- #1
cseil
- 28
- 0
[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
Kp = 1/2 mv^2
v = qBr/m (uniform circular motion).
So
Kp = 1/2 q^2B^2r^2/m
proton: Kp = \frac{1}{2} \frac{e^2B^2r^2}{u}
particle 1: K = \frac{1}{2} \frac{e^2B^2r^2}{2u} = Kp/2
particle 2: K = \frac {1}{2} \frac{4e^22B^2r^2}{4u} = Kp
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
Kp = 1/2 mv^2
v = qBr/m (uniform circular motion).
So
Kp = 1/2 q^2B^2r^2/m
proton: Kp = \frac{1}{2} \frac{e^2B^2r^2}{u}
particle 1: K = \frac{1}{2} \frac{e^2B^2r^2}{2u} = Kp/2
particle 2: K = \frac {1}{2} \frac{4e^22B^2r^2}{4u} = Kp
Is that right?
Thank you
Last edited by a moderator: