- #1
mbrmbrg
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Homework Statement
A magnetic field CANNOT:
[a]exert a force on a charge
accelerate a charge
[c]change the momentum of a charge
[d]change the kinetic energy of a charge
[e]exist
Note: The answer options are radio buttons (rather than check boxes).
1a. Question
I found that the answer was [d] (see bottom for my fun, tedious thought process), but I'm not sure why that must be so.
Given a three-dimensional system, with a magnetic field acting along one of the axes, say the x-axis, and a moving charged particle.
Any x-component of velocity will not affect the magnetic force, so let's forget about the x-component (possibly a dangerous thing to forget about, but I will disregard it nonetheless).
A y- or z-component of velocity will produce a force along the z- or y-axis respectively.
If the velocity vector has one non-zero component that points along either the y- or z- axis, then the magnetic field will indeed affect only the direction of the velocity, and not its magnitude. This is consistent with magnetic field not affecting (scalar) kinetic energy.
But if velocity has both y- and z- components, then the resulting force has both y- and z- components, as well. Thus, the force should change both the direction and magnitude of velocity, and thus kinetic energy as well.
So is my reasoning flawed, or is the problem incorrect/imprecise?
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HOW I GOT PART D AS THE CORRECT ANSWER
Homework Equations
\vec{F_B}=q\vec{v} \times \vec{B} (equation 1)
\vec{F}=m\vec{a} (equation 2)
\vec{a}=\frac{\Delta \vec{v}}{\Delta t} (eqation 3)
\vec{p}=m\vec{v} (equation 4)
K=\frac{1}{2}mv^2 (eqation 5)
The Attempt at a Solution
Radio buttons tell me that there is one (real), unique correct answer. None of the options (with the exception of [e]) make sense unless there is an implicit "ed particle" after the word "charge"--so I inserted it and analyzed as follows:
[a] False. Equation 1 says "insert magnetic field (and non-zero velocity) and get a force."
False. We have concluded that a magnetic field produces a force, and equation 2 says that a force produces acceleration (assuming a non-zero mass. Which I am.)
[c] False. Since there's an acceleration, equation 3 states that the velocity much change. If the velocity changes, then by equation 4 the momentum changes.
[d] Process of Eliminatin. See a, b, c, and e.
[e] Ummmm... Well, I wouldn't know, really. But seeing how my entire study of magnetism seemed to be predicated on the existence of magnetic fields, I'll assume that a magnetic field is a useful model, and as such, exists.
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Yeah; I was trying to avoid the problem of existence. Hell, does it think? Then since I want to use Descartes' axes, the field's existence really is called into question!
Oh. Good Lord. Right. Thenk you. No need for crazy thinking, just look at the cross product operator.
). 
