I need help finding the solution to a vector problem from my calc 3 lab.

Question

I've tried to answer the problem but I'm pretty sure that everything I did was wrong.

Here's the problem.

 

This fun problem relates to physics. In case you’re not a physics major, we urge you to follow the immortal advice of Douglas Adams: Don’t Panic! You don’t need to be Marie Curie to solve this one. Let M be a vector in three-space.

 

A constant magnetic field of value M exists at all points of three-space.

(a) Suppose M = <2, 1, 5>. What is the value of the magnetic field experi- enced at the point (3, 2, −3)? [Hint, this is not a trick question, it’s a warm-up]

 

A particle of unit mass and unit electric charge is in the field and it experiences velocity along the vector v. The field acts on the particle via the Lorentz force F = v × M. Since this force is perpendicular to v and M, by the laws of physics it does no work, and thus it can change the direction of the particle but not its speed. If v is perpendicular to M then the particle will circle in a plane perpendicular to M. The radius of the circle is equal to |v|/|M|.1

(b) Suppose the particle is at the point (3, 2, −3) and v = <0, 1, 0>. What is the particle’s charge and in which direction is it going to move? [Hint, this is also meant to be a softball question.]

(c) Suppose that the particle is observed to be travelling along the circle (cos t, sin t, 0) in the x,y-plane. What is the velocity v of the particle at time t? Use Mathematica to plot in a single graph the circle and the velocity vector at some point on the circle. What magnetic field M could be acting on the particle?

(d) What sort of trajectory do you expect if the particle acquires a velocity component parallel to M? For example, what sort of trajectory do you expect if M = <0, 0, 1> and v = <1, 1, 1>? A qualitative answer is accept- able here.

Answer 1

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!