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Tricky Magnetic Field Question... (Proton Enters Magnetic Field.....) 
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#1
Mar3112, 07:56 AM

#2
Mar3112, 09:23 AM

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P: 11,805

At the outset you also might want to convince yourself that the effect of gravity will be negligible. What's the initial kinetic energy of the particle in the vertical direction, and what's the change in energy that can occur? 


#3
Mar3112, 07:39 PM

P: 167

Because.. I assumed that the radius would kind of moderately proportional to the diagram, which i should never think, but honestly i showed another student, and he's like that makes no sense... So i kind took his opinion.. which i shouldn't have... 


#4
Mar3112, 09:48 PM

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P: 11,805

Tricky Magnetic Field Question... (Proton Enters Magnetic Field.....)



#5
Apr112, 06:38 PM

P: 167

since this is a uniform magnetic field, and the charge enters at a right angle,
the charge experiences circular motion. There is a constant force perpindicular to its motion. therefore.. A) Fm = Fc R = mv/qb R= 5.14m b) now i was thinking in terms of that angle, if my original radius is correct, i should be able to make a trianglee... with the Radius R, and the distance (d) of the magnetic field and solve.. SinA=d/R A=2.79 degrees now.. i don't really want to move on, yet because i'm not positive if this is an okay strategy to solve for this angle. and in terms of energy, i did a calculation... Et1= 1/2MVo^2 (initial point) potential energy lvl zero.. The change in energy i believe would be, no change.. There are no non conservative forces acting upon the charge... and in circular motion work done on an object is = 0 because the force is perpindicular to the velocity.. so therefore the change in energy is 0...? and this can be proven i noticed when i calculated Et2 = 1/2Mvf^2 + mgd If i make energy totals the same and solve for Vf Et1 = Et1 1/2mvo^2 = mgd + 1/2mvf^2 Vf = sqrtall(mv^2 2gd) Vf = 145000 which is its original velocity can you tell me how i'm doing :O 


#6
Apr112, 06:54 PM

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P: 11,805

For the check to make sure that gravity is an insignificant factor, you might have just compared the initial kinetic energy with the maximum change in potential energy across the extent of the apparatus. 


#7
Apr112, 09:40 PM

P: 167

ughhhhhhhh. alright... umm i'm gonna have to think about this one some more then...
but i have to solve from a time t1, and the x1 is the angle of deflection... i thought that kinematics would be required in that case ( b, and c) 


#8
Apr112, 10:02 PM

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P: 11,805

I don't understand why you're using formulas for motion under gravitational acceleration when you've decided that gravity can be ignored.
The particle's path curves while it is within the magnetic field, after which it travels in a straight line. After the "bend" in the trajectory due to the magnetic field, the problem boils down to geometry. 


#9
Apr312, 10:31 PM

P: 167

okay so..using the angle i had originally 2.79
i can use that angle to solve for x1 becaues the angle is the point which the proton deviates. tanA = d/x1 x1=d/tanA now in order to get t1 i need the hypoteneuse where it travels.. cosa = d/z z= d/cosA z = vot t= z/vo 


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