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Initial velocity of a charged particle acted upon by magnetic and electric fields

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Weistber
#1
Sep17-11, 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: Vx,Vy,Vz = 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 6.2510−3 N in the positive x direction, find the components of the particle's velocity. Assume the particle's velocity is in the x-y plane.

Vx = ? Vy =? Vz = ?


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*10-3N

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)

Fx1 along x axis = -7.8*103N

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 x-y plane. Since the net force is in the x-direction, it should be along the x-axis. This was also the reasoning that led me to using sin90 when determining the force on the charge due to the magnetic field.

Fx2 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.

Vx= (Net Force x + Fx1) / (Fx2 without velocity term)
= ((6.2510−3 )+(7.8*103)) / (6.63*10-6)
= 2.12*107 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
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Andrew Mason
#2
Sep17-11, 09:33 AM
Sci Advisor
HW Helper
P: 6,671
Quote Quote by Weistber View Post
Force on particle due to Electric Field = E*q
= (1200N/C)(-6.5*10-3 C)
= -7.8*10-3N
Answer is correct, along the x axis. So the force is in the -x direction. Charge is -6.5*10-6 C. - check typo.

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)


Fx1 along x axis = -7.8*103N
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 x-y plane. Since the net force is in the x-direction, it should be along the x-axis. This was also the reasoning that led me to using sin90 when determining the force on the charge due to the magnetic field.

Fx2 along x axis = (6.63*10-6)(v)
But which direction along the x axis? Positive or negative? Since the net force is in the positive direction and the magnitude of the Coulomb force is negative, this force must be in the positive x direction - ie. opposite to the Coulomb force.

To find the direction of v then, apply the cross-product (right-hand) rule.
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.

Vx= (Net Force x + Fx1) / (Fx2 without velocity term)
= ((6.2510−3 )+(7.8*103)) / (6.63*10-6)
= 2.12*107 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.
You understand the problem. You just have to keep a few things straight. You have to determine the direction of v first and that will help you solve the problem. Be careful with the numbers. You are forgetting the - sign in some of the powers.

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
Weistber
#3
Sep17-11, 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.

Weistber
#4
Sep17-11, 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:
Fx1 = E*q
= (1200N/C)(-6.5*10-6)
= -7.8*10-3

Force on particle due to magnetic field:
Fx2 = |q|*v*B*sin tither
= (6.5*10-6)(1.02)(sin90)v
= (6.63*10-6)(v)

Fx2 is in the positive x direction as Fx1 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 y-axis. 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 Fx1 used to determine the value of velocity. I added Fxnet to Fx1 instead of subtracting Fxnet by Fx1.

Fxnet - Fx1 = Fx2

(6.25*10-3) - (7.8*10-3) = (6.63*10-6)(v)

v = (Fxnet - Fx1) / (Fx2 /v)
= ((6.25*10-3) - (7.8*10-3)) / (6.63*10-6)
= (-233.7) m/s

vy = -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.
Andrew Mason
#5
Sep17-11, 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 half-way 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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