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
v=qBRm=1.6∗10−19∗0.3∗0.21.673∗10−27=5.7∗105v=\frac{qBR}{m}\\=\frac{1.6*10^{-19}*0.3*0.2}{1.673*10^{-27}}\\=5.7*10^5v=mqBR=1.673∗10−271.6∗10−19∗0.3∗0.2=5.7∗105 m/s
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