Field region beams penetrate


by warnexus
Tags: beams, field, penetrate, region
warnexus
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#1
Dec21-12, 08:15 AM
P: 90
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)
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mfb
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#2
Dec21-12, 09:12 AM
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Which path do the electrons describe in the field?
This will help to solve the problem.
rude man
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#3
Dec21-12, 03:45 PM
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I would start by writing F= ma equations for all three axes. (Might two suffice?).
Good problem.

warnexus
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#4
Dec21-12, 03:48 PM
P: 90

Field region beams penetrate


Quote Quote by rude man View Post
I would start by writing F= ma equations for all three axes. (Might two suffice?).
Good problem.
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
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Dec21-12, 03:51 PM
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Quote Quote by warnexus View Post
all 3 axes, even if z axis is not mentioned. we have to resolve components using a force diagram?
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
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#6
Dec21-12, 03:54 PM
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Quote Quote by rude man View Post
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.
magnetic force perpendicular to the magnetic field. the y -axis is perpendicular!
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#7
Dec21-12, 03:55 PM
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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.
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Dec21-12, 03:56 PM
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Quote Quote by warnexus View Post
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
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.
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Dec21-12, 03:59 PM
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Quote Quote by mfb View Post
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.
Request not granted . And the answer is in x, not z.
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#10
Dec21-12, 04:01 PM
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Quote Quote by warnexus View Post
magnetic force perpendicular to the magnetic field. the y -axis is perpendicular!
Right. So there are only two equations, in x and in z.
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#11
Dec21-12, 04:02 PM
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Quote Quote by rude man View Post
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.
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..
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Dec21-12, 04:10 PM
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Quote Quote by warnexus View Post
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..
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.
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#13
Dec21-12, 04:22 PM
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Quote Quote by rude man View Post
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.


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?
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Dec21-12, 04:33 PM
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Quote Quote by warnexus View Post
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?
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 ...
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#15
Dec21-12, 04:50 PM
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Quote Quote by rude man View Post
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 ...
i had two semester of calculus: differential and integral. taking 3rd and last semester of calculus in the spring
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Dec21-12, 04:55 PM
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Quote Quote by warnexus View Post
i had two semester of calculus: differential and integral. taking 3rd and last semester of calculus in the spring
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).
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#17
Dec21-12, 05:08 PM
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Quote Quote by rude man View Post
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).
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?
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Dec21-12, 05:23 PM
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Quote Quote by warnexus View Post
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?
1 Tesla = 10,000 Gauss. Your teach is trying to make life even more miserable than it has to be ... tell him you won't do any more problems in any other than SI units...

Actually, pure physicists tend to use the cgs system which includes the Gauss. Applied physicists and engineers gravitate towards the SI system, aka 'rationalized mks" system ("meter, kilogram, second").
.


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