1.

What is the radius of the path of an electron (mass 9xx10^(-31)kg and charge 1.6xx10^(-19)C) moving at a speed of 3xx10^(6)m//s in a magnetic field of 6T perpendicular to it? What is its frequency ? Calculate its energy in keV. (1eV=1.6xx10^(-19)J)

Answer»


Answer :R = 52 cm, v = 910 kHz, E = 47 KEV


Discussion

No Comment Found

Related InterviewSolutions