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

The radius of the circular path is

$r=\frac{mv}{Bq}$

The speed of the proton is

$v=\frac{Bqr}{m}$

$=\frac{(0.3)(1.60\times10^{-19})(0.20)}{1.673\times10^{-27}}$

$=5.738\times 10^6 m/s$