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Proton in magnetic field

by TheDoorsOfMe
Tags: field, magnetic, proton
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TheDoorsOfMe
#1
Feb18-10, 10:29 PM
P: 46
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

F=q(v x B)


3. 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.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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danago
#2
Feb19-10, 07:05 AM
PF Gold
P: 1,125
Quote Quote by TheDoorsOfMe View Post
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

F=q(v x B)


3. 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.
Id take your first answer, zero.

By the looks of it, you have taken cross product to be the same thing as scalar multiplication, which is not true. Have a look at these links for more about the vector cross product:

http://cnx.org/content/m13603/latest/
http://tutorial.math.lamar.edu/Class...ssProduct.aspx

If you were to define the velocity vector as <0, 3.01x10^7, 0> m/s and the magnetic field vector as <0, -5, 0> T, and then proceeded to apply the cross product operation, you would end up with a zero vector, which would agree with your first answer.


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