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Magnetic Fields and Acceleration

  1. Oct 8, 2006 #1

    A particle with mass [tex]1.81 \times 10^{-3}\;{\rm kg}[/tex] and a charge of [tex]1.22 \times 10^{-8}\;{\rm C}[/tex] has, at a given instant, a velocity [tex]\vec{v} = (3.00 \times 10^{4}\;{\rm m}/{\rm s})\hat{j}[/tex].

    What is the magnitude of the particle's acceleration produced by a uniform magnetic field [tex]\vec{B}=(1.63\;{\rm T})\hat{i} + (0.980\;{\rm T})\hat{j}[/tex]?

    I use the equation F = qv (CROSS) B.

    I rewrite this as F = q*v*B*sin().

    I plug in to get F = (1.22*10^(-8))*(3.00*10^(4))*(1.90) and get (6.96*10^(-4)).

    I divide this by the mass, but my answer is wrong.

    Obviously my procedure is not correct. What's up?
    Last edited: Oct 8, 2006
  2. jcsd
  3. Oct 8, 2006 #2


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    Homework Helper

    You forgot the sine term.
  4. Oct 8, 2006 #3
    So the angle is not 90 degrees?
  5. Oct 8, 2006 #4


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    Homework Helper

    Most certainly not. You have two vectors, one with only j component, but the other has i and j components. You could just compute the cross product, or you could figure out the angle. Or you could break the problem into two components: B parallel to v and B perpendicular to v.
  6. Oct 9, 2006 #5
    I just figured it out. Thanks!
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