Now they are even more different. Anything else?

Last edited: Apr 24, 2010
22. Apr 24, 2010

### Dunnis

"formula you wrote down"? Say it properly please so we do not end up arguing semantics.

Are you saying the Lorentz force: F= I*L x B, applies only to infinitely long wires?

4Pi got canceled with magnetic constant, and r I forgot about.

Last edited: Apr 24, 2010
23. Apr 24, 2010

### Born2bwire

You're still using the wrong equation for the magnetic field but dimensionally that is correct now.

24. Apr 24, 2010

### Dunnis

No, I'm using Biot-Savart law. Either of these two will do just fine:

http://en.wikipedia.org/wiki/Biot-savart_law

..and this for Lorentz Force, either will do fine (where B= Biot-Savart from above):

$$F = q*v \times \mathbf{B}$$

http://en.wikipedia.org/wiki/Lorentz_force

...together they produce this:

http://en.wikipedia.org/wiki/Ampère's_force_law

...which is integral form of this equation given by 'Le Bureau international des poids et mesures' (BIPM), the most general and exact definition for magnetic force:

d2F = mu0/(4pi*r^3) * i1*dl1 x (i2*dl2 x r)

http://www.bipm.org/en/si/si_brochure/chapter1/1-2.html

These equation above is what defines the unit of ampere and along with it all the rest of electrics and electronics, and any other equation that has anything to do with any el. currents. These two are Biot-Savart and Lorentz force equations in their full integral form, where arbitrary segments and relative geometry can be integrated. That's what "general" means.

This all IS magnetostatics anyway, we are talking about steady currents and the definition of ampere unit. There is no such thing as "special case" of Ampere's Force Law. -- This is Ampere's Force Law: F = 2*mu0/(4Pi*r)*I1*I2; there is no other, or special version, and that particular equation is not general. General is something you can apply to any scenario, with arbitrary geometry and arbitrary wire, or wire-segment, lengths.

Code (Text):

CASE A:                        CASE B:

+                   -          +          -
| I2= 1A            |          | I2= 1A   |
|===================|          |==========|
|                          |
|r=1m                      |r=1m
|                          |
|===================|          |==========|
| I1= 1A            |          | I1= 1A   |
|<------- 9m ------>|          |<-- 1m -->|
+                   -          +          -

CASE A: dl1 || dl2, i1= 1A, i2= 1A, r= 1m, L= 9m
======================================

Mathematica - WIRE 1:
D[10^-7Newtons/1^2 * 9*{l1,0,0} cross 9*{l2,0,0} cross {0,-1,0}, l1,l2] = [0, + 8.1*10^-6 N, 0]

Mathematica - WIRE 2:
D[10^-7Newtons/1^2 * 9*{l1,0,0} cross 9*{l2,0,0} cross {0,1,0}, l1,l2] = [0, - 8.1*10^-6 N, 0]

Execute on-line, here: www.wolframalpha.com/input

CASE B: dl1 || dl2, i1= 1A, i2= 1A, r= 1m, L= 1m
======================================

* L = int(dl)= unit length = 1m
* d2F(single wire) = mu0/(4pi*r^3) * i1*dl1 x (i2*dl2 x r)

=> mu0/(4pi*|1m|^2) * 1A*1m x 1A*1m = 1*10^-7 N

Mathematica - WIRE 1:
D[10^-7Newtons/1^2 * 1*{l1,0,0} cross 1*{l2,0,0} cross {0,-1,0}, l1,l2] = [0, + 1*10^-7 N, 0]

Mathematica - WIRE 2:
D[10^-7Newtons/1^2 * 1*{l1,0,0} cross 1*{l2,0,0} cross {0,1,0}, l1,l2] = [0, - 1*10^-7 N, 0]

Execute on-line, here: www.wolframalpha.com/input

Last edited by a moderator: Apr 25, 2017
25. Apr 24, 2010

### Staff: Mentor

I don't know how you got your number for the Lorentz force, but it is wrong. Your number for the Ampere's force is correct. In any case, you can show that they are equivalent algebraically.

F = I*L x B
F/L = I1*B = I1*mu0*I2/(2Pi*r) = 2*mu0/(4Pi*r)*I1*I2