Motion of a particle in a magnetic field

  • #1
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Homework Statement:
A particle q > 0, mass = m, v = ##\vec{v}##

Magnetic field ##\vec{B} = B\hat{u}##

friction ##\vec{F} = -k\vec{v}##

At t = 0
##x = x_0 , y = y_0, z = z_0##
##v_x = v_{0x} , v_y = v_{0y}, v_z = v_{0z}##
Relevant Equations:
##\vec{B} = qvB sin \theta##
Hi,
I have to find the motion of a particles ##(x,y,z)##. However, I'm not sure where to begin.

Is it correct to split the problem and first find what's the motion in the x direction then y and z.

For exemple,

##m \frac{d^2x}{dt^2} = -kv_{0x} + qv_{0x}B sin 90 ##

##m\int\int \frac{d^2x}{dt^2} = \int \int -kv_{0x} + qv_{0x}B sin 90 ##
 

Answers and Replies

  • #2
kuruman
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What is the direction ##\hat u## of the magnetic field? If it's arbitrary, then you need to write three equations, one for each cartesian axis. You can make your life simpler if you rewrite the velocity vector in terms unit vectors one of which is long ##\hat u## and two are perpendicular to it. That simplifies the problem because you have circular motion in the plane perpendicular to the field and drift in the direction of the field. I think that's the best way to split the motion with or without friction.
 
  • #3
berkeman
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Relevant Equations:: ##\vec{B} = qvB sin \theta##
It's better to work with the vector Lorentz Force equation. Have you learned that yet? BTW, the equation you show above has a vector on the left hand side (LHS) and a scalar on the RHS...
 
  • #4
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Sorry,
The magnetic field ##\vec{B} = B\hat{u}_z##

And I haven't learned yet The Lorentz force equation.

I tried all the weekend, but I don't even get the correct answer for x which is
##x = -\tau V_{0x} cos \beta e^{-t/\tau} cos(\omega t + \beta) + x_0 + \tau cos^2 \beta V_{0x}##

Where ##\tau = m/k##, ##\omega = qB/m##, ##tan(\beta) = \omega t##

kuruman, I'm not sure to understand when you say that I should rewrite the velocity vector in terms unit vectors.

I tried to integrate

## m \int d^2x =\int -kv_{0x} + qv_{v0x}B sin 90 dt^2##

But I'm not even sure if this is correct.
 
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  • #5
kuruman
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@kuruman, I'm not sure to understand when you say that I should rewrite the velocity vector in terms unit vectors.
I mean write ##\vec v = v_x~\hat x+v_y~\hat y+v_z~\hat z##. Then if, as you say, the magnetic field is in the ##+z##-directon, ##\vec B=B~\hat z## and the magnetic force is ##\vec F_M=q \vec v\times \vec B=qB(\vec v\times \hat z)## which you can write out in terms of the unit vectors ##\hat x##, ##\hat y## and ##\hat z##.

Add the result to the friction force ##\vec f=-kv_x~\hat x-kv_y~\hat y-kv_z~\hat z## and you have the net force. Then write three equations for Newton's second law, one for each principal axis.
 
  • #6
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If the result is ##x = -\tau V_{0x} cos \beta e^{-t/\tau} cos(\omega t + \beta) + x_0 + \tau cos^2 \beta V_{0x}##

I shouldn't write the magnetic force as ##F = qvB sin 90## ?

If ##v = v_{0x}## and B is in the +z direction then we have sin 90 ?

Or

##m\ddot{x} = qB(\vec{v} \times \hat{z}) - kv##

##=> \ddot{x} = \frac{qB}{m}(v_{0x} - v_{0y}) +(- kv_{0x} - kv_{0y} - kv_{0z})/m ##
 
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  • #7
kuruman
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If the result is ##x = -\tau V_{0x} cos \beta e^{-t/\tau} cos(\omega t + \beta) + x_0 + \tau cos^2 \beta V_{0x}##

I shouldn't write the magnetic force as ##F = qvB sin 90## ?

If ##v = v_{0x}## and B is in the +z direction then we have sin 90 ?
No. The velocity is along the ##x##-axis only at some instances.
##m\ddot{x} = qB(\vec{v} \times \hat{z}) - kv##
##=> \ddot{x} = \frac{qB}{m}(v_{0x} - v_{0y}) +(- kv_{0x} - kv_{0y} - kv_{0z})/m ##
That's not how it works. You have three differential equations that form a system and must be solved simultaneously. These are
##m\ddot x=qB(\vec{v} \times \hat{z})_x-kv_x##
##m\ddot y=qB(\vec{v} \times \hat{z})_y-kv_y##
##m\ddot z=qB(\vec{v} \times \hat{z})_z-kv_z##
and can be summarized in one vector equation as
$$m\dfrac{d^2 \vec r}{dt^2}=qB(\vec{v} \times \hat{z})-k\vec v$$ where ##\vec r=x~\hat x+y~\hat y+z~\hat z## is the position vector. This is the three equations above written compactly as one. The vector on the left is the same as the vector on the right. If you set the components of the vector on the left equal to the components of the vector on the right, you get the three equations that I listed.

