# Finding Magnetic Field for Bending Magnet - Help

1. Jun 19, 2012

### Wolfman29

Hey everyone. I'm new here, but I figured I would jump right in.

First of all, if this forum is anything like the forum I regularly participate in, then if this thread is in the wrong section, go ahead and move it to where it belongs, mods!

I am a young (very young) researcher at FermiLab and, in my project, the problem I am currently encountering is one of calculating the magnetic field necessary to bend a k+ beam 45° through a 38° sector bending magnet.

More in depth, the issue stems from the change in radius of curvature. Our magnet has an arclength of .8m along the central trajectory, however this only forms a 38° sector. (This yields a radius of curvature of 1.206m.)

Due to the nature of k+ decay, the experimental design we are working with dictates a 3.5° pole-face rotation on both sides of the bending magnet in order to achieve focusing. However, because we are only using a 38° magnet, we also need to increase the strength of the magnetic field in order to achieve the correct 45° bend required to meet the 3.5° + 3.5° + 38° = 45° bending requirement.

The problem, then, is calculating the B-field necessary to push our k+ through the magnet with the correct trajectory.

My partner and I have worked for several days on this problem and keep on running into a similar problem.

The best way we have figured out to attempt to solve this problem is working with a reference trajectory. Our reference k+ starts at the left end of the sector bend, angled 3.5° upwards. The particle then travels through the magnet, its trajectory changing by 45 degrees within the magnet, and comes out angled downward 3.5°, thereby achieving a focusing effect downstream.

One basic assumption we made previously, which I am now questioning, is the following: assuming the correct magnetic field for the bend described above, will distance between the entry and exit points be the same as if the same k+ went through the 38° bending magnet straight on with a B-field tuned to bend it exactly 38°?

If that is true, then we may be on the right track - that implies that the chord formed by the entry and exit of the magnet in both cases are the same, despite turning through different angles and having different radii of curvature - we could then do this (assuming the above is correct):

$\left(2\right) \left(1.206\right) \sin(\frac{38°}{2}) = \left(2\right) \left(R\right) \sin(\frac{45°}{2})$

This would allow us to calculate the radius of curvature necessary for the particle travelling through the magnet to achieve a 45° bend, thus allowing us to calculate the B-field necessary to make this bend (using simple centripetal acceleration and the Lorentz force).

I wouldn't be posting this if that worked, though!

I think there may be something we are missing because of the entry and exit angles... but this is where we are stuck.

Just as a reference, I'll throw up a picture I drew in paint (I multiplied all angles by 2 to make it easier to depict).

https://dl.dropbox.com/u/50345936/38%20sector%20bend.png [Broken]

Green is what I want, where the reference particle, while not passing through the center of the magnet, comes out at the exact center, not off axis at all. However, notice that it does enter the bending magnet with an angle of 7 degrees relative to the pole-face, which is required for the focusing effects of the bending magnet to take place. Red is what would happen in the case where the magnet's B-field was tuned to turn it exactly 84 degrees. The question raised above is depicted here - would the paths really intersect at the end of the sector bend assuming they entered at the same location?

Thanks everyone for taking the time to help a noobie out!

Last edited by a moderator: May 6, 2017