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