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In a particle accelerator... 
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#1
Jan405, 09:40 PM

P: 42

Heres one more question
In a particle accelerator, an alpha particle with a mass of 6.64 x10^27 kg is moving with a speed of 2.50x10^7 m/s. It is moving perpendicularily through a magnetic field of intensity 0.150 T. Using appropriote equations and method find the radius of curvature of its path. I'm completly stumped on this HW question due tomorrow, I'm looking through my notes for an equation for this. Also I have no idea how to find the radius of the curvature. PLEASE HELP. 


#2
Jan405, 09:59 PM

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Daniel. 


#3
Jan405, 10:29 PM

P: 42

umm...
whats the formula for the lorentz magnetic force? I thought it was like... F=qv+B I dont see how that helps. Could you help me out a bit more? I need to get this one before tonight. 


#4
Jan505, 12:37 AM

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P: 905

In a particle accelerator...
The point is that it is a force prependicular to velocity. That means that it compels the particle to travel in a circle. Equate it to the formula for the centrifugal force.



#5
Jan505, 07:24 AM

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[tex] \vec{F}_{mag,L}=q\vec{v}\times\vec{B} [/tex] The centripetal force is: [tex] \vec{F}_{cp}=\frac{m\vec{v}^{2}}{r^{2}}\vec{r} [/tex] Daniel. 


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