Speed of charged particle moving in E and B fields

In summary, a particle with mass m and charge q is released at the origin in a region with given magnetic and electric fields. The speed of the particle as a function of z coordinate is found by calculating the net force in both x and z directions, using the equations Fx= qBvz and Fz= q(E-Bvx). The solution involves using the equations for acceleration in both directions and solving for the z-dependence of speed.
  • #1
[moderator note: Thread title changed to make it descriptivie of the problem]

Homework Statement


A particle having mass m and charge q is released at the origin in a region in which magnetic field and electric field are given by
B= -B' j and E= E' k
where j and k are unit vectors along y-axis and z axis respectively.
Find the speed of the particle as a function of z co ordinate.

2. Relevant questions:


The Attempt at a Solution


The particle is released at the origin so at t=0 only electric force will act on the particle along z axis. As the particle gains some velocity, magnetic force will start acting and will push the particle in xz plane. Here the velocity has two components, x and z, so magnetic force will act on both these components. I calculted the net force in both x and z directions as:
Fx= qBvz
Fz= q(E-Bvx)
As can be seen from the equations, the acc. in the both the directions is variable , so how come in the solution(picture), he has used the equation v2 - u2= 2as?
 

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  • #2
AnwaarKhalid said:
The particle is released at the origin so at t=0 only electric force will act on the particle along z axis. As the particle gains some velocity, magnetic force will start acting and will push the particle in xz plane. Here the velocity has two components, x and z, so magnetic force will act on both these components. I calculted the net force in both x and z directions as:
The particle gains velocity in direction of the electric field and the magnetic field acts as cross product of velocity vector and the magnetic field which will be normal to the plane containing velocity acquired and the B field. therefore the change in velocity can be calculated using acceleration in E field direction. but naturally that's not the path taken by the particle.the path will be affected by B-field.
 
  • #3
drvrm said:
The particle gains velocity in direction of the electric field and the magnetic field acts as cross product of velocity vector and the magnetic field which will be normal to the plane containing velocity acquired and the B field. therefore the change in velocity can be calculated using acceleration in E field direction. but naturally that's not the path taken by the particle.the path will be affected by B-field.
You meant change of speed, which is not influenced by B.
 
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  • #4
AnwaarKhalid said:

The Attempt at a Solution


The particle is released at the origin so at t=0 only electric force will act on the particle along z axis. As the particle gains some velocity, magnetic force will start acting and will push the particle in xz plane. Here the velocity has two components, x and z, so magnetic force will act on both these components. I calculted the net force in both x and z directions as:
Fx= qBvz
Fz= q(E-Bvx)
As can be seen from the equations, the acc. in the both the directions is variable , so how come in the solution(picture), he has used the equation v2 - u2= 2as?
The solution is correct, but you can solve it with a less intuitive way.
The two equations give the components of accelerations.
##a_x=\frac{dv_x}{dt}=\frac{qB}{m}v_z##
##a_z=\frac{dv_z}{dt}=\frac{q}{m}(E-Bv_x)##
As the z-dependence of speed is the question, you can change to z as independent variable. Applying chain rule,
##\frac{dv_x}{dt}=\frac{dv_x}{dz}v_z=\frac{qB}{m}v_z## and ##a_z=\frac{dv_z}{dz}v_z=\frac{q}{m}(E-Bv_x)##
vz cancels from the first equation, so vx is proportional to z. Knowing vx as function of z, the second equation is easy to integrate.
 

1. What is the equation for calculating the speed of a charged particle moving in E and B fields?

The equation is v = E/B, where v is the speed of the particle, E is the electric field strength, and B is the magnetic field strength.

2. How do the electric and magnetic fields affect the speed of a charged particle?

The electric field accelerates the particle in the direction of the field, while the magnetic field causes the particle to move in a circular path perpendicular to the field.

3. Can the speed of a charged particle in E and B fields ever exceed the speed of light?

No, according to the theory of relativity, the speed of light is the maximum speed at which any object can travel. Therefore, the speed of a charged particle in E and B fields cannot exceed the speed of light.

4. What factors can influence the speed of a charged particle in E and B fields?

The speed of the particle can be affected by the strength and direction of the electric and magnetic fields, as well as the mass and charge of the particle.

5. How do E and B fields affect the trajectory of a charged particle?

The electric field affects the direction of the particle's velocity, while the magnetic field affects the curvature of its path. Together, these fields determine the overall trajectory of the particle.

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