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-   -   Magnetism in a current carrying wire (http://www.physicsforums.com/showthread.php?t=660750)

 DiracPool Dec24-12 08:55 PM

Magnetism in a current carrying wire

If we use the right hand rule, we see that a hole current, or positive ion flow, creates a magnetic field around the wire where the flux moves in a clockwise direction. Counteractively, electron flow through a wire produces a magnetic flux that flows counterclockwise over the wire. Three questions:

1. Do I have this right?
2. Why on earth would a positive charge cause a circular flux in one direction whereas a negative charge cause one in the opposite direction? Is there some characteristic of the EM field that is responsible for this?
3. Do we get the same effect if the charges are flowing through free space and not a wire? Like cosmic rays or just shooting a beam of electrons through free space?

 Sdtootle Dec24-12 10:24 PM

I recommend reading up Amperes law especially the differential form

Quote:
 Quote by DiracPool (Post 4207919) If we use the right hand rule, we see that a hole current, or positive ion flow, creates a magnetic field around the wire where the flux moves in a clockwise direction. Counteractively, electron flow through a wire produces a magnetic flux that flows counterclockwise over the wire. Three questions: 1. Do I have this right? 2. Why on earth would a positive charge cause a circular flux in one direction whereas a negative charge cause one in the opposite direction? Is there some characteristic of the EM field that is responsible for this?
EM field theory is taught with the standardization of how Positive (commonly) charges are effected in a given scenario. The induced magnetic field due to a current carrying wire depends not only the charge of the current (positive vs negative) but also the charge you use as a reference to observe the effects of a magnetic field. If you have a wire with a current of negative charge and utilize a positive charge particle for observation, you will see it getting deflected in the counter clockwise direction. However, if you have a wire with a current of negative charge, but put a negative test particle in the region of the induced magnetic field, it would be clockwise rotation just as you saw with a positive charge current and positive test particle. Positive charge test particles are the usual convention. I hope I didn't miss your true question.

Quote:
 3. Do we get the same effect if the charges are flowing through free space and not a wire? Like cosmic rays or just shooting a beam of electrons through free space?
Once you take a look at amperes law, you'll quickly see that the dependence is on the current density (velocity and density of charges). The wire will effect the mobility of charged particles and depending on where you're measuring your magnetic field, the material can effect the magnetic field as well (magnetic permeability).

 DiracPool Dec25-12 12:34 AM

Re: Magnetism in a current carrying wire

Hmm..I'm not sure I got what I was lookin for for question 3, but your response to question two is interesting:

Quote:
 The induced magnetic field due to a current carrying wire depends not only the charge of the current (positive vs negative) but also the charge you use as a reference to observe the effects of a magnetic field.
So then, the right hand rule is more of a "guideline" rather than a rule and depends upon the test charge. I'm still wondering why, though, one trajectory (clockwise vs counterclockwise) is preferred over the other. Its the asymmetry that bugs me. That is, if we fire a negative test charge at a positive "target" charge, the trajectory of the negative charge will bow inwards toward the positive charge symmetrically on both sides due to any error in the aiming of the test charge. Similarly, if we fire a positive test charge at a positive target charge, the trajectory of the test charge will bow away from the target charge symmetrically. That makes sense. This asymmetry of the magnetic "loop" around the wire doesn't make sense. What is it about the magnetic field that produces this asymmetry. I'm not finding that by looking at Ampere's Law or any of the Maxwell equations.

 DiracPool Dec25-12 02:08 AM

Re: Magnetism in a current carrying wire

Oh yeah, there's a perhaps importantly-related number 4 question too:

4) Is magnetism a relativistic effect as asserted in this video:

whereby the sideways force I'm curious about has to do with increased electron charge density due to length contraction? I've heard about this model in a couple different places but it doesn't seem to be widely discussed. Is this relativistic model generally accepted or is it a fringe theory?

 vanhees71 Dec25-12 05:33 AM

Re: Magnetism in a current carrying wire

A formula says more than thousend words:biggrin:. The formula in question here is Ampere's Law, i.e., one of the Maxwell equations, specialized for stationary currents. It says
$$\vec{\nabla} \times \vec{B}=\vec{J}.$$
Further you need the fact that there are no magnetic charges ("monopoles"):
$$\vec{\nabla} \cdot \vec{B}=0.$$
This together Ampere's Law tells you that only some current density can cause a magnetic field (in this special case of stationary, i.e., time-independent situations). This tells you that it is only this current density determines the direction and magnitude of the magnetic field. The current density of a single particle is given by
$$\vec{J}=q \int \mathrm{d} t \dot{\vec{y}}(t) \delta^{(3)}[\vec{x}-\vec{y}(t)],$$
where $\vec{y}(t)$ is the trajectory of the particle. This tells you that the sign of the charge, of course included in the current density, determines wether the current at any instance of time is pointed in (positive charge) or against (negative charge) the particle's velocity. What goes into the magnetic field is still only the current density and that's it.

