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Force as a Function of Velocity (Lorentz Force)

  1. Oct 9, 2006 #1
    A singly charged lithium ion is accelerated by an electric field E=10i V/m applied between two plates ten centimeters apart. After passing through a small hole in the negative plate, the ion enters a region of space where there is a magnetic field B=5j mT. The ion moves in a circular path. What is the radius of this circle? The answer to this problem is 7.06 cm. How is this found?
     
  2. jcsd
  3. Oct 9, 2006 #2
    If it moves in a circular path, it means all the force exerted on the particle is doing centripetal acceleration on it. Do you know the formulae relating radial acceleration and linear acceleration and the one for the Lorentz force?

    During the first part of the particle's trajectory, it is simply accelerated with an E-field, and the B-field is zero. Then, having gained some velocity, it enters a non-zero magnetic field, but a zero electric field, and this is where it starts to go round in circles. You need to relate the force exerted on the particle by the B-field to the centripetal acceleration it's experiencing by going in a circle. You should find that the equations are only consistent if the radius is your stated answer.
     
  4. Oct 11, 2006 #3
    How do you prove the following?

    force=-(del velocity)


    please any one answer me.
     
  5. Oct 11, 2006 #4

    Astronuc

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    Staff: Mentor

  6. Oct 11, 2006 #5

    Astronuc

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    Staff: Mentor

    As masudr indicated, the Li+ ion is accelerated across the potential, where F = qE, and the energy gained is just F*d. Then in the magnetic field, the particle of velocity v is subjected to a force q(v x B) where v x B is the cross product of the velocity and magnetic field. The resulting Lorentz force equals the centripetal force.
     
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