Trajectory of a charged particle in the magnetic field (variable)

In summary, the problem reads as follows: An electron travels parallel to an current carrying wire with an initial velocity of 10^5m/s (y) and its position being x=0.2 (meters) and y=0 at t=0. Determine the trajectory of the electron and plot it in geogebra.f
  • #1
Homework Statement
Find the trayectory(x,y) of an electron traveling in a magnetic field created by an an infinite current carrying wire current with intensity 5A. At x=0.2 v=10^5 m/s (y)
Electron charge q=-1.6x10^-19
u0/2π=2*10^-7

The magnetic field varies with the distance of the particle to the wire.

I have been working on this problem but I just simply can work this out, i will link some articles that I have read:
https://aip.scitation.org/doi/pdf/10.1063/5.0063755
https://www.sciencedirect.com/science/article/pii/S2211379719302268

At the start we only have j component of the velocity, but as time passes on we could imagine (at least myself) that Vi starts to exist because of the acceleration created by the force.

This would create another component of force, in this case this force would point downwards.

Fx is created by Vy and Bz
Fy is created by Vx and Bz
Relevant Equations
F=q(VxB)=qVB V(y) perpendicular to B(z)
B=(u0*I)/(2πR)

If i'm not wong, this are the ecuations of acceleration.

ax=(q*Vy*u0*I)/(2πRm)
ay=(q*Vx*u0*I)/(2πRm)
electron.png
 
  • #2
You don’t specify the orientation of the wire, nor the initial y of the particle.
I could deduce these from your diagram, but that comes from your interpretation. I'd like to see exactly how the problem was stated.
 
  • #3
You don’t specify the orientation of the wire, nor the initial y of the particle.
I could deduce these from your diagram, but that comes from your interpretation. I'd like to see exactly how the problem was stated.
Thanks for replying,
1) The wire rests on the y axis, and the initial y of the particle is 0.
2) The problem reads as follows:
An electron travels parallel to an current carrying wire with initial velocity being 10^5m/s (y) and it's position being x=0.2 (meters) and y=0 at t=0 Determine the trajectory of the electron and(optional) plot it geogebra.
 
  • #4
Thanks for replying,
1) The wire rests on the y axis, and the initial y of the particle is 0.
2) The problem reads as follows:
An electron travels parallel to an current carrying wire with initial velocity being 10^5m/s (y) and it's position being x=0.2 (meters) and y=0 at t=0 Determine the trajectory of the electron and(optional) plot it geogebra.
Ok, thanks.
Shouldn't one of your acceleration equations have a minus sign? Check using the constant field version.
 
  • #5
Ok, thanks.
Shouldn't one of your acceleration equations have a minus sign? Check using the constant field version.
1) Sorry, i don't understand why there would be a minus sign
As my current understanding of the Lorentz Force Fx=q(vyxBz)=qvyBz (because v is perpendicular to B)
Doing this vectorial product gives us a force pointing towards -x, but because we are working with an electron, (-q) points towards +x


2) I have tried to simplify the problem to the constant field version but still, I am unable to solve it, it gets complicated really fast and goes beyond my current understanding (even if i have been working on this for about 1-2 weeks)
 
  • #6
1) Sorry, i don't understand why there would be a minus sign
As my current understanding of the Lorentz Force Fx=q(vyxBz)=qvyBz (because v is perpendicular to B)
Doing this vectorial product gives us a force pointing towards -x, but because we are working with an electron, (-q) points towards +x


2) I have tried to simplify the problem to the constant field version but still, I am unable to solve it, it gets complicated really fast and goes beyond my current understanding (even if i have been working on this for about 1-2 weeks)
For the constant field, you know it should go in a circle, so convert to polar and set ##r, \dot\theta## as constants.

Edit: you also know that even in the general case the speed should be constant. I can only deduce that from your equations if I flip the sign of one.
 
Last edited:
  • #7
I was able to get a third order ODE in x(t) only, but none of the online solvers I've found can find a solution.
There may be a way to find the trajectory as y(x) but not as (x,y) as a function of t.
Maybe start by plotting it.
 
  • #8
I just asked my professor via email and she told me that x,y(t) and y(x) are both worth it

So let's use the easiest of both of them
 
  • #9
I just asked my professor via email and she told me that x,y(t) and y(x) are both worth it

So let's use the easiest of both of them
Unfortunately I cannot see how to obtain an ODE not involving time.
 
  • #10
Unfortunately I cannot see how to obtain an ODE not involving time.
What field of mathemathics should I learn in order to solve this problem?( x,y(t) or y(x) )
Thanks!

And could you tell me how did you obtain x(t)? thanks!
 
Last edited:
  • #11
The basic equations are $$\ddot x = \frac{c}{x}\dot y \,\,\,\,\, \,\,\,\,\,\,\,\,\,\, \ddot y = -\frac{c}{x}\dot x$$ where ##c## is a positive constant. The y-axis runs along the current and the origin is chosen such that the initial position of the electron is ##(x_0, y_0) =## (0.2 m, 0).

You can immediately integrate the second equation to obtain $$\dot y = f(x)$$ where you will find that ##f(x)## can be written in terms of ##c##, ##v_0##, and ##\log(x/x_0)##.

Using the constant speed condition ##{\dot x}^2 + {\dot y}^2 = v_0^2##, $$\dot x = \pm \sqrt{v_0^2-f(x)^2}$$ where the minus sign is used for the parts of the trajectory where ##x## is decreasing. In principle, this equation could be integrated to obtain ##t## as a function of ##x## and then inverted to obtain ##x(t)##. But the integration looks to me to be too difficult to do except numerically.

