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Calc. field gradient of Stern-Gerlach magnet

  1. Sep 4, 2009 #1

    From Townsend "Modern Approach to Quantum Mechanics", problem 1.1:

    "Determine the field gradient of a 50-cm long Stern-Gerlach magnet 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=1500K. Assume the detector is 50 cm from the magnet. Note...the emitted atoms have average kinetic energy 2kT. The magnetic dipole moment of the silver atom is due to the intrinsic spin of a single electron.

    Use Gaussian units.

    2. Relevant equations


    [tex]F_z=\mu_z \del B[/tex]

    3. The attempt at a solution

    Deriving an equation:

    Assume the electron is moving in the x+ direction towards the detector, and that it is deflected in the [tex]\pm z[/tex] direction. It feels a force [tex]F_z=\mu_z \del B[/tex] for a time [tex]t=\frac{d_1}{v}[/tex] where d1 is the length of the magnet along the x-axis.

    Using Newton's laws, we can derive the acceleration, velocity, and thus displacement of the particle:

    [tex]a_z=\frac{\mu_z \nabla B}{M}[/tex] where M is the mass of a silver atom.

    [tex]v_z = a_z t=\frac{\mu_z \nabla B d_1}{M v}[/tex] where v is the magnitude of the total velocity of the atom and [tex]v_z[/tex] is the velocity in the z-axis direction.

    [tex]s_z^\prime=\frac{a_z t^2}{2}=\frac{\mu_z \nabla B d_1^2}{2Mv}[/tex]

    [tex]s_z=\frac{v_z d_z}{v}+s_z^\prime=\mu_z (\nabla B) (\frac{d_1^2 + 2d_1 d_2}{2Mv^2})[/tex] where [tex]s_z[/tex] is the total deflection in the z-axis direction at the detector.

    Solving the average kinetic energy for velocity...

    [tex]2kT=\frac{Mv^2}{2} \implies v^2=\frac{4kT}{M}[/tex]

    Now we can solve...

    [tex]s_z=\frac{\mu_z \nabla B (d_1^2 + 2d_1 d_2)}{2Mv}=\frac{\mu_z \nabla B (d_1^2 + 2d_1 d_2)}{2M(\frac{4kT}{M})}=\frac{\mu_z \nabla B (d_1^2 + 2d_1 d_2)}{8kT}[/tex]

    [tex]\therefore \nabla B=\frac{8kT s_z}{u_z (d_1^2 + 2d_1 d_2)}[/tex]

    Here is my problem...

    I understand that [tex]\nabla B[/tex] means to differentiate B according to the rules of vector calculus, but I'm not sure how to solve this equation for B as a number.

    Do I need to integrate both sides? If so, with respect to what? Space? 3D space or just 1D space?

    I can plug in all the numbers on the right hand sand, but I'm not sure what to do with it after that.

    All help is much appreciated.
  2. jcsd
  3. Sep 4, 2009 #2


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    Homework Helper
    Gold Member

    The potential energy of a dipole μ in a magnetic field is

    [tex]U = - \vec{\mu}\cdot \vec{B}[/tex]

    The force on this dipole is

    [tex]\vec{F}=-\vec{\nabla}U=+\vec{\nabla}(\vec{\mu}\cdot \vec{B})[/tex]

    After applying some vector identities and exploiting the symmetry of the design, one ends up with the product of the magnetic moment and field gradient:

    [tex]\vec{F}=F_{z}\hat{k}=\mu \frac{\partial B_{z}}{\partial z}[/tex]

    As always, μ takes the two allowed magnetic moment values


    Start from there.
  4. Sep 4, 2009 #3
    Thanks for the response.

    I realized this morning that the problem is asking for the "field gradient" which actually is [tex]\nabla B[/tex], so I don't need to isolate B. I have the formula for the answer there and I just need to plug in the values.

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