Field region beams penetrate

warnexus

1. The problem statement, all variables and given/known data



2. Relevant equations

magnetic force = charge * velocity * magnetic field


they ask for distance but this was the only relevant equation for this problem to corresponding section

3. The attempt at a solution

veloctity = 8.4* 10 ^ 6 m/s
magnetic field = 190 Gauss
perpendicular = sin (90) = 1

magnetic force = (1.6 * 10 ^ -19 C) (8.4 * 10 ^ 6 m/s )(190 G)

mfb

Which path do the electrons describe in the field?
This will help to solve the problem.

rude man

I would start by writing F= ma equations for all three axes. (Might two suffice?).
Good problem.

warnexus

all 3 axes, even if z axis is not mentioned. we have to resolve components using a force diagram?

okay here goes:
F_x = m * a_x
F_y = m * a_y
F_z = m * a_z

rude man

What I alluded to is that perhaps one of the axes never sees any motion. Since mag force is always perpendicular to the B field, which axis do you suppose that might be?

PS certainly the z axis is in on this.

warnexus

magnetic force perpendicular to the magnetic field. the y -axis is perpendicular!

mfb

Z-axis is the most important one...
And please, do not use cartesian coordinates to solve equations of motions. This is just overkill for this problem.

rude man

OK, now what are F_x, F_y and F_z in terms of q, m, velocities, etc? Also, write x_double-dot in lieu of a_x, etc.

rude man

Request not granted . And the answer is in x, not z.

rude man

Right. So there are only two equations, in x and in z.

warnexus

F_x is force in the x or force sub x(probably should have done that)

you mean like this:

F_subx = m* a..
F_subz = m* a..

rude man

No. Look at what I wrote.
You need to introduce q, B, and velocity components in lieu of forces F_x etc.

For a_x you should write d²x/dt² etc.

warnexus

oh I see.

charge* velocity* magnetic field = mass * (d^2 * x)/(d*t^2)

may you explain how you got acceleration to be d^2 * x)/(d*t^2). I went through the chapter called amperes law where this question was associated with and don't see the equation like that of (d^2 * x)/(d*t^2)

d as in distance, t as in time, and what is x and we need to find time?

rude man

Question: have you had calculus incl. elementary differential equations? I never know what level of student I am dealing with.

This is a pretty sophisticated problem, involving coupled motions. I.e. the initial x velocity generates a z velocity, and the z velocity in turn generates an x velocity. I wouldn't try to solve this in a non-calculus manner. Maybe mfb knows how ...

warnexus

i had two semester of calculus: differential and integral. taking 3rd and last semester of calculus in the spring

rude man

OK. Then I can say that I meant d²x/dt² to be "the second derivative of x with respect to time". Aka "acceleration in the x direction". Etc.

My way you wind up with two second-order linear differential equations in x and z. Solve for x and there's your answer. The equation in x is the same you get from a spring-mass oscillator.

(You're not asked for the z solution which is similar but not exactly the same as a spring-mass oscillator).

warnexus

oh i see. interesting.

btw how come magnetic field is given in Gauss, the previous questions that involved with magnetic field were given in Tesla? Wouldn't Tesla have to be used?