Question

protons in uniform magnetic field of .3T follow a circular trajectory with 20cm radius. Determine the speed of the protons. (q=1.6×10 -¹⁹ C and

m=1.673×10 -²⁷ kg)

Answer 1

Answer

Speed of protons

$V=\frac{qBr}{m}\\=\frac{1.6*10^{-19}*0.3*0.20}{9.1*10^{-27}}\\=10.50*10^5m/sec$