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Distance from one period to the next, ##e## in helical path.

  1. Dec 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A source injects an electron of speed ##v## into a uniform magnetic field of magnitude ##B##. The velocity of the electron makes an angle ##\theta## with the direction of the magnetic field. Find the distance ##d## from the point of injection at which the electron next crosses the field that passes through the injection point.

    2. Relevant equations
    ##F_B=m\frac{v^2}{r}##
    ##T=\frac{2\pi r}{v}##

    3. The attempt at a solution

    We want to find how much vertical displacement there is in one period essentially. To find ##T## in terms on known quantities:

    $$
    F_B=q(v\times B)=qvB\sin\theta=\frac{v^2}{r}\Longrightarrow r=\frac{mv}{qB\sin\theta} \\
    T=\frac{2\pi r}{v}=\frac{2\pi m}{qB\sin\theta}
    $$

    Now the vertical displacement would be given by ##v\cdot\cos\theta## multiplied by the time (period in this case) Therefore:
    $$
    d=v\cos\theta\cdot \frac{2\pi m}{qB\sin\theta}=\frac{2\pi m v \mbox{ cotan }\theta}{qB}
    $$

    In my solution manual however they have completely neglected ##F_B=q(v\times B)## and just used the result for when ##\theta=90## (##F_B=qvB##), in the textbook the angle given is not equal to 90 so this isn't the case. Is there something I'm missing here or did they solve it incorrectly?

    Edit: Some errors in my derivation:

    $$F_B=qvB\sin\theta=m\frac{v^2}{r}$$

    the velocity ##v^2/r## in this term is the velocity that is perpendicular to the force so it becomes: ##qvB\sin\theta=m\frac{(v\sin\theta)^2}{r}## This solves the problem.
     
    Last edited: Dec 9, 2015
  2. jcsd
  3. Dec 9, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    Closed: resolved by OP.
     
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