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I was thinking, what happens to the force on a charged particle when it is moving along the magnetic field lines?

I am familiar with the right-hand-rule and it seems to me that RHR does not apply in this situation. Since the RHR, the field lines, direction of particle moving and the force on the particle are all perpendicular to each other.

But the question I am asking is, what happens to the force when the direction of a particle moving is parallel to the field lines or along the field lines?

Here is a picture of what I mean:

enter image description here

The electron, moving down or along the field lines with a velocity of v.

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Because $\mathbf F=q\mathbf v\times\mathbf B$, if $\mathbf v$ is parallel to $\mathbf B$ then $\mathbf F=0$ and the particle experiences no magnetic force.

This (which uses a RHR to get the directions right) is always valid.

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    $\begingroup$ If the OP hasn't seen the cross product in their courses, it might also be worth including the form for the magnitudes in this answer (i.e. $F = q v B \sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.) In the given case, $\theta = 0$, and so $F = 0$. $\endgroup$ – Michael Seifert Aug 13 '20 at 13:20
  • $\begingroup$ I should have known the formula gives the answer to this question. It totally makes sense. I also see why the force is perpendicular to v and B. Thank you for your quick answer and time. Also, I agree with @MichaelSeifert about including F=qvBsinθ in your answer. It will make it more complete. $\endgroup$ – Tachyon Aug 13 '20 at 13:24

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