# Magnetic field produced by a short wire element (on 360 degree basis)

1. Jun 23, 2010

### kmarinas86

In consideration of a lone wire element of differential length dL carrying a current I:

Two things concern me here:

1) Is a magnetic field generated by current I really limited to only spaces that exist orthogonally to the line between the two ends of the wire element? If not, what does the magnetic field generated to the sides of these ends (and outward) look like?
2) How does one draw the equipotential surface of the magnetic field around such a lone wire element? What shape is it? Is it normally accepted that lines of magnetic force can be parallel to the magnetic equipotential surface, or does the conventional wisdom say that this does not happen? Normally, I thought that the lines of force must be perpendicular to the equipotential surface, but perhaps that is not really true since the circular magnetic field lines around a wire do not appear to cut equipotential surfaces of the magnetic field as far I am aware of. If equipotential surfaces correspond only to a scalar potential, then it relates only to line density of the magnetic field, which I know decreases with distance from the wire. I cannot see how conventional wisdom can take that as given and yet say that the lines of force are perpendicular to the equipotential of the magnetic field. So what do they really say?

2. Jun 24, 2010

### netheril96

1)What does the "spaces that exist orthogonally to the line between the two ends of the wire element" mean?How could space be orthogonal to something?
In fact,the magnetic field exists everywhere (of course it fades when one reaches infinity)
You should see Biot-Savart law

2)Magnetic field has no potential
Or,more specifically,no scalar potential as you have used in electrostatics

3. Jun 24, 2010

### kmarinas86

Stick a pencil through the binding of a spiral notebook. Let's pretend that the pencil is a metallic conductive rod and the spiral is actually a set of concentric rings which are circular the magnetic lines surrounding it.

The length of the pencil is the differential length dl. Let's imagine that a static discharge on one end of our "pencil" conductor creates a current I across the length of the pencil. Let's say that the current is the same throughout the pencil.

Problem: Is there a perfectly sharp, non-differentiable cutoff of the magnetic field strength beyond the ends of the pencil? Or does the magnetic field weaken gradually off those ends, making the magnetic field differentiable?

What if we all we have is a current that is just as long as its wire is wide? And I ask again, what is the shape of this field going to be? Instead of being flat on the ends like a cylinder, wouldn't they have a round or pointy cutoff at the ends?

You were referring to this:

I read an article (here) and assumed at first that equipotential surfaces required a scalar potential, but I guess that is not really true.

So the second thing I really want to ask is, "How you do draw the equipotential surface of a magnetic field due to just one extremely short current element? Is it more like a dartboard or more like an onion?"

4. Jun 24, 2010

### netheril96

I have no patience to answer each of your questions in detail.
Just see Biot-Savart law and you will get answers to the first three questions.

As for the fourth one,magnetic field has a vector potential
How could you draw a vector equipotential surface?

5. Jun 24, 2010

### K^2

By no means. The equation is actually very similar to the one from electrodynamics.

$$dB = d^3r \frac{\mu_0}{4\pi} \frac{J \times r}{r^3}$$

J is the current density here. If you want to talk about infinitely thin wire, d³r becomes dl and J becomes I. Note that r³ in denominator partially cancels with magnitude of r in numerator, so except for angle-dependence, it's the same 1/r² behavior as that of electric field.

You don't. Magnetic potential is a vector potential, and it is only defined as far as its rotor, which means that even if you could draw equipotential surface, it would depend on the gauge.

6. Jun 25, 2010

### kmarinas86

Thanks!

Well okay then.

I am still wondering if anyone has ever developed a contour map for the B field that varied simply by the line density of that field, regardless of field's orientation. I am now sure that this is not a equipotential surface. I'd imagine that such a contour map would have contour lines which would surround a bar magnet to form a peanut shape. Is there a name for this kind of contour map?

7. Jun 25, 2010

### K^2

Because B-fields do not diverge, the most useful "contour" map is the map of fields themselves.

8. Jun 26, 2010

### kmarinas86

I'll take that as a "no" to my question.

9. Jun 26, 2010

### Staff: Mentor

Note that this setup is non-physical as it violates conservation of charge and therefore Maxwell's equations. The closest you could come is a dipole antenna with an oscillating charge. This will have both an E field and a B field and the wire will charge and discharge as the current oscillates.

10. Jun 26, 2010

### Taturana

The magnetic field is a consequence of moving charges. When you take a small segment of a wire (it will have two ends), you consider that the current is the flux of positive electric charges and these charges are moving from one end to the other. For that reason we don't have moving charges in the ends of the wire, thus we don't have any significant magnetic field generated by the ends of the wire.

