Electron spin problem

  • Thread starter Benzoate
  • Start date
  • #1
421
0

Homework Statement



the force on a magnetic moment with z component u(z) moving in an inhomogenous magnetic field is given by Equation 7-51: F(z)=u(z)*(dB/dz). If the silver atoms in the stern gerlach experiment traveled horizontally 1 meter through the magnet and 1 meter in a field free region at a speed of 250 meters/second, what must have been the gradient of B(z) , dB(z)/dz, in order that the beams each be deflected a maximum of .5mm from the central , or no field, position?


Homework Equations



F(z)=u(z)(dB/dz)
u(z)=-m(Ag)*u(B)
u(B)=9.27e-24 Joules/tesla
v=250m/s
x=1 meter
distance in field free region=1m
U=u(z)*B

The Attempt at a Solution



F(z)=u(z)*(dB/dz)=> m(Ag)*gravity=-m(Ag)*u(B)*dB/dz.

masses cancel , so I'm left with::
gravity=-u(B)*dB/dz

-gravity/u(B)=dB/dz

to find B(z), I apply the equation U=u(z)*B. U=mgh and I alrealdy know what u(z) is equal to from the first part of the problem. THEREFORE, mgh/u(z)=B(z) . I don't know what relevancy the velocity and the deflected maximum distance served in finding B(z) and dB(z)/dz
 

Answers and Replies

  • #2
421
0
Will someone please help me?
 
  • #3
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,083
18
First of all, ignore gravity - it is small and it affects both spins identically.

The separation between the spin up and spin down atoms comes from the force due to the field gradient. This force is oppositely directed for the oppositely oriented moments. Since you know that the force is proportional to the field gradient, which exists over a region that is 1m long, you can find the z-component of the velocity after this 1m has been traversed (the initial z-component of velocity is 0), in terms of dB/dz and the magnetic moment (don't forget the Lande' g-factor). From here it's a kinematics problem.

Did that help?
 

Related Threads on Electron spin problem

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
956
  • Last Post
Replies
8
Views
2K
Replies
1
Views
5K
  • Last Post
Replies
0
Views
4K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
3K
Top