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Particle in Magnetic Field: Conceptual

  1. May 8, 2007 #1
    1. The problem statement, all variables and given/known data

    In Figure 28-28 (see attached), a charged particle enters a uniform magnetic field B with speed v0, moves through a half-circle in time T0, and then leaves the field.

    Which statements are true? (Select all that apply.)

    1. The charge is positive.
    2. The charge is negative.
    3. The final speed of the particle is greater than v0.
    4. The final speed of the particle is less than v0.
    5. The final speed of the particle is equal to v0.
    6. For an initial speed 0.5v0, T > T0.
    7. For an initial speed 0.5v0, T < T0.
    8. For an initial speed 0.5v0, T = T0.
    9. For an initial speed 0.5v0, the path is more than a half-circle.
    10. For an initial speed 0.5v0, the path is less than a half-circle.
    11. For an initial speed 0.5v0, the path is also a half-circle.

    2. Relevant equations

    [tex]F_B=qv\times B[/tex]

    [tex]F_c=\frac{mv^2}{r}[/tex]

    3. The attempt at a solution

    I'm fine until I'm asked to halve the velocity. I combined my two equations and found that when v'=0.5v, r'=0.5r. but I'm not sure how that affects the shape of the trajectory. Will the particle still make a semicircle, only now with a smaller radius? If so, it would make sense that the t'>t. But on the other hand, a smaller velocity should result in a longer time!
    I guessed 2, 5, 7, & 11, then I tried 2, 5, 6, & 11. Both combinations are incorrect.
    I'd appreciate any help in understanding how velocity affects the trajectory and time. Thanks!
     

    Attached Files:

  2. jcsd
  3. May 8, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good. More generally, v/r is a constant.
    Yes. Since the magnetic force is always perpendicular to the velocity, it must take a circular path. By symmetry, the path will be a semicircle, only smaller.
    What's your reasoning here? Figure it out! Hint: What distance does it travel? How long does it take?
     
  4. May 8, 2007 #3
    Cool! The particle either travels 2*pi*r with velocity v or 2*pi*0.5r with velocity 0.5v. So time is the same even though velocity changes.
    Thanks, Doc!
     
  5. May 8, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    You've got the idea. (Although a semicircle is only pi*r. :wink: )
     
  6. May 8, 2007 #5
    :redface: That's me. It seems like when I get the physics, I lose the algebra; when I get the algebra, I lose the arithmetic.
    Thanks!
     
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