# Is the wire necessary to create a B field?

1. Sep 7, 2010

### idea2000

Hi,

In the Purcell book on E&M that we used during college (berkeley physics course volume 2), the b field of a current carrying wire was explained as the result of a relativistic transformation between two frames of reference...

The explanation in purcell seems to require the existence of the wire, but is the wire actually needed to create the B field? Or can any moving point charge in vacuum also create a B field? (And if so, is there an alternative explanation for where the B field comes from besides one given in purcell?)

2. Sep 7, 2010

### Petr Mugver

A single particle moving does create a magnetic field, but this one is of order v/c times the electric field, so you only notice it for very fast particles, or when their number is big.Try googling "Liénard-Wiechert potentials". In most elementary e.m. textbooks the Maxwell equations are derived from other equations (for example the B produced in wires, and then they consider "Infinitesimally small wires" to "deduce" the Maxwell equations). This is historically how magnetism has developed, but from a modern standpoint the Maxwell equation are the fundamental equations, from which you deduce all others equations (including the B field produced by wires).

3. Sep 8, 2010

### bcrowell

Staff Emeritus
Purcell is a great book. I learned E&M from it, and it was a wonderful experienece. Too bad it used cgs. If they'd publish an mks edition maybe it would be used more widely. Realistically, nobody has a multimeter that reads in statvolts and abamps.

No.

Yes.

The point of using a wire rather than a point charge is only that the field of a wire has a higher degree of cartesian symmetry than the field of a point charge.

To demonstrate that we need the concept of a B field, all we need is *one* situation where we can show that a pure E field is insufficient. If you believed that all there was was an E field, and then you transformed to a frame of reference that was in motion relative to your original frame, you would want to know whether you could still successfully describe everything using only an E field. To convince you that that *won't* work, all we need is one example.

Here's a treatment in the spirit of Purcell but without the vector calculus: http://www.lightandmatter.com/html_books/0sn/ch11/ch11.html [Broken]

Last edited by a moderator: May 4, 2017
4. Sep 8, 2010

### nicksauce

Many stars have large magnetic fields. Do stars have wires?

5. Sep 9, 2010

### idea2000

Thanks for the links and recommended searches...I did try googling Liénard-Wiechert potentials and I found a lot of useful information. However, I wanted to make sure that I competely understand what is going on...

The B field of a moving point charge was derived entirely from Maxwell's independently by Liénard (1898) and Wiechert (1900), which predates relativity (1905). There is no way to take a moving point charge in vacuum and derive it's B field from it's E field by using only relativity. We can only accept that the B field of a moving point charge in vacuum comes from Maxwell's equations, which were derived based on observation. Is this correct?

And, currently, is any deeper understanding of where the B field of a moving point charge in a vacuum comes from other than just based on Maxwell's (and observation)? Thanks!

Last edited: Sep 9, 2010
6. Sep 9, 2010

### Petr Mugver

Well, you can always move to the frame in wich the particle is at rest. If the particle moves in uniform motion, this frame is inertial. Since the particle is at rest, there is no B field in this frame, but only the Coulomb E field. Then you can transform the fields into the original frame, and you'll find a B field.

7. Sep 9, 2010

### bcrowell

Staff Emeritus
I'm not sure I'm completely following you here, but this sounds incorrect to me. If you know that a particle has a certain E field in a frame where the particle is at rest, then you can certainly find its B field in another frame, simply by using relativity.

Yes, it comes from relativity, and Purcell's treatment is a good example of how to do that without assuming all of Maxwell's equations. Purcell's argument shows that if you only knew the electric parts of Maxwell's equations, then using relativity you could prove that it was necessary to add in the magnetic parts of Maxwell's equations.

8. Oct 2, 2010

### idea2000

Hi,

Thanks for all your posts, I think i finally understand what is going on...

I had one last question, is it possible to do this for a photon as well? Meaning, is it possible to use relativity to transform the sinusoidal E field of a photon to get its corresponding B field? Thanks for any help in advance...

9. Oct 3, 2010

### bcrowell

Staff Emeritus
If you transform the E and B fields of a wave into another frame, you get E' and B' that describe the wave in that frame. For example, if you transform into a frame moving along the direction of propagation of the wave, the E' and B' will describe a Doppler-shifted wave.

10. Oct 3, 2010

### starthaus

There is no difference between (E,B) "for an electron" and (E,B) "for a photon". There is just (E,B).