gnome
Dec12-03, 11:11 AM
Given that the electromagnetic power radiated by a nonrelativistic moving point charge q having acceleration a is:
P = \frac {q^2a^2}{6 \pi \epsilon_0 c^3}
(This formula was presented without discussion or explanation in the problems section of my textbook.)
If a proton is placed in a cyclotron with a radius of 0.500 m and a magnetic field of magnitude 0.350 T, what electromagnetic power is radiated by this proton?
This looks like one of those dumb problems where you just plug numbers into a formula, but ... what's a in this situation? We are not given the voltage difference between the dees of the cyclotron. So I thought maybe I can use the exit velocity of the proton of the proton given by
v = \frac {qBr}{m}
Edited:
OK. It turns out that was the right approach. Only I went off talking about angular acceleration, when I should be dealing with centripetal acceleration. So, the following lines of tex are garbage. The corrected stuff is in the next post.
a = \frac{v^2}{r} = \frac{q^2B^2r^2}{m^2r}
a^2 = \frac{q^4B^4r^2}{m^4}
P = \frac {q^2a^2}{6 \pi \epsilon_0 c^3}
(This formula was presented without discussion or explanation in the problems section of my textbook.)
If a proton is placed in a cyclotron with a radius of 0.500 m and a magnetic field of magnitude 0.350 T, what electromagnetic power is radiated by this proton?
This looks like one of those dumb problems where you just plug numbers into a formula, but ... what's a in this situation? We are not given the voltage difference between the dees of the cyclotron. So I thought maybe I can use the exit velocity of the proton of the proton given by
v = \frac {qBr}{m}
Edited:
OK. It turns out that was the right approach. Only I went off talking about angular acceleration, when I should be dealing with centripetal acceleration. So, the following lines of tex are garbage. The corrected stuff is in the next post.
a = \frac{v^2}{r} = \frac{q^2B^2r^2}{m^2r}
a^2 = \frac{q^4B^4r^2}{m^4}