Determine the magnitude of the force on an electron traveling 3.00×105 horizontally to the east in a vertically upward magnetic field of strength 0.69.

I found the magnitude. I'm just having a hard time finding the direction of the force of the electron?

Question

Determine the magnitude of the force on an electron traveling 3.00×105 horizontally to the east in a vertically upward magnetic field of strength 0.69.

I found the magnitude.  I'm just having a hard time finding the direction of the force of the electron?

yrelhan4 · Answer

you know lorentz force?

yrelhan4 · Answer

can you tell me the formula?

yrelhan4 · Answer

you have to be quick. i have to go.

ghazi · Answer

$$F= q(V * B)$$ you know the value of charge, you know magnetic field and you have the velocity so you can have force now :D

ghazi · Answer

$$F= 1.6*10^{-19}(3*10^{5}*0.69) N$$

anonymous · Answer

I found the force but what would the direction of the force be? Upward, Downward, north, south, or East and West?

ghazi · Answer

well its cross product of velocity and magnetic field

anonymous · Answer

East or west rather.  I chose south and downward and that was wrong

yrelhan4 · Answer

he wants the answer. :/

ghazi · Answer

just apply cross product rule

anonymous · Answer

What do you mean by cross product?

ghazi · Answer

|dw:1362846186168:dw| vector cross product of velocity and magnetic field also you can flemings rule :)