How Do You Solve Particle Motion Equations in an Electromagnetic Field?

[akinoshigure]

So I'm stuck on the second part of this problem and really don't know hwere to go from here... let me type it up and show where I got stuck at.

3. The motion of particle of chage q in an electromagnetic field is governed by the Lorentz force (for low velocities v<<c): F=qE + qvxB.
With both constant B=B k and E=Ey j + Ez K show that:
z(t)=z(sub-o)+v(sub-zo)t+qE(sub-z)t^2/2m
vx(t)=Asin(omega-t)+E(sub-y)/B
vx(t)=+Acos(omega-t)

I did F=qE+qVxB=m (dv/dt)

dvx/dt= q/m(vyBz)
dvy/dt= q/m(Ey-vxBo)
dvz/dt= q/m (Ez)

I think I'm suppose to now take a second derivative and find the second order differential equation but I'm not too sure how to approach that.

 

[HallsofIvy]

So I'm stuck on the second part of this problem and really don't know hwere to go from here... let me type it up and show where I got stuck at.

3. The motion of particle of chage q in an electromagnetic field is governed by the Lorentz force (for low velocities v<<c): F=qE + qvxB.
With both constant B=B k and E=Ey j + Ez K show that:
z(t)=z(sub-o)+v(sub-zo)t+qE(sub-z)t^2/2m
vx(t)=Asin(omega-t)+E(sub-y)/B
vx(t)=+Acos(omega-t)

I did F=qE+qVxB=m (dv/dt)

dvx/dt= q/m(vyBz)
dvy/dt= q/m(Ey-vxBo)
dvz/dt= q/m (Ez)

One slight error here: the strength of the magnetic field is just "B", not
"Bz" or "Bo".

I think I'm suppose to now take a second derivative and find the second order differential equation but I'm not too sure how to approach that.

What you are "supposed" to do is solve those equations. Since the last one (for dvz/dt) does not involve the other two components, you can solve it directly. The other two are "entwined". One method of solving a pair of equations is to differentiate the first (so that you have dvy/dt on the right side) and then replace dvy/dt from the second equation. That will give you one second order differential equation for vx/

 

[quicknote]

As a theoretical mechanic, I can provide some guidance on how to approach this problem. It looks like you have correctly identified the equations of motion for the particle in an electromagnetic field. To solve for the position and velocity of the particle, we need to integrate these equations.

First, let's focus on the z-direction. We can integrate the equation dvz/dt = q/m(Ez) to get vz(t) = vz0 + qEz(t)/m, where vz0 is the initial velocity in the z-direction. Then, we can integrate again to get z(t) = z0 + vz0t + qEz(t)^2/2m, where z0 is the initial position in the z-direction.

Next, let's look at the x-direction. We can integrate the equation dvx/dt = q/m(vyBz) to get vx(t) = vx0 + qBz(t)/m, where vx0 is the initial velocity in the x-direction. Then, we can use the trigonometric identity sin^2(x) + cos^2(x) = 1 to write Bz(t) as Bsin(ωt), where ω is the frequency of the oscillation. Substituting this into the equation for vx(t), we get vx(t) = vx0 + qBsin(ωt)/m. Finally, we can use the initial condition vx(0) = vxo + qEy/m to solve for vx0, giving us vx(t) = (vxo + qEy/m) + qBsin(ωt)/m.

Similarly, we can integrate the equation dvy/dt = q/m(Ey - vx0B) to get vy(t) = vy0 + qEy(t)/m - qvx0Bt/m. We can then use the initial condition vy(0) = vyo to solve for vy0, giving us vy(t) = vyo + qEy(t)/m - qvx0Bt/m.

In summary, the equations of motion for the particle in the electromagnetic field are:

z(t) = z0 + vz0t + qEz(t)^2/2m

vx(t) = (vxo + qEy/m) + qBsin(ωt)/m

vy(t) = vyo + qEy(t)/m - qvx0Bt/m

These equations can