That's how it works. Do you know how to find the Cartesian components of a cross product? If not, look here.
 
  • #8
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##x \times z = -y##, but I'm not sure how it works the way you write it.Is it the same?

##(v_x,v_y,v_z) \times (0,0,z) = v_y \times z = v_x##

thus,
##\ddot{x} = \frac{qB}{m}(v_x) -kv_x##
##\ddot{x} -\frac{qB}{m}(v_x) + kv_x = 0##

I'm not sure how to integrate this.

It is close to ##\ddot{x} + (b/m)\dot{x} + (k/m) x = 0##
Is it the same as a pendulum. I mean, I have to guess x = ?, to find ##\dot{x}## then I plug those values in the equation above.

I don't see how to find x. I don't have any of that in my book. And I only have the guess for ##\ddot{x} + (b/m)\dot{x} + (k/m) x = 0## in my waves and oscillations book, but I don't find any explanation on how to get that guess.

Am I on the right way?
 
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  • #9
kuruman
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You can integrate the third equation in z after you separate variables, but you cannot do it with equations in x and y because they are coupled. You need to decouple them first. It seems that this is the first time you encounter coupled differential equations. You need to see what is involved when you try to solve them. To that end I suggest that you do some research on the web and then solve the simpler problem with no friction. Tricks and substitutions are involved and you have to get your feet wet as it were. Solving the problem first without friction will help you imagine what might be involved when you turn the friction on.
 
  • #10
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Alright,

I didn't find a lot of things about coupled second order differential equation.

However, I'll try to find the solution for z which is not coupled.

I found that for ##\ddot{z} + (k/m) \dot{z}= 0##

##z = e^{ct}## then
##\dot{z} = ce^{ct}##
##\ddot{z} = c^2e^{ct}##

Then I plug the z's in the equation above to find the value of c.
I should have 2 different values because this is a second order equation.

I get something like
##e^{ct}(c^2 +k/m) = 0##
so ##c^2 + k/m = 0##
##c^2 = - k/m## ... Ouch
 
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  • #11
kuruman
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Alright,

I didn't find a lot of things about coupled second order differential equation.

However, I'll try to find the solution for z which is not coupled.

I found that for ##\ddot{z} + (k/m) \dot{z}= 0##

##z = e^{ct}## then
##\dot{z} = ce^{ct}##
##\ddot{z} = c^2e^{ct}##

Then I plug the z's in the equation above to find the value of c.
I should have 2 different values because this is a second order equation.

I get something like
##e^{ct}(c^2 +k/m) = 0##
so ##c^2 + k/m = 0##
##c^2 = - k/m## ... Ouch
It’s a first order diff eq in ##v_z##. Find that as a function of time then integrate once more.
 
  • #12
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I'm still stuck. I didn't make any progress.

I try to learn how to solve a first order differential equation, but I don't understand what I'm doing.. I never did it before.
I don't know if the way I write the newton's second law for z is correct.

Is it the same process for a homogeneous first order differential equation as for a "regular" first order differential equation?

For exemple I found this link,but in my case k/m is not a function of t.

##\ddot{z} = -k/m \cdot \dot{z}##

##\frac{dv}{dt} = -k/m \cdot v##

##\int \frac{dv}{v} = \int -k/m \cdot dt##

##ln |v| + C_1 = -k/m \cdot t + C_1##

##v =Ae^{(-t/\tau) + C}## where ##\tau = m/k##
 
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  • #13
kuruman
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I'm still stuck. I didn't make any progress.

I try to learn how to solve a first order differential equation, but I don't understand what I'm doing.. I never did it before.
I don't know if the way I write the newton's second law for z is correct.

Is it the same process for a homogeneous first order differential equation as for a "regular" first order differential equation?

For exemple I found this link,but in my case k/m is not a function of t.

##\ddot{z} = -k/m \cdot \dot{z}##

##\frac{dv}{dt} = -k/m \cdot v##

##\int \frac{dv}{v} = \int -k/m \cdot dt##

##ln |v| + C_1 = -k/m \cdot t + C_1##

##v =Ae^{(-t/\tau) + C}## where ##\tau = m/k##
I would recommend that you set limits to the integrals on the two sides of the equation. This is a physics problem. When time t is zero the z component of the velocity has some value ##v_{0z}.##. These are the lower limits. At some later time ##t## the z-component will be what you’re looking for ##v_z##. Do the integrals using those limits and you will not need integration constants.
 
  • #14
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I tried something else and I think it works.

