Theoretical Mechanics

  • #1
akinoshigure
2
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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.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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970
akinoshigure said:
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/
 

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