# Speed of charged particle moving in E and B fields

1. Apr 27, 2016

### AnwaarKhalid

[moderator note: Thread title changed to make it descriptivie of the problem]
1. The problem statement, all variables and given/known data
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:

3. 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. Apr 27, 2016

### drvrm

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 thats not the path taken by the particle.the path will be affected by B-field.

3. Apr 27, 2016

### ehild

You meant change of speed, which is not influenced by B.

4. Apr 27, 2016

### ehild

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.