Question

protons in a uniform magnetic field 0f .3 T follow a circular trajectory with 20 com radius. determine the speed of protons. ( q = 1.6 x 10^-19 and m = 1.673 x 10^-27 K )

Answer 1

Answer

$v=\frac{qBR}{m}\\=\frac{1.6*10^{-19}*0.3*0.2}{1.673*10^{-27}}\\=5.7*10^5$ m/s