magnetic moment of charged particles

Feb 2010
5
0
im having difficulty understanding how physicists measure the magnetic moment of charged particles. Some articles i have read state that a charged particle will turn in an external magnetic field so that its magnetic poles align with the opposite poles of the external magnetic field. But, other articles state that charged particles are niether attracted to nor repelled by the poles of an external magnetic field. Clearly, both of these cant be true. Do physicists know for sure that charged particles possess magnetic dipoles or is this purely theoretical? im very confused.
(Thinking)
 
Dec 2009
209
121
Yes and No

im having difficulty understanding how physicists measure the magnetic moment of charged particles. Some articles i have read state that a charged particle will turn in an external magnetic field so that its magnetic poles align with the opposite poles of the external magnetic field. But, other articles state that charged particles are niether attracted to nor repelled by the poles of an external magnetic field. Clearly, both of these cant be true. Do physicists know for sure that charged particles possess magnetic dipoles or is this purely theoretical? im very confused.
(Thinking)
It's not that a charged particle has or is a magnetic dipole, a magnet by its very nature is a dipole in that it has two poles, a N and a S pole, hence the label of magnetic dipole.

Charge is influenced by a magnetic field under certain conditions. If the charged particle is simply at rest within a magnetic field it will have no force applied upon it by the field.

If the charged particle is moving through the magnetic field at any angle other than moving directly parallel with the field lines, then the magnetic field acts to deflect the charge in a direction that can be found by the cross product of VxB or BxV, its one of those two I can't remember. If you look up the Lorentz Force law you will find that the force acting upon a charged particle is f = q(E + BxV), again it may be VxB I can't remember for sure.

As you can see from the Lorentz force law, the force that acts upon some amount of charge, q, is due to either or both, an electric field it may be in or by moving with some velocity, V, through a B field. An E field applied to some charge will apply a force upon that charge whether the charge is moving or not, but for the B field to apply a force upon the charge it must be moving with some velocity, V, through that field. Also since the force applied upon a moving charged particle by some B field is directly related to the cross product of that field and the velocity vector for that charged particle, then if the charge is moving parallel with the field, the cross product is zero and no force is applied upon the charge. It must be moving with some non-zero velocity at any angle other than parallel with the field for the field to apply a displacement force upon that charge.

So when you read in one book that a B field can apply a force upon some amount of charge, it can, as long as that charge is moving through the field at some non parallel angle to that field.

When you read in a different book that a B field has no effect on some amount of charge, well that too is true. Again as long as the charge is not moving within this B field, or is moving in a direction that is parallel with the field lines, then the B field applies no force upon that charge.

So both books are correct and I believe if you go back and look at what each book has to tell is that the book that claims a B field can apply a force upon some charge, they should state that the charge is moving through that B field. The book that said the B field will apply no force upon some charge should also state that the charge is either at rest or moving parallel with the field lines.

I hope this helps.

Many Smiles,
Craig (Smile)
 
Feb 2010
5
0
Thanks craig. thats cleared up some confusion for me. I had been visualising the charge as being an actual dipole and couldnt understand why it wasnt pulled towards one pole or the other in a B field. im a little wiser now (i think. Ha Ha). much appreciated.(Hi)