Determine the magnitude of the force on an electron traveling 7.05×105m/s horizontally to the east in a vertically upward magnetic field of strength 0.55T .

I got Fmax = 6.2×10−14 N

now I am confused on this part

Determine the direction of the force on an electron.
to the North
to the South
to the East
to the West

Question

Determine the magnitude of the force on an electron traveling 7.05×105m/s horizontally to the east in a vertically upward magnetic field of strength 0.55T .

I got Fmax = 6.2×10−14 N

now I am confused on this part


Determine the direction of the force on an electron.
	 to the North
	 to the South
	 to the East
	 to the West

roadjester · Answer

$$\huge \vec F = q\vec v X\vec B$$
It's a cross product so you need to use the right-hand rule.
The direction your thumb points is the direction of the force. on the electron.

anonymous · Answer

I did that and I got south but that is the incorrect answer, I am ppointing my index finger to the east horizontally and the 3 last fingers are pointing upward, when I do this the thumb is pointing toward me

roadjester · Answer

You're forgetting one minor detail. Electrons are negative so the direction flips. It's supposed to be North. I kind of thought that was impllied. Sorry about the confusion.

Your thumb points towards you so thus it is the force on the "electron" and therefore you need to flip it in the other direction.

anonymous · Answer

ohh thank you very much that makes more sense ahaha =)