Question

A) Determine the magnitude of the force on an electron traveling 6.35×10^5 m/s horizontally to the east in a vertically upward magnetic field of strength 0.75 T.

B) Determine the direction of the force on an electron.

     1. to the North

     2. to the South

     3. to the East

     4. to the West

Answer 1

The force experienced by the charged particle is given by : F = q ( x 8) where, q= 1,6 X 10-19 C No . 6.35 x 109 m/s î Ř - 0.75 T Ê A) F - (1.6x10-19)/(6.35x10°) x 0.75 Â ] 7,62 x 1014Ne êx ) [Cross product] => Ě 7.62x10-14 N (-) :. Fl17.62x10-14 N ] B) Direction is ŷ , i.e. towards south e to the south is the correct option. 3