# Determing field gradient of Stern-Gerlach magnet

1. Oct 5, 2012

### jay ess

1. The problem statement, all variables and given/known data
Determine the field gradient of a 50-cm-long Stern-Gerlach magnet (d1) that would produce a 1-mm separation at the detector between spin-up and spin-down silver atoms that are emitted from an oven at T=1500. Assume the detector is located 50 cm from the magnet (d2). Note: While the atoms in the oven have average kinetic energy 3kBT2, the more energetic atoms strike the hole in the oven more frequently. Thus the emitted atoms have average kinetic energy 2kBT2, where the kB is the Boltzmann constant. The magnetic dipole moment of the silver atom is due to the intrinsic spin of the single electron. The Bohr magneton, e[STRIKE]h[/STRIKE]/2mec≈5.788×10-9eV/G

2. Relevant equations
Fz=μ⋅∂B/∂zμ∂Bz/∂z

3. The attempt at a solution
The setup: the magnetic field gradient is oriented in the z-direction while the initial velocity of the atoms is in the x-direction.
The separation between the silver atoms is 1-mm, therefore, the distance traveled in the z-direction by the silver atoms is 0.5 mm=5x10-2cm which I call dz.
∂Bz/∂z=∇B
Fz=maz=μ∂Bz/∂z=μ∇B
Average kinetic energy of the particles: 1/2mvx2=2kBTvx=√(4kBT/m)
vx=d1/t1→t1=d1/vx=d1√(m/4kBT)
I know that while the atoms are in the field gradient they will be experiencing a force that causes them to move in the z-direction (I'm just sticking with the positive z-direction for simplicity) and once they're out of the field gradient they will be traveling at a constant velocity in both the x- and z-directions. I also know that I need to somehow relate all of this to the kinematic equations but I'm kind of at a loss right now.
vz=azt1 (since there's no initial velocity in the z-direction and acceleration is constant).

2. Oct 5, 2012

### TSny

Hi, Jay ess.

You've got a good start. Can you find an expression for the z- deflection of an atom while traveling through the magnet in terms of t1? As the atom leaves the magnet it will have a z-component of velocity as you indicated as well as the x-component. Can you use these two components of velocity to determine the direction that the atom is traveling as it leaves the magnet? Then see if you can find the additional z-deflection as it travels to the detector.

3. Oct 6, 2012

### jay ess

That's the bit that's giving me the most trouble, relating the kinematics equations in all of this mess. I know that since the atom started out at the origin and without any initial velocity in the z-direction that I can find the position of the particle as it is leaving the gradient (let's call it dz1) by dz1=1/2az*t12=1/2μ∇B/m(m/4kBT)=μ∇B/(8kBT). But, after that I'm still kind of lost because everything else I try just seems to fall apart.

4. Oct 6, 2012

### TSny

That look's good except for leaving out a factor of d12.
OK. You just have to calculate the additional displacement in the z direction after the atom leaves the magnet. This will be determined by the z-component of velocity and the time to get from the magnet to the detector. Since no force acts on the atom as it travels to the detector, the z-component of velocity will be constant and will equal whatever the z-component of velocity was at the instant the atom emerged from the magnet.