Question

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 ?

Answer 1

Larmor radius is given by a formula:

$R = \frac {mv}{qB}$

The energy of electron:

$E = \frac{mc^2}{1-(\frac v c)^2}$

In this case we can express a speed of electron:

$v=c\sqrt{\frac{E-mc^2}{E}} = 534976 \frac m s$

$\frac m q = 10^{-3}$

$R = 668 m$