Electron accelerator build

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To determine the magnetic field strength required for a 20kV @ 60mA electron beam to navigate a 90° turn, it's essential to know the speed of the electrons, which can be calculated using the energy equation E = V * e. The calculated speed is approximately 8.4 x 10^7 m/s, without considering relativistic effects. The dimensions of the magnet poles must be optimized to enclose the beam while maintaining a high magnetic field, possibly necessitating multiple magnets along the path. The radius of curvature and the centripetal force required for the desired path can then be derived using Lorentz's force law. Understanding these parameters is crucial for the successful design of the electron accelerator.
Rob Hoff
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How strong of a magnetic field do I need to get a 20kv @ 60ma beam around a 90° turn? / Do I need more information to find out?
 
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The electrons go around the corner independently. They will need to travel 90° around a circular path. How much time do you have? What radius do you require?

You will need to determine the dimension of the poles of your magnet. The poles will need to be big enough to enclose the electron beam during the turn, yet small enough to have a high field.

The optimum solution may be to have several magnets along the path. That may explain why accelerators are usually circular.

http://en.wikipedia.org/wiki/Lorentz_force
http://en.wikipedia.org/wiki/Gyroradius
 
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First you need to know the speed of the electrons. Applying a voltage to a free moving electron will impart kinetic energy to it.

E = V * e = 20kV * 1.6*10^-19 = 3.2*10^-15

and 1/2 * m v^2 = E

so v = (E*2/ m)^0.5 = 8.4*10^7 m/s

Note that I didn't factor in relativity. I'll leave that for you to figure out.

Now you can pick a radius of curvature that you would like, the centripetal force needed for that curvature, and the magnetic field needed to achieve that force from Lorentz's force law and the speed of the electron in your experiment.
 
Hey thanks you guys, and sorry if that sounded like I wanted you guys to do my homework. It really wasn't meant that way. Thanks for the equations. :)

And Baluncore, I have as much time as it takes!
 
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I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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