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cseil
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[Note by mentor: This thread does not use the homework template because it was moved here from a non-homework forum.]
Hello,
I have a problem with this exercise. I don't know if my solution is right! Could you check it, please?
A proton with kinetic energy Kp is moving in a uniform magnetic field on a circular orbit.
I have to calculate the energy of two particles moving on the same orbit knowing that:
particle 1 : q=+e, m=2u
particle 2: q=+2e, m=4u
[itex]Kp = 1/2 mv^2[/itex]
[itex]v = qBr/m[/itex] (uniform circular motion).
So
[itex]Kp = 1/2 q^2B^2r^2/m[/itex]
proton: [itex]Kp = \frac{1}{2} \frac{e^2B^2r^2}{u}[/itex]
particle 1: [itex]K = \frac{1}{2} \frac{e^2B^2r^2}{2u} = Kp/2[/itex]
particle 2: [itex]K = \frac {1}{2} \frac{4e^22B^2r^2}{4u} = Kp[/itex]
Is that right?
Thank you
Hello,
I have a problem with this exercise. I don't know if my solution is right! Could you check it, please?
A proton with kinetic energy Kp is moving in a uniform magnetic field on a circular orbit.
I have to calculate the energy of two particles moving on the same orbit knowing that:
particle 1 : q=+e, m=2u
particle 2: q=+2e, m=4u
[itex]Kp = 1/2 mv^2[/itex]
[itex]v = qBr/m[/itex] (uniform circular motion).
So
[itex]Kp = 1/2 q^2B^2r^2/m[/itex]
proton: [itex]Kp = \frac{1}{2} \frac{e^2B^2r^2}{u}[/itex]
particle 1: [itex]K = \frac{1}{2} \frac{e^2B^2r^2}{2u} = Kp/2[/itex]
particle 2: [itex]K = \frac {1}{2} \frac{4e^22B^2r^2}{4u} = Kp[/itex]
Is that right?
Thank you
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