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Initial velocity of a charged particle acted upon by magnetic and electric fields 
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#1
Sep1711, 06:12 AM

P: 4

Hello all,
I got this question off mastering physics and was unable to understand it. I eventually gave up to concentrate on other assignments and asked for the answer. I was unable to come up with more than one answer I thought possible without resorting to trial and error. Answer: V_{x},V_{y},V_{z} = 0,234,0 m/s 1. The problem statement, all variables and given/known data A 6.50 negative microC particle moves through a region of space where an electric field of magnitude 1200 N/C points in the positive x direction, and a magnetic field of magnitude 1.02 T points in the positive z direction. If the net force acting on the particle is 6.25×10^{−3 }N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the xy plane. V_{x} = ? V_{y} =? V_{z} = ? 2. Relevant equations Force = electric field * charge or F=e*q Force = magnitude of charge * velocity * magnetic field * sin tither or F=q*v*B*sin tither 3. The attempt at a solution Force on particle due to Electric Field = E*q = (1200N/C)(6.5*10^{3} C) = 7.8*10^{3}N Force on charge due to magnetic field = q*v*B*sin tither = (6.5*10^{6} C)(1.02T)(v)(sin 90) = (6.63*10^{6})(v) F_{x1} along x axis = 7.8*10^{3}N I'm sure the force on charge due to magnetic field lies between axis of positive Y and axis of negative X. The force on the particle needs to be perpendicular to the positive z direction of the magnetic field and the velocity of the particle which is one the xy plane. Since the net force is in the xdirection, it should be along the xaxis. This was also the reasoning that led me to using sin90 when determining the force on the charge due to the magnetic field. F_{x2} along x axis = (6.63*10^{6})(v) At this point, I thought I understood the problem statement fully. I took the net force from these two vectors I had come up with and seperated velocity to LHS and the constants to the right. V_{x}= (Net Force x + F_{x1}) / (F_{x2} without velocity term) = ((6.25×10^{−3 })+(7.8*10^{3})) / (6.63*10^{6}) = 2.12*10^{7} m/s The value was ridiculously large and I was not surprised it was rejected by mastering physics. I think I have a huge misunderstanding of the concept or method behind this question. Please explain to me how I should go about dealing with a question of this type or point out my mistakes. I do not need a worked out solution, coming up with that myself would probably be better for me. Thank you 


#2
Sep1711, 09:33 AM

Sci Advisor
HW Helper
P: 6,671

To find the direction of v then, apply the crossproduct (righthand) rule. I suggest you work out the direction of v from the physics to avoid being confused by all the + and  signs. Also, it is better to work out the solution algebraically and then plug in numbers. AM 


#3
Sep1711, 10:31 AM

P: 4

Thank you for the swift and verbose response. I'll work through the question again and pay attention to the signs and powers. I apologise, this was obviously sloppy work on my part.



#4
Sep1711, 11:31 AM

P: 4

Initial velocity of a charged particle acted upon by magnetic and electric fields
Thank you, I got the answer. Here is how I got it.
Force on particle due to electric field: F_{x1} = E*q = (1200N/C)(6.5*10^{}6) = 7.8*10^{3} Force on particle due to magnetic field: F_{x2} = q*v*B*sin tither = (6.5*10^{6})(1.02)(sin90)v = (6.63*10^{6})(v) F_{x2} is in the positive x direction as F_{x1} is in the negative x direction while net force is in the positive x direction. My first mistake was in the direction of the particle's velocity, I'm now more familiar with the right hand rule. Magnetic field is in the positive Z direction, net force is in the positive x direction. According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the yaxis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction. My second mistake was my sign of F_{x1} used to determine the value of velocity. I added F_{xnet} to F_{x1} instead of subtracting F_{xnet} by F_{x1}. F_{xnet}  F_{x1} = F_{x2} (6.25*10^{3})  (7.8*10^{3}) = (6.63*10^{6})(v) v = (F_{xnet}  F_{x1}) / (F_{x2} /v) = ((6.25*10^{3})  (7.8*10^{3})) / (6.63*10^{6}) = (233.7) m/s v_{y} = 234 m/s Thank you very much for your help, I'm glad I learned something instead of just walking away from mastering physics without any idea. 


#5
Sep1711, 01:21 PM

Sci Advisor
HW Helper
P: 6,671

Good job. To do physics you have to persevere. If you master perseverance, you are be halfway there. And by perseverance I don't mean just plodding along until you get the right answer. I mean working at it until you really understand what is going on. Good luck!
AM 


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