# Moving charge in a magnetic field, magnetic-magnetic interaction?

1. Nov 23, 2010

### henxan

When a charged particle moves through a magnetic field, the field excerts a force in the normal-direction (90 degrees)..

Is this caused by magnetic-magnetic interaction? I.e. a charged moving particle sets ut a circular magnetic field, is it this field which interacts with the magnetic field?

Or, is it a megnetic electric interaction?

2. Nov 23, 2010

### rl.bhat

It is magnetic-magnetic interaction. Force of magnetic field interaction is maximum when they are parallel. When the charge is moving in the perpendicular direction to the external magnetic field, concentric magnetic field is parallel to the external field. So the force on the charge is perpendicular to both the external field and the direction of the electron.

3. Nov 24, 2010

### henxan

But, does not the forces, by the external homogenous field on the concentric field, cancel out?

Do you know of any illustrations which can clarify this?

4. Dec 7, 2010

### henxan

I never got a good enough answer to my enquiry. I will try to express the question more clearly:

Figure 1. The magnetic field of a moving charge segment. This is in principle equal to the field set up by a moving charge. Here we have a moving positive particle traveling to the left. The magnetic field set up by the charged particle is illustrated for 3 concentric fields.

Figure 2. An object enters a homogenous magnetic field.

What i was wondering about was the case when we combine figs.1 and 2.

• The charged particle is setting up a magnetic field
• The charged particle moves into the magnetic field
• The charged particle moving in the magnetic field is subjected to a force
So my question, again, is:
Why is the particle bending?
1. The magnetic field excerts a force on the moving point charge, in a magnetic-charge interaction.
2. The magnetic field excerts a force on the magnetic field SET UP by the moving particle, in a magnetic-magnetic interaction.
Which is correct? 1 or 2?

5. Dec 7, 2010

### Bob S

You can have it either way. The magnetic-magnetic interaction approach requires an integration over the magnetic field energy density of the combined vector fields. On one side, the vector fields add, and on the other side, they subtract, as rl.bhat pointed out. So using the Lorentz qv x B force method is much simpler. If you do a Lorentz transform into the charged particle rest frame, The static magnetic field transforms into an electric field, so all the magnetic fields disappear..

Bob S

6. Dec 8, 2010

### henxan

at: Bob S
Thanks! I sure hope it is possible to show a magnetic-magnetic interaction :).. but how??

I have inserted an image:

At the top left: A proton moving through a homogenous magnetic field. The red, poorly illustrated, circles with dots and crosses, illustrates the cross section of the magnetic field created by the particle.
Top right: The same as in the top left image, but now our angle of view is 90degrees. The particle is moving into the screen.
Bottom image: To get an understanding of the top right image, lets illustrate the concentric magnetic field by 8 magnets. If we look at the forces excerted on the individual magnets, they only give a torque on the individual magnets. For the whole, they all add up.

So, how can there be a force directed left?

7. Dec 8, 2010

### Bob S

In the attached thumbnail, a current Ix out of the page (in the x direction) crossed into the magnetic field By (in the y direction) produces a force Fz in the z direction) (see x y z coordinate system in thumbnail).

Calculate the magnetic energy density everywhere for a unit length conductor

$$W = \frac {1} {2\mu_0} \int B_{zw}^2 + (B_y+B_{yw})^2) dy dz$$

where Byw and Bzw represent the field due to the wire. In particular, note that there is a higher energy density on the right hand side, where the two magnetic fields are parallel, than on the left side, where the two magnetic fields are antiparallel. Note that there is a gradient along z in (By + Byw)2 at z = 0.

$$F_z = \frac {dW} {dz} = \frac {1} {2\mu_0} \int \frac{d} {dz}(B_{zw}^2 + (B_y+B_{yw})^2)dy dz$$

Bob S

#### Attached Files:

• ###### v x B force.jpg
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Last edited: Dec 8, 2010