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Proton in magnetic field 
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#1
Feb1810, 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 antiparallel 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 


#2
Feb1910, 07:05 AM

PF Gold
P: 1,125

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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