Note that ##\large \frac{dy}{dx} = \frac{\dot y}{\dot x}##. Thus, you can obtain $$ \frac{dy}{dx} = \pm \frac{f(x)}{\sqrt{v_0^2-f(x)^2}} \equiv h(x) $$

Then $$y(x) = \int_{x_0}^x h(\bar x) d{\bar x}$$

Again, the integrand looks to me to be too complicated to evaluate. I did check that doing this integral numerically yields the same trajectory ##y(x)## as obtained by numerically integrating the initial equations of motion for ##\ddot x## and ##\ddot y## to get a numerical solution for ##x(t)## and ##y(t)##.
 
  • #12
First of all, thank you so much, this really helps me a lot!
I have a few questions:
1) How could I numerically integrate the expression? y(x)? or x(t)/y(t)
2) I'm having real trouble when trying to integrate y(x), online solvers fails too.
 
Last edited:
  • #13
1) How could I numerically integrate the expression? y(x)? or x(t)/y(t)
2) I'm having real trouble when trying to integrate y(x), online solvers fails too.
I used Mathematica to do the numerical integrations. It didn't seem to have any trouble. I have not studied numerical techniques very much. To me, Mathematica is magic in a black box. I'm not familiar with any online software that you could use. Has your professor indicated whether or not she would accept solutions in the form of graphs obtained from numerical integration?
 
  • #14
I used Mathematica to do the numerical integrations. It didn't seem to have any trouble. I have not studied numerical techniques very much. To me, Mathematica is magic in a black box. I'm not familiar with any online software that you could use. Has your professor indicated whether or not she would accept solutions in the form of graphs obtained from numerical integration
The task that my professor asked consists on drawing the trajectory in Geogebra.
I have to draw a point(x,y) that represents the electron, and then, by changing the x, y get an y, wich gives me a point(x,y) then, by changing the value of x automatically, the trajectory of the electron is apreciated.
When I try to integrate the expression cln(x) / sqrt(v² - (cln(x))²), on geogebra, it just doesn't work (maybe I am wrong somewere?)

f(x)=-cln(x)
1638914578059.png

i tried integrating h(x) with 0,10, but the program returned an ? as an answer.
I will ask my professor if she can accept the solution in the form of graphs but the answer would probably be no :(
Thanks!

edit: trying with any limits of integration, for example 8,10 also gives error.
 
Last edited:
  • #15
I'm not familiar with GeoGebra. But, from a quick search I found some examples of using GeoGebra to plot a numerical solution to a first-order differential equation. For example, one of the simplest numerical methods is "Euler's method", which is not great but might be sufficient. See this video example.

f(x)=-cln(x)

This is not quite correct. Note that earlier we had ##\dot y = f(x)##.
When ##x = x0##, we know ##\dot y = v_0##. So, we must have ##f(x_0) = v_0##. Your expression for ##f(x)## does not satisfy this initial condition.

i tried integrating h(x) with 0,10, but the program returned an ? as an answer.

The lower limit of integration cannot be ##x = 0##. This would place the electron at the location of the current. The starting position is ##x = x_0 = .2\,##m and ##y = 0##.

View attachment 293751

edit: trying with any limits of integration, for example 8,10 also gives error.

The problem here could be that your incorrect expression for ##f(x)## is producing a negative argument of the square root. I don't know what software you are using here.
 
  • #16
This is not quite correct. Note that earlier we had ##\dot y = f(x)##.
When ##x = x0##, we know ##\dot y = v_0##. So, we must have ##f(x_0) = v_0##. Your expression for ##f(x)## does not satisfy this initial condition.
Sorry, maybe I am missing something out, but, when I integrate aceleration (y) I get as aresult Cln(x) +C, and then i can find the constant of integration by aplying f(0.2)=105.

What am I missing out over here?
 
  • #17
Sorry, maybe I am missing something out, but, when I integrate aceleration (y) I get as aresult Cln(x) +C,

Don't confuse the two C's here. They do not represent the same constant.

Initially, I wrote ##\ddot x = c \frac{\dot y}{x}## and ##\ddot y = -c \frac{\dot x}{x}##. Here, ##c## is a positive constant that you can express in terms of ## \mu_0, I, m## and ##e##. (##e## is the magnitude of the electron charge; i.e., ##e## is a positive number.) When you do the integration of ##\ddot y##, you should get ##\dot y = -c\ln(x) + C_1##, where ##C_1## is the constant of integration. Note the negative sign in front of the log term. Find ##C_1## from the initial conditions.

and then i can find the constant of integration by aplying f(0.2)=105.
I would not plug in numbers at this stage. Express ##C_1## in terms of the symbols ##v_0, x_0## and ##c##.

When it does come time to plug in numbers, I suggest that you take the microsecond as the unit of time. The unit of length can be kept as the meter, or you could choose centimeters if you like. The initial velocity will then be ##v_0 = 0.1## m/μs ##= 10 ## cm/μs. This avoids the messy powers of 10 for the velocities that you would have to carry around if you used m/s for the velocity unit. Also, with time in μs, the constant ##c## will have a nicer numerical value compared to using seconds as the time unit. (You should be able to show that ##c## has the dimensions of velocity.)
 

Suggested for: Trajectory of a charged particle in the magnetic field (variable)

Back
Top