# Find the magnetic moment

1. May 28, 2015

### Maximtopsecret

1. The problem statement, all variables and given/known data
Here we have two electrons rotating around z axis with angular speed w in a circle of radius R. They are on the same straight line (have difference in phase π). Find the magnetic moment.

2. Relevant equations
Magn. moment m=(1/2c)*∫dV [ r j ]
3. The attempt at a solution
Current J=dq/dt=qw/π;
m=(qw/2cπ)*∫dr∫dφ r*r*δ(r-R)=qwR2/c
Am I right?

2. May 29, 2015

### Simon Bridge

Is there a way you can verify this?
i.e. have you compared this answer with the magnetic moment for just one electron, same circle and speed?

3. May 29, 2015

### Maximtopsecret

Oh yes, I looked up a similar problem with 1 electron. There J=q/T; therefore m=0.5qwR2
What about 1/c? Is it referred to the fact that I used CGS system of units?

4. May 29, 2015

### Simon Bridge

You should certainly compare like to like.
Where does the c come from in the derivation?

You can also check:
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magmom.html
... and derive the equation for a charge q going in a circle radius R.
Note: the current I is the amount of charge passing a point on the circuit every second.

5. May 30, 2015

### Maximtopsecret

Yes, I checked your link above. From their formula m=I*S it follow for 2 electrons m=qwR2 - no 1/c. This result corresponds to the result of 1 electron.
But initially I tried to apply the general formula for m=(1/2c)*∫dV [r*j]; this formula was used in class where all tasks were done using CGS units.
So, I would stick to my opinion that 1/2c comes from CGS...

6. Jun 6, 2015

### rude man

In SE units the c in the denominator would not be there. You on cgs or ???

Nice fancy equation but you could have just said m = IA, I = current, A = area.

7. Jun 6, 2015

### Maximtopsecret

Yes, I do use CGS here.

8. Jun 6, 2015

### rude man

As real physicists do! (I'm not one of them, I'm afraid)
(I meant SI of course, not SE).
rm