Answer to Question #300261 in Electricity and Magnetism for Sree

Question

An electron moves with an energy of 6.0×10^-16J across a magnetic field of 4.0×10^-3T perpendicularly. Calculate the velocity of electron, force on the electron,orbital radius of the electron and the period of motion.

Answer 1

Answer

velocity of electron is given by

$v=\sqrt{\frac{2E}{m}}\\=\sqrt{\frac{2*6*10^{-16}}{9.1*10{-31}}}\\=0.36*10^7m/s$

Force

$F=qvB\\=1.6*10^{-19}*0.36*10^7*4*10^{-3}\\=2.304*10^{-15}N$

Radius

$R=\frac{mv}{qB}\\=\frac{9.1*10^{-31}*0.36*10^7}{1.6*10^{-19}*4*10^{-3}}\\=0.51cm$

Frequecy

$\omega=\frac{qB}{m}\\=\frac{1.6*10^{-19}*4*10^{-3}}{9.1*10^{-31}}\\=0.70*10^{-8}rad/sec$