Position of particle in a magnetic field

In summary, the conversation discusses finding the position of a particle moving in a magnetic field as a function of time. The relevant equations are the Lorentz Force and F=Ma. The attempt at a solution involves using the equations to find the positions in terms of time. However, the person is unsure of how to proceed.
  • #1
gabrown
8
0

Homework Statement



I am wanting to get position (x and y) of a particle moving in a magnetic field as a function of time.
So, we have a particle of mass m, charge q, velocity (Vx,0,0) moving in a magnetic field (o,o, Bz) starting at t=0 as (0,0,0), what is X(t) and Y(t).

Homework Equations



Lorentz Force F=q(VxB)
F=Ma



The Attempt at a Solution



Ax = q Vy Bz
Ay = - q Vx Bz

follows that

X(y,t) = (q.Y.Bz + A1)t + A2
Y(x,t) = (-q.X.Bz +A3)t + A4

but not sure where to go from here


cheers

Gareth
 
Physics news on Phys.org
  • #2
No ideas?
 
  • #3




To determine the position of a particle in a magnetic field, we can use the Lorentz force equation F=q(VxB) and the equation for acceleration F=ma. From the given information, we can see that the particle is moving in the x-direction with a constant velocity in the y and z directions. This means that the only force acting on the particle is in the x-direction, given by F=q(VyBz).

Using the equation F=ma, we can set this force equal to the mass of the particle times its acceleration in the x-direction, giving us ma=q(VyBz). We can rearrange this to find the acceleration in the x-direction as a=(q/m)(VyBz).

Next, we can use the equation for position x(t)=x0+v0t+(1/2)at^2, where x0 is the initial position and v0 is the initial velocity. Since the particle starts at (0,0,0) with a velocity of (Vx,0,0), we know that x0=0 and v0=Vx. Substituting in our expression for acceleration, we get x(t)=(Vx)t+(1/2)(q/m)(VyBz)t^2.

Similarly, we can use the equation for position y(t)=y0+v0t+(1/2)at^2, where y0 is the initial position and v0 is the initial velocity. Since the particle starts at (0,0,0) with a velocity of (Vx,0,0), we know that y0=0 and v0=0. Substituting in our expression for acceleration, we get y(t)=(1/2)(q/m)(VyBz)t^2.

Therefore, the position of the particle in the x-direction is given by x(t)=(Vx)t+(1/2)(q/m)(VyBz)t^2 and the position in the y-direction is given by y(t)=(1/2)(q/m)(VyBz)t^2. These equations show how the position of the particle changes over time in the magnetic field.
 

1. What is the equation for calculating the force on a particle in a magnetic field?

The equation for calculating the force on a particle in a magnetic field is F = q(v x B), where F is the force in Newtons, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength.

2. How does the direction of the magnetic field affect the motion of a particle?

The direction of the magnetic field affects the motion of a particle by causing it to experience a force perpendicular to both the direction of the magnetic field and the particle's velocity. This results in circular motion for charged particles.

3. Can a particle's velocity affect its position in a magnetic field?

Yes, a particle's velocity can affect its position in a magnetic field. This is because the force on a charged particle in a magnetic field is dependent on its velocity.

4. How does the strength of the magnetic field impact the position of a particle?

The strength of the magnetic field impacts the position of a particle by affecting the magnitude of the force experienced by the particle. A stronger magnetic field will result in a greater force on the particle, causing it to move in a larger radius.

5. What is the relationship between the charge of a particle and its position in a magnetic field?

The charge of a particle has a direct relationship with its position in a magnetic field. This is because the force on a charged particle in a magnetic field is directly proportional to its charge. Therefore, a particle with a higher charge will experience a greater force and move in a larger radius in the same magnetic field.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
5K
  • Introductory Physics Homework Help
Replies
3
Views
697
  • Introductory Physics Homework Help
Replies
1
Views
684
  • Introductory Physics Homework Help
Replies
31
Views
972
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
589
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top