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Field region beams penetrate 
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#1
Dec2112, 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) 


#2
Dec2112, 09:12 AM

Mentor
P: 12,113

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


#3
Dec2112, 03:45 PM

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PF Gold
P: 4,917

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


#4
Dec2112, 03:48 PM

P: 90

Field region beams penetrate
okay here goes: F_x = m * a_x F_y = m * a_y F_z = m * a_z 


#5
Dec2112, 03:51 PM

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PF Gold
P: 4,917

PS certainly the z axis is in on this. 


#6
Dec2112, 03:54 PM

P: 90




#7
Dec2112, 03:55 PM

Mentor
P: 12,113

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


#8
Dec2112, 03:56 PM

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PF Gold
P: 4,917




#9
Dec2112, 03:59 PM

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PF Gold
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#10
Dec2112, 04:01 PM

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PF Gold
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#11
Dec2112, 04:02 PM

P: 90

you mean like this: F_subx = m* a.. F_subz = m* a.. 


#12
Dec2112, 04:10 PM

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PF Gold
P: 4,917

You need to introduce q, B, and velocity components in lieu of forces F_x etc. For a_x you should write d^{2}x/dt^{2} etc. 


#13
Dec2112, 04:22 PM

P: 90

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? 


#14
Dec2112, 04:33 PM

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PF Gold
P: 4,917

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 noncalculus manner. Maybe mfb knows how ... 


#15
Dec2112, 04:50 PM

P: 90




#16
Dec2112, 04:55 PM

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PF Gold
P: 4,917

My way you wind up with two secondorder 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 springmass oscillator. (You're not asked for the z solution which is similar but not exactly the same as a springmass oscillator). 


#17
Dec2112, 05:08 PM

P: 90

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? 


#18
Dec2112, 05:23 PM

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PF Gold
P: 4,917

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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