By guessing z
##z = e^{rt}##
##\dot{z} = re^{rt}##
##\ddot{z} = r^2e^{rt}##
thus,
##re^{rt}(r + \frac{1}{\tau}) = 0##

r must be 0 or ##-1/\tau##


##z = C_1e^{0t} + C_2e^{-t/\tau}##
##z(t=0) z_0 = C_1 + C_2##
##v(t=0) = v_0##

##\dot{z} = (1/\tau) c_2^{-t/\tau}##

##v_0 = (-1/\tau) C_2##

then,

##C_2 = -v_0\tau##
##C_1 = Z_0+ v_0\tau##

Finally,
##z(t) =z_0 +\tau v_0(1-e^{-t/\tau})##

Unfortunately, I still need to solve for x and y.
 
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  • #15
kuruman
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You are not done with z yet. You need to define constant ##\tau## in terms of the given quantities. Doing x and y is more involved. Can you guess what the motion looks like and describe it?
 
  • #16
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You are not done with z yet. You need to define constant ##\tau## in terms of the given quantities. Doing x and y is more involved. Can you guess what the motion looks like and describe it?
##\tau = m/k##
it's given in the question, sorry.

I had to find that ##z(t) = z_0 + \tau v_0(1-e^{-t/\tau})##

for y I have

##\ddot{y} = -\omega \dot{x} - \frac{k}{m} \dot{y}##

Then I have to use the separation of variables method?
 
  • #17
kuruman
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I would solve the v˙x and v˙y equations first then integrate, just like you did with z. These two equations need to be solved simultaneously. You cannot say that you will solve one first and then the other. What are the two velocity equations?

Note that
##\vec v\times \hat z=(v_x~\hat x+v_y~\hat y+v_z~\hat z)\times \hat z=\dot x~(\hat x\times \hat z)+\dot y~(\hat y\times \hat z)+\dot z~(\hat z\times \hat z).## Can you simplify the expression?
 
  • #18
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##\vec{y} \times \hat{z} =-\dot{x}\hat{y} + \dot{y}\hat{x}##

Can I isolate ##\dot{y} ## from the first equation then plug ##\dot{y}## and ##\ddot{y}## in the second equation ?
 
  • #19
kuruman
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##\vec{y} \times \hat{z} =-\dot{x}\hat{y} + \dot{y}\hat{x}##

Can I isolate ##\dot{y} ## from the first equation then plug ##\dot{y}## and ##\ddot{y}## in the second equation ?
That equation should be ##\vec{v} \times \hat{z} =-\dot{x}\hat{y} + \dot{y}\hat{x}##, not what you show. This is only a piece of the three differential equations that are shown in posting #7. You have already solved the third equation in ##z##. The other two equations in ##x## and ##y## are coupled. Can you write them both, one below the other?
 
  • #20
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That equation should be ##\vec{v} \times \hat{z} =-\dot{x}\hat{y} + \dot{y}\hat{x}##, not what you show.
Yeah I made a typo error... It was 4 am.


##\ddot{x} = \omega \dot{y} - \frac{k}{m} \dot{x}##
##\ddot{y} = - \omega \dot{x} - \frac{k}{m} \dot{v_y}##
It is not that? I don't understand.

Otherwise,
Is it correct to say:
from the first equation
##\dot{y} = \frac{m\ddot{x} + k \dot{x}}{\omega m}##

##\ddot{y} = \frac{m\dddot{x} + k\ddot{x}}{\omega m}##

Then replace both of them in the second equation and solve for ##p = \dot{x}##

If this is correct, I'm not sure how to find the solution for ##p##.
 
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  • #21
kuruman
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You are not helping me help you. You say you cannot solve the equation ##p=\dot x##. That's OK, but what is ##p##? You assume that, if I followed the procedure you describe, I would get what you got for ##p##. But I don't know that you did it right if you don't show me step by step what you did and what you got as a result. You should also be more specific about what you cannot do and why. So please post the final differential equation you get and explain to me why you cannot solve it and what kind of differential equation you can solve. Then I will be in a better position to help you (if I can).

Also ##\ddot{y} = - \omega \dot{x} - \frac{k}{m} \dot{v_y}## is incorrect. It should be ##\ddot{y} = - \omega \dot{x} - \frac{k}{m} \dot{y}##. Consider whether your difficulties arise from being careless and/or working late or from lacking a basic understanding. Only you can address this issue.
 
  • #22
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I type and retyping and then writing on a paper and rewriting using different method and I just mix everything, Sorry.


I'll type from scratch. Hopefully I'll not make any error.

