Proton in magnetic field

  • #1

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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
danago
Gold Member
1,122
4

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.

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/Classes/CalcII/CrossProduct.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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