The right-hand rule to determine the direction of the $\vec{B}$ thus applies to this current density and nothing else: Point the thumb of your right hand in direction of the current density the fingers tell you the direction of the magnetic field. That's what's written in Ampere's Law, becaus it is customary to define the vector operations in three-dimensional Euclidean space always according to right-hand rules.

To see this quantitatively at work, we have to solve for the magnetic field in terms of the current density. The vanishing of magnetic charge implies that there exists a vector potential for the magnetic field, i.e.,
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Further we can use Cartesian coordinates to calculate the curl:
$$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla}(\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}.$$
Now the vector potential is only determined up to the gradient of a scalar field, and thus we can impose one scalar constraint on it. In modern slang the physicists call this "fixing a gauge". Here, it's most convenient to use "Coulomb Gauge", i.e.,
$$\vec{\nabla} \cdot \vec{A}=0.$$
With this Ampere's Law simplifies to
$$\Delta \vec{A}=-\vec{J}.$$
This is just the same law (for each component) as that for the electric potential in electrostatics, and thus we can write down the solution immediately. Behind this is of course the Green's function of the Laplace operator:
$$\vec{A}(\vec{x})=\frac{1}{4 \pi} \int \mathrm{d}^3 \vec{x}' \frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
It is easy to check that this solution indeed fufills the Coulomb-gauge condition, because from Ampere's Law we must impose the constraint
$$\vec{\nabla} \cdot \vec{J}=0,$$
which is nothing else than the conservation of electric charge for the here considered special case of stationary currents.

Now we evaluate the magnetic field. For this we need to take the curl of the above solution. We can lump the differentiation operator under the integral and then use
$$\vec{\nabla} \times \frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|} = -\vec{J}(\vec{x}') \times \vec{\nabla} \frac{1}{|\vec{x}-\vec{x}'|}=+\vec{J}(\vec{x}') \times \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$
This gives you finally Biot-Savart's Law
$$\vec{B}(\vec{x})=\frac{1}{4 \pi} \int \mathrm{d}^3 \vec{x}' \vec{J}(\vec{x}') \times \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$
This formula only deals with vector manipulations and is thus valid in any coordinate system (not only Cartesian coordinates).

More intuitive as an example than the magnetic field of a point charge is a steady current along an infinite very thin straight wire. Let's choose the $z$ axis to be directed in the direction of the current densitiy. For this situation we have (in Cartesian coordinates)
$$\vec{J}(\vec{x}')=\vec{e}_z I \delta(x') \delta(y').$$
Then we have
$$\vec{B}(\vec{x})=\frac{1}{4 \pi} \int \mathrm{d} z' I \vec{e}_z \times \frac{x \vec{e}_x+y \vec{e}_y}{\sqrt{(z-z')^2+x^2+y^2}^{3}}.$$
The evaluation of the integral gives
$$\vec{B}(\vec{x})=\frac{I}{2 \pi} \vec{e}_z \times \frac{\vec{x}}{x^2+y^2}.$$
It's more intuitive to write this in terms of cylinder coordinates
$$\vec{B}(\vec{x})=\frac{I}{2 \pi \rho} \vec{e}_{\varphi}.$$
As you see, the right-hand rule works: The magnetic field curls around the current-conducting wire counter-clockwise when looking against the current density.

Note that we do not have to think about the motion of the electrons in the wire at all. Of course, they move in negative $z$ direction. Together with their negative charge the current density points into positive $z$ direction.

 DiracPool Dec25-12 06:37 AM

Re: Magnetism in a current carrying wire

Quote:
 As you see, the right-hand rule works

Thanks for the thoroughness of your reply Vanhees. I see some light in there although its going to take me a little time to digest it all. :approve:

 andrien Dec26-12 08:12 AM

Re: Magnetism in a current carrying wire

you can use either hand,it does not change any physics in most cases.At least in electromagnetism.

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