# Field region beams penetrate

by warnexus
Tags: beams, field, penetrate, region
 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)
 Mentor P: 12,113 Which path do the electrons describe in the field? This will help to solve the problem.
 HW Helper Thanks PF Gold P: 4,917 I would start by writing F= ma equations for all three axes. (Might two suffice?). Good problem.
P: 90
Field region beams penetrate

 Quote by rude man 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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PF Gold
P: 4,917
 Quote by warnexus 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.
P: 90
 Quote by 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.
magnetic force perpendicular to the magnetic field. the y -axis is perpendicular!
 Mentor P: 12,113 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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PF Gold
P: 4,917
 Quote by 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
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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PF Gold
P: 4,917
 Quote by 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.
Request not granted . And the answer is in x, not z.
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PF Gold
P: 4,917
 Quote by warnexus magnetic force perpendicular to the magnetic field. the y -axis is perpendicular!
Right. So there are only two equations, in x and in z.
P: 90
 Quote by 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.
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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PF Gold
P: 4,917
 Quote by 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..
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.
P: 90
 Quote by 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.

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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PF Gold
P: 4,917
 Quote by 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?
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 ...
P: 90
 Quote by 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 ...
i had two semester of calculus: differential and integral. taking 3rd and last semester of calculus in the spring
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PF Gold
P: 4,917
 Quote by warnexus 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).
P: 90
 Quote by 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).
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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PF Gold
P: 4,917
 Quote by 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?
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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