For you try to figure out what the magnetic field generated by the ends of the wire looks like (if there is any moving charge there) we have the postulate of electromagnetism that says something like: "A charge moving in a straight line (a wire for example) will generate a circular magnetic field. The field generated will be contained in a plane that is perpendicular to the line that the moving charge is moving and the (direction? clockwise, counter-clockwise) of the magnetic field is determined by the right-hand rule.

That helps you?

11. Jun 27, 2010

### kmarinas86

12. Jun 27, 2010

### kmarinas86

Well it certainly does not answer my question. However, it does clarify my concern. They say, "A charge moving in a straight line (a wire for example) will generate a circular magnetic field." However, if the field drops to zero as soon as you go off to the side, that is an extreme cut off in the strength of the magnetic field as it would be non-differentiable. You could have a current generating 1 Tesla, and in one nanometer beyond that the wire element (let's call it an antenna) the field strength is 0 Tesla. That does not make sense.

13. Jun 27, 2010

### Relena

lets not talk about a short wire or charge, just a infinitesimal current element di
when you try to find the magnetic field at any point P in space apply the biot-savart law you'll find a magnetic field from this current element at ANY point in the space.

however, when you integrate this current element into a long wire, you'll get that the magnetic field at any point surrounding the wire is the summation of the effect of these infinitesimal current elements, so, when you are moving parallel to the straight wire within it's boundary the magnetic field will be constant, however, when you reach the cliff end of the wire the magnetic field will begin to decrease gradually, not a sharp cut off

so, the circular shape of the magnetic field is NOT because the points of calculation have a distance perpendicular to the wire, but because the summation of the effects of the current elements keeps the field constant as long as:1- you are moving parallel to the wire and 2- you're within the wire limits, you can verify that simply using the bio-savart law

this is applicable to charges as well, the magnetic field generated by a moving charge will be circular only in the plane that contains this point charge AND perpendicular to the direction of motion, in any other plane the field will decrease gradually, the contour or the equipotential surface will no longer form a cylinder.

14. Jun 27, 2010

### Staff: Mentor

No, it doesn't, you just must not know what a dipole antenna is. The current in a dipole antenna is non-uniform because it must be 0 at the ends of the antenna. The charge also is non-uniform in order to properly conserve charge. You simply cannot have an isolated uncharged wire carrying current. Either it must be in a loop or it must be charging or discharging.

15. Jun 28, 2010

### AJ Bentley

That's the nub of it.

16. Jun 30, 2010

### kmarinas86

Sorry about the misunderstanding then. Still, your example clearly has nothing to do with the problem I originally posed. I am talking about a current from a static discharge (which in reality will merely approximate DC current) not an AC standing wave where current values reach 0 at the ends. If I don't know what dipole antenna is, and if I am not asking about a dipole antenna, doesn't it make sense then that my problem should have nothing to do with a dipole antenna?

The phrase "static discharge" better refers to a "discharge of static electricity".

Now for someone who understands my problem (or "my nub" so to speak):

Thank you!

I did not know that the Biot-Savart Law could be used outside that plane. I will have the examine the vector notation more closely.

As for you DaleSpam, you responded:

This was in response to my scenario:

You then imagined a different scenario:

This is not what I am talking about.

I responded:

In retrospect, this was an irrelevancy, not a contradiction. It had nothing to do what I was talking about. So I got caught off guard. I actually believed you were on topic, but I was wrong.

Relena understood. That's the "nub".

As for the rest of the problem (i.e. the exact shape of the magnetic fields off the ends), I believe I can work that on my own.

Last edited: Jun 30, 2010
17. Jun 30, 2010

### AJ Bentley

I still don't understand.

Taking an element to perform an integration is a mathematical device, it bears no relation to physically extracting a piece of wire in the lab and measuring the fields around it.

If you did that, and somehow contrived to get a pulse of current through it in isolation - that is, without wires, without moving charge externally to it and without using an external field to induce a current. (Any of those things would make a field that would be as large as anything the wire produced - invalidating the experiment)
Then, you would get some sort of EM field...

I wouldn't speculate on it's shape because such an experiment would be physically impossible.

18. Jun 30, 2010

### Dickfore

Use the formula:

$$B = \mu_{0} \frac{I}{2\pi \, d} \, (\cos \alpha_{1} - \cos \alpha_{2})$$

The relevant variables are denoted in the attached figure:

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19. Jun 30, 2010

### kmarinas86

It works well. :)

Below is the graph of abs(cos(α_1)-cos(α_2)).

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• ###### Biot-Savart Law (differential wire element).xls
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20. Jun 30, 2010

### Dickfore

The formula is wrong by a factor of 1/2:
$$B = \mu_{0} \frac{I}{4\pi \, d} \, (\cos \alpha_{1} - \cos \alpha_{2})$$