What will be the radius of curvature of the path of a 3.0 KeV proton in a
perpendicular magnetic field of magnitude 0.8 T ?
Larmor radius is given by a formula:
R=mvqBR = \frac {mv}{qB}R=qBmv
The energy of electron:
E=mc21−(vc)2E = \frac{mc^2}{1-(\frac v c)^2}E=1−(cv)2mc2
In this case we can express a speed of electron:
v=cE−mc2E=534976msv=c\sqrt{\frac{E-mc^2}{E}} = 534976 \frac m sv=cEE−mc2=534976sm
mq=10−3\frac m q = 10^{-3}qm=10−3
R=668mR = 668 mR=668m
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