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Magnetic force and electric field, analytic problem about Lorentz Force.

  • Thread starter AdamP
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  • #1
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Homework Statement



Question is E=(x,y,z) B=(x,y,z) v=(x,y,z) (in vector form)
For all E,B,v find v(t) and r(t) v and r of course have their vector arrows.


Homework Equations


F=ma= qE + qv x B

There is a hint to take second derivatives across and some terms will clear up. I was thinking more about the integrating it back to velocity and position functions.


The Attempt at a Solution



dv/dt - (q/m)v x B= qE

V= Vxi+Vyj+Vzk
B=Bxi+Byj+Bzk
E=Exi+Eyj+Ezk

so VxB = (VyBz-VzBy)i-(VxBz-VzBx)j+(VxBy-VyBx)k

dvx/dt - q/m(VyBz-VzBy) = q Ex

So I got the cross product, and put the Lorentz equation in to its differentiate form, but I am stuck here since I don't understand what I am supposed to do when the question is asking "for all v(t) and r(t)". I guess I am more stuck on the format of the answer, like how is it supposed to look like, or how many vectors/equations there needs to be by the time I am done. At the end am I supposed to get just two equations that are the derivative/integral of each other that describe the position and velocity of any given charge if data plugged in right? I would assume so since it is all analytic.

Appreciate the help.
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi AdamP ! Welcome to PF! :smile:

(on this forum, you can use bold letters for vectors, and try using the X2 tag just above the Reply box :wink:)

Yes, you need to find equations for r (and v) in terms of r0 and v0.

There's two ways you could go about this …

i] transform to a frame of reference in which either E or B = 0, solve that much easier equation, then transform back again :wink:; or

ii] try various simple algebraic manipulations, such as "dotting" or "crossing" the Lorentz force equation with r0 or v0. :smile:

(and remember eg a2 = a.a)
 
  • #3
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Hello and thank you for your help.

As far as transferring to a frame of reference where E or B are 0, could you elaborate on that a little? I am in over my head in this course and lack the necessary background to truly succeed, which is why I am struggling with this question so much. However I am willing to try as much as possible to get it done.

Also, why would you cross or dot r and v? I think I don't understand why you would dot and cross the position function with the velocity function and what would you gain out of that.

let me try a2 I did it yay :)
 
  • #4
tiny-tim
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Hello AdamP! :wink:
As far as transferring to a frame of reference where E or B are 0, could you elaborate on that a little?
oh, if you haven't covered that, just forget about it …
Also, why would you cross or dot r and v? I think I don't understand why you would dot and cross the position function with the velocity function and what would you gain out of that.
Well, for example, if you "dot" the Lorentz equation with v, the v x B goes to zero, and the dv/dt becomes v.dv/dt, which is d/dt of … ? :smile:
 
  • #5
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Ok ok I am slowly getting it. I think i struggled because I didint know about the (AxB).C = (CxA).B. So ok (VxB).V =0 from VxV=0 . If I did that I am getting vdv/dt=qEv .
Then I played around a little v=dx/dt and dv/dt=d2x/dt=a . Probably vdv/dt equals something else..hmmm

I mean can I now divide by V and just get dv/dt=qE=a . It doesnt make any sense. Would you seperate the variables and integrate dv to v and qEdt, to qEt? I dont know :(
 
  • #6
tiny-tim
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I mean can I now divide by V and just get dv/dt=qE=a . It doesnt make any sense. Would you seperate the variables and integrate dv to v and qEdt, to qEt? I dont know :(
No, you can't divide v.dv/dt by v

you can't divide a scalar by a vector!! :wink:
Then I played around a little v=dx/dt and dv/dt=d2x/dt=a . Probably vdv/dt equals something else..hmmm
Yes … v.dv/dt = 1/2 d(v.v)/dt :smile:
 
  • #7
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Yes … [B said:
v[/B].dv/dt = 1/2 d(v.v)/dt :smile:
I was away due to other exams and did not do any additional work on this problem, but I still need to solve it.

It looks like we are making progress all though you are doing most of the work. :) Anyhow, I flat out don't get that last part.

I don't see how you got 1/2 d(v.v)/dt . after that v dot v is v squared magnitude, but how does that help me. dv/dt is a...

Sorry man I am not trying to make you solve this, its just I don't get it :(
 
  • #8
tiny-tim
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I don't see how you got 1/2 d(v.v)/dt . after that v dot v is v squared magnitude, but how does that help me. dv/dt is a...
If you have an equation with a (like F = ma), you can dot it with v

then v.a = v.dv/dt = 1/2 dv2/dt, which is easy to integrate.
 
  • #9
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Oh excellent tnx tim.
So dv/dt is a, and we dotted that with v, which gave us vdv/dt and got rid of the -(q/m)vXB.

Then vdv/dt integrated to 1/2dv^2 /dt and so then it is equal to intg(qEv).

1) Now by doing all this are we closing in on the E? Like single out E on the right hand side.

2) How is the integral on the right hand side of the equation is going to work? ( I am assuming this since you integrated v(dv/dt) to 1/2 dv^2 /dt, so both sides must get the same operator.
 
  • #10
tiny-tim
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So dv/dt is a, and we dotted that with v, which gave us vdv/dt and got rid of the -(q/m)vXB.

Then vdv/dt integrated to 1/2dv^2 /dt and so then it is equal to intg(qEv).
(Not Ev, but E.v.)

So you have a energy equation … 1/2 mv2 = q∫E.v dt + constant …

that's progress, isn't it? :smile:

(and you can similarly get an angular momentum equation by differentiating r x v)

btw, in this question, are E and B constants, or do they depend on position and time?
 
  • #11
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Well, they are given as vectors with (x,y,z) coordinateness but no function of (t) is implied. However velocity and position r, are supposed to be found as functions of time v(t), r (t). I am not sure if that forces the E B and v to be as functions of time also.


How did 1/2dv^2/dt become 1/2 mv^2?

RxV ( u mean position crossed with velocity?) Is this the position function r(t) that we are looking for?
When are we going to use the VXB, that I have all the way on top? Do we even need it to go forward?
 
  • #12
tiny-tim
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Well, they are given as vectors with (x,y,z) coordinateness but no function of (t) is implied. However velocity and position r, are supposed to be found as functions of time v(t), r (t). I am not sure if that forces the E B and v to be as functions of time also.
The fact that they have (x,y,z) coordinates doesn't mean that they depend on x y and z, it only means that they are vectors.

I'm hoping they're constants, since that makes everything much easier. :smile:
How did 1/2dv^2/dt become 1/2 mv^2?
By integrating.
RxV ( u mean position crossed with velocity?) Is this the position function r(t) that we are looking for?
Yes, and no.
When are we going to use the VXB, that I have all the way on top? Do we even need it to go forward?
We dotted with v to get rid of it, but we'll need it back later, or the result won't depend on B at all, will it? :smile:
 
  • #13
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Ok the assignment states for partial credit you can assume Bi=B(i) for all i, Bx=B(x) By=(y)
I guess that is what you are talking about since supposedly this makes the problem much easier, however does not provide full credit.
 

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