# Speed of the magnetic field around a wire.

1. Jul 18, 2004

### grounded

Sending current through a wire creates a magnetic field around the wire.

I have a couple questions about the above that I was hoping someone here may know about?

How do we know the magnetic field is traveling a specific direction around the wire (clockwise or counter clockwise viewed from above)?

If it has a direction, it must have a speed, what is the speed?

Does the field (concentric circle) right next to the wire travel at the same speed around the wire as the field located a further distance from the wire?

If the wire is spinning in the same direction as the magnetic field (the power supply and the observer are at rest), will it alter the speed of the field? Will it alter the resistance of the wire? What if the wire spins in the opposite direction?

I can't find any information on the above, any thoughts or links will be appreciated.

Thank you...

2. Jul 18, 2004

### Alkatran

The magnetic field is expanding outward from the wire, it isn't actually moving around the wire, we just apply a direction so the calculations are possible (the electrons in the field do the moving).

3. Jul 18, 2004

### reilly

Magnetic fields from stationary currents do not move at all -- actually there's a bit of dynamics when the curent is turned on, but that goes away at the speed of light. Check any text book on E&M. The problem of the field of a spinning wire is totally dependent on the wire's current; you pays your money and takes you chances.

Regards,
Reilly Atkinson

4. Jul 19, 2004

### krab

Why would you say that? If I draw an arrow on a piece of paper, it has a direction, but no speed.

5. Jul 19, 2004

### Antonio Lao

The concentric circles as flux lines indicate that magnetic field can only form closed loops (nonexistence of magnetic monopoles). These circles are used to defined the magnetic field strength. larger circle indicates lesser field intensity.

If (a big IF) magnetic field has a speed, this speed would be the speed of light in vacuum. And this speed would be less for the magnetic field in matter.

The reason why the speed can never be found by experiment is because of the nonexistence of magnetic monopoles, the source and sink of the field similar to the electric field. The speed of a charged particle in an electric field can always be determined but it is not what you are looking for. But to generate a magnetic field, all it takes is a constant speed of electric charges in motion (an electric current).

But speed as absolute motion can never be detected (Newton’s 1st law of motion) except the speed of light (special relativity). But the change in speed as absolute acceleration can be. The existence of acceleration implies a force (Newton’s 1st and 2nd law of motion). The existence of magnetic field implies the existence of a magnetic force given by $\vec{F_B} = i \vec{dl} \times \vec{B}$ where i is $\frac{dq}{dt}$, time rate of change of electric charge and $dl$ is the length and direction of the conductor carrying the current. The strength of the magnetic field does follow an inverse square law of distance from the current carrying conductor. But if ever the magnetic field has a speed, it would still depend on the existence of electric charge given by

$$\vec{v} = \frac{1}{q} \frac{dq}{dt} \vec{dl}$$

where $q$ is the electric charge variable. When the magnetic force is zero, the direction of the magnetic field is the same as the direction of the conductor. When the magnetic force is at a maximum (one concentric circle), the direction of the magnetic field is perpendicular to the conductor. There seems to be a generalized force given by

$$F = \frac{E^2}{q^2}$$

6. Jul 19, 2004

### Tom Mattson

Staff Emeritus
Sure it has a speed: 0 m/s

Just like the magnetic field around a wire!

7. Jul 19, 2004

### Antonio Lao

If we are allowed to define an electric dipole moment as $\vec{u}=\frac{1}{2}q \vec{dl}$ then the velocity can also be given by

$$\vec{v} = \frac{1}{q} \frac{d \vec{u}}{dt} + \frac{\vec{u}}{q^2} \frac{dq}{dt}$$

Since the electric dipole moment or its time derivative is zero for a conducting wire, the velocity is zero.

8. Jul 20, 2004

### reilly

The devil is always in the details. The best way to figure out why a steady current generates a static magnetic field is to start with a particulate model of electric current. At this atomic level individual electrons collide with other electrons, and with the metallic lattice of nucleii. Strictly speaking, colliding electrons will generate some radiation, which acts in a resistive fashion. In the steady state, this radiation will, of course, interact with electrons and the lattice, but will not greatly influence the external magnetic field. That is, some computations should show that you can neglect the radiation. What's left is the current of the electrons, which will be generated by some EMF, and under normal circumstances, the distribution of electron speeds will be very narrow. So, to a good approximation, the external electric and magnetic fields will be those generated by uniformly moving electrons -- the magnetic field will, if course, be due to the Lorentz transformed stationary electric field of an electron at rest.(Such transformation of fields is a truth of SR that the anti-Einstein crew completely ignore.)

To be sure, to carry out the program above in full mathematical glory would be a good PhD dissertion topic.

Regards,
Reilly Atkinson

9. Jul 20, 2004

### Antonio Lao

Magnetic field can also come from the partial time derivative of electric flux density called displacement current.

$$\vec{J_d} = \frac{\partial D}{\partial t}$$

10. Jul 20, 2004

### reilly

If you look carefully at the role of the displacement current, you will find that it's role is primarily as a generator of radiation, which, you will find, is quite negliable in the steady state curent problem. In fact, if you think about it, the detailed investigation I suggested necessarily involves the full set of Maxwell's equations. If you want a real challenge, do the problem with Quantum Electrodynamics.
Regards,
Reilly Atkinson

11. Jul 20, 2004

### Tom Mattson

Staff Emeritus
Not so! In a parallel plate capacitor, for instance, the displacement current between the plates and the conduction current in the wire on either side of the plates have precisely the same value.

edit:

I mean "not so" in response to the comment on the displacement current being negligible, not on primarily being involved as a generator of radiation. I agree with that part.

Last edited: Jul 20, 2004
12. Jul 20, 2004

### reilly

Tom -- I'm talking a basically static situation, DC, not AC. DC current does not radiate, AC current does radiate. Think about a capacitor in a dC circuit. No current will flow. But in an AC circuit, current does flow, and you are correct that, as Maxwell noted, the displacement current effectively transmits the current across the gap -- due to the changing electrostatic charge on the capacitor plates. In the atomic level description of a DC current, the displacement current is, for all practical purposes, negliable -- not so for an AC current.

If I recall correctly, Sommerfeld and Pauli had the first approach to the basics of electrical resistance based on Fermi-Dirac statistics. And, this and other early approaches worked with DC, and no displacement current. (See, for example, the old Bible, Joos's Theoretical Physics for details.

Anyway, good point indeed.
Regards,
Reilly Atkinson

13. Jul 21, 2004

### Tom Mattson

Staff Emeritus
If you have a DC source with an RC circuit, a current will flow, and the displacement current between the capacitor plates will be equal to the current in the circuit. Given that, I still don't see how you can say that the displacement current is in general negligible in a DC circuit.

14. Jul 21, 2004

### reilly

Tom -- You are right, but remember that the current in an RC circuit driven by a steady voltage is transient, with a "1/2" life of 1/RC. Etc.So, the current will flow, but it will damp out.
Regards,
Reilly