Initial Velocity of Charged Particle in Electric/Magnetic Fields

In summary, the conversation discusses a question from mastering physics involving a charged particle moving through an electric and magnetic field. The net force on the particle is given and the components of the particle's velocity are to be found. The conversation includes calculations and reasoning for determining the forces and velocity, but the final answer is rejected by mastering physics. The individual is seeking clarification and understanding of the problem.
  • #1
Weistber
4
0
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

Homework Statement


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 x-y plane.

Vx = ? Vy =? Vz = ?

Homework 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

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 separated velocity to LHS and the constants to the right.

Vx= (Net Force x + Fx1) / (Fx2 without velocity term)
= ((6.25×10−3 )+(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
 
Last edited:
Physics news on Phys.org
  • #2
Weistber said:
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 separated velocity to LHS and the constants to the right.

Vx= (Net Force x + Fx1) / (Fx2 without velocity term)
= ((6.25×10−3 )+(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
 
Last edited:
  • #3
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
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.
 
  • #5
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
 

1. What is the initial velocity of a charged particle in an electric field?

The initial velocity of a charged particle in an electric field depends on the strength of the electric field and the charge of the particle. It can be calculated using the equation v = E/m, where v is the initial velocity, E is the electric field strength, and m is the mass of the charged particle.

2. How does the initial velocity of a charged particle change in a magnetic field?

In a magnetic field, the initial velocity of a charged particle will change direction but not magnitude. This is because the magnetic force acts perpendicular to the particle's velocity, causing it to move in a circular path. The magnitude of the initial velocity will remain constant unless acted upon by an external force.

3. How does the charge of a particle affect its initial velocity in an electric/magnetic field?

The charge of a particle affects its initial velocity in an electric and magnetic field in different ways. In an electric field, a greater charge will result in a greater initial velocity as the electric force is directly proportional to the charge. In a magnetic field, the charge does not affect the magnitude of the initial velocity but can impact the direction of the particle's motion due to the magnetic force acting on it.

4. Can the initial velocity of a charged particle in an electric/magnetic field be controlled?

Yes, the initial velocity of a charged particle in an electric/magnetic field can be controlled by adjusting the strength of the field or the charge of the particle. By manipulating these variables, scientists can change the initial velocity of a charged particle and study how it affects the particle's motion.

5. How is the initial velocity of a charged particle in an electric/magnetic field measured?

The initial velocity of a charged particle in an electric/magnetic field can be measured using various methods such as particle accelerators, particle detectors, and magnetic deflectors. These instruments use principles of electromagnetism to measure the initial velocity and track the motion of charged particles in electric and magnetic fields.

Similar threads

  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
149
  • Introductory Physics Homework Help
Replies
31
Views
971
  • Introductory Physics Homework Help
Replies
4
Views
177
  • Introductory Physics Homework Help
Replies
18
Views
714
  • Introductory Physics Homework Help
Replies
7
Views
932
  • Introductory Physics Homework Help
Replies
12
Views
983
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
484
  • Introductory Physics Homework Help
Replies
5
Views
730
Back
Top