# Force of magnetic moment

1. Nov 25, 2006

### Mindscrape

I feel really stupid that I can't solve this problem, but here goes...

The force on a magnetic moment $$\mu_z$$ in a nonuniform magenetic field $$B_z$$ is given by
$$F_z=\mu_z \frac{dB_z}{dz}$$
If a beam of silver atoms travels a horizontal distance of 1 m through such a field and each atom has a speed of 100 m/s, how strong must the field gradient $$dB_z / dz$$ be in order to deflect the beam 1 mm?

I guess I am not really sure where to start. I would have used potential energy to solve this, and I don't know how forces are going to make sense because even with the force gradient it doesn't seem like I can come up how far the beam should deflect. The force of gravity and the force of the moment will be in equilibrium and will have to equal each other. So my thought would be that I could just say:

$$\frac{mg}{\mu_z}= \frac{dB_z}{dz}$$

That doesn't incorperate any of the information given though, so I must be missing something.

Another thing I am not really sure about is extracting the quantum numbers out of silver. n=4 so l=3,2,1,0 and $$m_l$$=3,2,1,0,-1,-2,-3

Since
$$\mu_z = \frac{-e}{2m_e} L_z = \frac{-e \hbar}{2m_e}m_l$$

Which value of $$m_l$$ would I use?

Last edited: Nov 25, 2006
2. Nov 25, 2006

### nrqed

If that was the case, there would be no deflection!

I am not completely sure, but it seems to me that here the force of gravity is probably completely negligible.

either way, what you do is to basically use the high school equations of projectile motion. Alonx x, you simply have $x = v_{xi} t$ to find the time. Then go along y and use $y= {1 \over 2} a_y t^2$ to find the acceleration. Plugging this in F=ma you find the force.

I thought that in that situation, the silver atoms had no orbital excitation (l=0). And that it was really the spin that was providing the dipole moment, so that in that case $m_s = 1/2$. But don't take my word for it

Hope this helps

Patrick

3. Nov 26, 2006

### OlderDan

4. Nov 26, 2006

### Mindscrape

Ahh, okay, I think I misinterpreted the problem. Both responses were helpful, thank you.