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How to create a magnetic field gradient

by smboyle
Tags: magnetic field
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smboyle
#1
Jul13-11, 10:01 AM
P: 4
I'm working at a university to build a low-energy electron detector and it requires that we construct a magnetic field gradient. We know, through computer models, what the gradient should be, but we don't know how to make it.

We would rather use rare-earth magnets as opposed to an electromagnet, but at this point we open to ideas.

So, any ideas?
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timthereaper
#2
Jul13-11, 10:26 AM
P: 343
I guess it all depends on how fine you'd like to set up your gradient. I guess you could use rare-earth magnets of differing strengths, but that would probably be less effective and more costly than just using a series of solenoids. The advantage of using solenoids is that you can control the strength of the field inside by controlling the current in the wire. You can also control the fineness of the field gradient by the number of solenoids. Unless there's something I'm missing, I would suggest going that route.
smboyle
#3
Jul13-11, 10:40 AM
P: 4
Well, how would we create it with electromagnets? We understand how they're made, but only in the context of one. Would you have to make multiple solenoids with varying current in some geometric configuration?

chrisbaird
#4
Jul13-11, 10:56 AM
P: 617
How to create a magnetic field gradient

A magnetic field gradient can be made by taking the two ends of a horseshoe magnet and making one end flat and the other pointy, like in the Stern-Gerlach experiment. But this may not be what you are looking for.
smboyle
#5
Jul13-11, 11:05 AM
P: 4
That's a good point, but I'm not sure how to make that with our design, it requires a gradient with a 2*(x^2 + y2) dependance if the image is like this:


y
|
|
|
| bent by gradient field in this x-y plane
|
|_______________ x
/\
\
\
^electron enters through collimator
Attached Thumbnails
tripleDetectors.jpg  
timthereaper
#6
Jul13-11, 02:14 PM
P: 343
I think see what you're trying to do. I assumed you were talking about a linear gradient with a constant cross-section (don't ask me how I came to that). I think chrisbaird was on to something with permanent magnets with different shaped ends.

I'm getting confused on the correlation between the graph and the equation you posted. The way I understand it, the electrons are traveling along the y-axis and then getting deflected by a magnetic field with field lines pointing in the z-direction. The strength of the field varies with 2(x^2 + y^2) or 2*r^2. If I'm understanding this correctly, the equipotential lines for the magnetic field are pretty circular. In that case, you could have a small, powerful solenoid near the opening of the collimator. The force from the magnetic field would decrease with r^2, which would fit your experiment.
smboyle
#7
Jul13-11, 02:30 PM
P: 4
Sorry, I should have explained the picture a little better. The equation corresponds with the arbitrary x and y axes seen in the picture. The colored lines are electron trajectories corresponding to different energies entering a magnetic field at, I believe, a 30 degree angle.

The horizontal black line denotes the y value of where the electrons enter the magnetic field and the apex of the collimator indicates where on the x axis the electrons enter.
You'll also see a circular line starting from that initial y value, any point inside that circular area and underneath that horizontal black line has a B-field of zero.

If we were to use a solenoid, it would be placed at the top right of this x-y plane, where we want the B-field to be strongest.
timthereaper
#8
Jul13-11, 03:10 PM
P: 343
Okay, so I wasn't too far off. I think the solenoid (or even a bar magnet) might work then. Orient the north pole of the solenoid/magnet in the positive z-direction (assuming your electrons are traveling in the positive y-direction) at a location in the upper right quadrant and I think it might work.

I would like some confirmation from some of the other PFers out there to see if I'm overlooking anything.


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