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Biot Savart, Electrons, and Cirucular Currents

  1. Oct 12, 2009 #1
    A circular wire ring of diameter 18 cm carries a current of 5.0 A directed as indicated in Figure P.51. A moving electron just happens to be passing through the neighborhood. When the electron is at the center of the circular ring and moving at speed 2.50 multiplied by 106 m/s in the direction indicated in the sketch, find the acceleration of the electron.
    p20-51.gif

    Equations I found relevant:
    The Biot Savart Law:
    B=([tex]\mu[/tex]/4[tex]\pi[/tex])(I)([tex]\int[/tex]dl x r / r^2)

    F=qv x B

    Biot Savart solves for the magnetic field, they I can plug this into the force equation, and once I know the force, I can divide by the mass to get the particles acceleration.
    I had some issues with the Biot Savart...

    I used the following numbers
    B=([tex]\mu[/tex]/4[tex]\pi[/tex])(5A)(circumference of the circle/r^2)
    dl= circumference of the circle which eqauls d[tex]\pi[/tex]. or .5654
    since r is the distance from the circle to the particle, it equal the circle's radius which is .9m.
    Therefore B=3.4906e-5
    I substitute this into the force equation giving me
    F=qv x B
    The v vector and B vector are perpendicular, so it is just scalar multiplication.
    I get F=1.3980e-17
    I can then divide this by the mass to get the acceleration.
    Thus a= 1.5347e13
    I used q= 1.602e-19 and m= 9.1093e-31

    All of this led me to a wrong answer...
     
  2. jcsd
  3. Oct 12, 2009 #2
    You got the method right, but the metric units are a bit off: the radius is .09m.
     
  4. Oct 12, 2009 #3
    that was a typo. my final answer used .09 as the radius, and I still got the problem wrong.
     
  5. Oct 12, 2009 #4
    I still haven't checked the calculations yet...but for the moment, maybe what is needed is a negative (electrons have a negative charge, so the acceleration vector is in the opposite direction when compared to positively charged particles, such as protons).
     
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