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Using Cross Products To Find Electric Field Vector

  • Thread starter kmj9k
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  • #1
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A charged particle moves through a region of space containing both electric and magnetic fields. The velocity of the particle is v = (3.3 x 10^3 m/s) x + (2.7 x 10^3 m/s) y and the magnetic field is B = (0.81 T) z. Find the electric field vector E necessary to yield zero net force on the particle.



Relevant equations: F = qV x B (magnetic force)


I know you should probably use cross products for this problem, but I'm unsure of how to use that method in the context of this problem. They're talking about zero net force, so first I tried to find the magnetic force using F = qV x B and got 2673 x + 2187 y. I'm guessing the electric force should cancel this out. I'm stuck on how to proceed from here, and I don't think I even started the problem correctly. Any help would be greatly appreciated!
 

Answers and Replies

  • #2
HallsofIvy
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Yes, FM= q VxB and FE= qE. But how did you get that value for VxB? Your B is is in the z direction while V is in the xy-plane. The vectors are orthogonal and their cross product is 0. If the values are as stated, then there is no magnetic force on the particle and E must be 0!
 
  • #3
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Hmm I'm more confused than ever. I realize that E will have x, y and z components...but how do I get there? I'm sorry, I think I'll need step-by-step assistance. I think the vectors aspect is the big stumbling block for me.
 
Last edited:
  • #4
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You started from the wrong place - the force law is

[tex] \vec{F} = q(\vec{V}\times\vec{B}) + q\vec{E}[/tex]

Set the LHS to zero and it's elementary (?)
 
  • #5
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Oooh I see now. I wasn't thinking of the cross products of vectors correctly (ie, I was thinking of the magnetic field as a scalar). Remembering that it's 0.81 *Z* I got -2187x + 2673y for the electric field using the equation above and got it right.

Thanks so much to all the posters for pointing me in the right direction!
 

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