Steven W. answered • 09/22/16

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Physics Ph.D., college instructor (calc- and algebra-based)

Hi Kiara!

I understand your confusion. This problem is not very well written, in my opinion. But here is how I look at it.

(a) If the magnetic field is a uniform vector of <2, 1, 5>, then it has that vector description everywhere in space (note that it does not depend on x, y, or z)

(b) It looks like they are describing a particle at a point where it is moving strictly in the +y direction, with a speed of v = 1. And the particle was already defined to have unit charge, so its charge is also 1... although, in the very next part, the question posits a new magnetic field from the one described in a), without being very clear about it; so when to use what was already given before, and when not to, is not made very clear, in my opinion. Perhaps the writer should spend less time being cutesy and more time editing.

(c) Since what is given is displacement <x, y, z>, then instantaneous velocity at time t is just the first time derivative of that, which would be (-sin(t), cos(t), 0). As the question describes, the charged particle will move in a circle if its trajectory is always perpendicular to the magnetic field. Since the particle is moving in the x-y plane, the magnetic field must be only in the z direction, so that the particle's trajectory will always be perpendicular to it.

(d) The magnetic field will exert no force on a charge moving along the line of the field. So if a charge has a component of motion perpendicular to the field, and a component along the field line, the particle will circle in the perpendicular direction while, at the same time, traveling undeflected, in a straight line, along the magnetic field. Can you think of what shape such a trajectory might make?

If you would like to talk more about it, just let me know. I hope this helps!