- #1

TheDoorsOfMe

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## Homework Statement

Proton moving in -y direction with a magnetic field pointing in the +y direction with a manitude of 5.0 T. The proton has a velocity of 3.01x10^7, What is the magnitude and direction of the magnetic force.

## Homework Equations

F=q(v x B)

## The Attempt at a Solution

The answer I think is either 0 or 1.5x10^8.

I say it might be zero cause the cross product of anti-parallel vectors is zero right? Or I say its 1.5x10^8 because I just plugged into the above equation a solved. Which one is it? Or is it neither.