##\vec{B} = B\hat{z}, \vec{F} = -k\vec{v}, \vec{v} = v_x\hat{x} + v_y\hat{y}##

##F = q(v_x\hat{x} + v_y\hat{y}) \times B\hat{z} -k(v_x\hat{x} + v_y\hat{y})##

##F = qB(-v_x\hat{y} + v_y\hat{x}) -k(v_x\hat{x} + v_y\hat{y})##

##m\ddot{x} = qB\dot{y} - k\dot{x}##

##m\ddot{y} = -qB\dot{x} - k\dot{y}##

##\omega = \frac{qB}{m}, \tau = \frac{m}{k}##

##\ddot{x} = \omega \dot{y} - \frac{1}{\tau}\dot{x}## (1)
##\ddot{y} = - \omega \dot{x} - \frac{1}{\tau}\dot{y}## (2)

From equation (1)
##\dot{y} = \frac{\tau \ddot{x} + \dot{x}}{\omega \tau}##
##\ddot{y} = \frac{\tau \dddot{x} + \ddot{x}}{\omega \tau}##

Then, I replace both of them in the second equation.
##\frac{\tau \dddot{x} + \ddot{x}}{\omega \tau} = -\omega \dot{x} -\frac{1}{\tau}(\frac{\tau \ddot{x} + \dot{x}}{\omega \tau})##

##\frac{\tau \dddot{x} + \ddot{x}}{\omega \tau} +\omega \dot{x} +\frac{1}{\tau}(\frac{\tau \ddot{x} + \dot{x}}{\omega \tau}) = 0##

##(\frac{\tau \dddot{x} + \ddot{x}}{\omega \tau} +\omega \dot{x} +\frac{1}{\tau}(\frac{\tau \ddot{x} + \dot{x}}{\omega \tau})) \omega \tau^2 = 0##

##(\omega^2\tau^2 + 1)\dot{x} + (\omega \tau + \tau)\ddot{x} + (\omega \tau^2)\dddot{x} = 0##

Thus, if ##p = \dot{x}##
##(\omega^2\tau^2 + 1)p + (\omega \tau + \tau)\dot{p} + (\omega \tau^2)\ddot{p} = 0##


Finally, If I didn't make any error, I get a second order differential equation for p.

At this point, Can I simply guess the same solution for p as I did for z ?
 
  • #23
kuruman
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This equation is dimensionally inconsistent.
##(\omega^2\tau^2 + 1)p + (\omega \tau + \tau)\dot{p} + (\omega \tau^2)\ddot{p} = 0##
You can see that because ##\omega \tau## is a dinesionless quantity. Then the first term must have dimensions of ##p## which is velocity or [LT-1]. The second term is messed up because the coefficient ## (\omega \tau + \tau)## is the sum of a dimensionless quantity and a quantity that has dimensions of time. The third term has a coefficient that has dimensions of [T] and the ##\ddot p## term has dimensions [LT-3] so that the third term has dimensions [ [LT-2] which is not the same as the dimensions of the first term. You need to fix this.

At this point, Can I simply guess the same solution for p as I did for z?
I'd rather you didn't simply guess. You need to solve the second-order homogeneous ODE formally using the characteristic polynomial. If you don't know how to do this, look here. Study at least the first two sections, "Basic concepts" and "Real roots".
 
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  • #24
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I don't understand what you mean by dimensionally inconsistent.


he second term is messed up because the coefficient (ωτ+τ) is the sum of a dimensionless quantity and a quantity that has dimensions of time.

##\tau = m/k## how come ##\tau## has dimension of time. I don't understand.


If you don't know how to do this, look here. Study at least the first two sections, "Basic concepts" and "Real roots".

This is the link I was using to find ##z##, but I have hard time to find the value of ##r##. It is not as simple as 3.
 
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  • #25
kuruman
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I don't understand what you mean by dimensionally inconsistent.
Do you understand what dimensional analysis is all about? If not, read this. In short, it is a method for making sure that you are adding not apples and oranges to get bananas. As I indicated in posting #23, the three terms in your equation have different dimensions. For that reason, you cannot add them together.

##\tau = m/k## how come ##\tau## has dimension of time. I don't understand.
Look at the argument of the exponential in the equation that you wrote: ##z(t) =z_0 +\tau v_0(1-e^{-t/\tau})##. What dimension do you think ##\tau## has?
This is the link I was using to find ##z##, but I have hard time to find the value of ##r##. It is not as simple as 3.
We are back to you telling me that you cannot do something without showing to me what it is that you cannot do.

This is what you have to do in this order
1. Derive the correct form of this equation: ##(\omega^2\tau^2 + 1)p + (\omega \tau + \tau)\dot{p} + (\omega \tau^2)\ddot{p} = 0##. Your method is on the right track but your algebra is flawed. I am not going to derive it for you, but I can tell you if your equation agrees with mine.
2. Get the characteristic polynomial and solve for ##r##.
3. Write down the most general solution and apply the initial conditions.
4. Integrate the expressions for the velocity components to get ##x(t)## and ##y(t)##.
 

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