Spring-mediated dipole in a magnetic field

[Hak]

Homework Statement

Two charges having equal magnitude and opposite sign are connected by a spring with rest length zero, initially inextended, within a region of space in which there is a uniform magnetic field $\vec B$. The charges begin to move with velocity $\vec v$ perpendicular to $\vec B$.

$a$. Describe the system's motion and the forces under examination.
$b$. How many degrees of freedom are there in the system?
$c$. The center of mass is desired to move in uniform rectilinear motion, what initial conditions must be imposed? And what appears in the frame of reference of the center of mass?

Relevant Equations

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Here is my attempt to answer the questions:

a. The system's motion is a combination of two types of motion: the translational motion of the center of mass and the rotational motion of the charges around the center of mass. The forces acting on the system are the Lorentz force, which is the force exerted by the magnetic field on the moving charges, and the spring force, which is the restoring force that tends to keep the charges at their equilibrium distance. The Lorentz force is given by $$\vec F = q \vec v \times \vec B$$, where $$q$$ is the charge, $$\vec v$$ is the velocity, and $\vec B$ is the magnetic field. The spring force is given by $$\vec F = -k \vec x$$, where $$k$$ is the spring constant and $$\vec x$$ is the displacement from the equilibrium position. The Lorentz force causes the charges to move in circular paths in planes perpendicular to $\vec B$, while the spring force causes them to oscillate along the line joining them.

b. The number of degrees of freedom of a system is the number of independent parameters that can be varied to describe its state. For this system, we can choose three parameters: the position of the center of mass, the angle of rotation of the charges around the center of mass, and the distance between the charges. Therefore, the system has three degrees of freedom.

c. For the center of mass to move in uniform rectilinear motion, it must have a constant velocity that is parallel to $$\vec B$$. This means that the initial velocity of each charge must also be parallel to $$\vec B$$, so that there is no component of velocity perpendicular to $$\vec B$$ that would cause a Lorentz force. In addition, the initial distance between the charges must be equal to their equilibrium distance, so that there is no spring force acting on them. In this case, there will be no net force on the system and it will move with constant velocity along $$\vec B$$. In the frame of reference of the center of mass, the charges will appear to be at rest relative to each other.

Since I am pretty sure the answers are all wrong, could you help me out?

 

[haruspex]

we can choose three parameters: the position of the center of mass, the angle of rotation of the charges around the center of mass, and the distance between the charges. Therefore, the system has three degrees of freedom.

A position in 3 space has 3 degrees of freedom and the orientation of the system has two, so that adds up to six. But it is not clear to me whether it only wants those degrees that will be exercised in this setup. What does it mean by "begins to move"? Is that some unknown imposed movement (as suggested by part c) or just the movement that would arise from being released from rest with the spring at relaxed length?

In your analysis, I don't see the electrostatic attraction of the particles.

 

[Hak]

A position in 3 space has 3 degrees of freedom and the orientation of the system has two, so that adds up to six. But it is not clear to me whether it only wants those degrees that will be exercised in this setup. What does it mean by "begins to move"? Is that some unknown imposed movement (as suggested by part c) or just the movement that would arise from being released from rest with the spring at relaxed length?

Thank you. I lean toward the first option, but I think both possibilities, although different from each other, should be examined.

 

[Hak]

In your analysis, I don't see the electrostatic attraction of the particles.

Okay, I'll try to add it.

a. The system consists of two charges, $q$ and $-q$, connected by a spring with spring constant $k$ and rest length zero. The system is placed in a uniform magnetic field $\mathbf{B}$ and the charges begin to move with velocity $\mathbf{v}$ perpendicular to $\mathbf{B}$. The forces acting on each charge are the Lorentz force, the spring force, and the electrostatic force. The Lorentz force is given by $\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$, where $x$ denotes the cross product. The spring force is given by $\mathbf{F} = -kx$, where $x$ is the displacement from the equilibrium position. The electrostatic force is given by $\mathbf{F} = k_e \frac{q^2}{r^2}$, where $k_e$ is the Coulomb constant and $r$ is the distance between the charges. The system’s motion can be described by two coupled differential equations for the coordinates of the charges along the direction of $\mathbf{v}$, denoted by $x_1$ and $x_2$. These equations are:
$$m \frac{d^2 x_1}{dt^2} = q v B - k (x_1 - x_2) + k_e \frac{q^2}{(x_1 - x_2)^2}$$
$$m \frac{d^2 x_2}{dt^2} = - q v B + k (x_1 - x_2) - k_e \frac{q^2}{(x_1 - x_2)^2}$$.
These equations can be solved by introducing the center of mass coordinate $X = \frac{(x_1 + x_2)}{2}$ and the relative coordinate $x = x_1 - x_2$. Then, the equations become:
$$m \frac{d^2 X}{dt^2} = 0$$
$$m \frac{d^2 x}{dt^2} = -k x + 2 k_e \frac{q^2}{x^3}$$.
The first equation implies that the center of mass moves with a constant velocity along the direction of $\mathbf{v}$. The second equation describes an oscillatory motion of the relative coordinate with an effective potential $$U(x) = k \frac{x^2}{2} - 4 k_e \frac{q^2}{x}$$.

b. I had thought that the number of degrees of freedom in the system is four, but it is wrong. However, I report my reasoning. We need four independent coordinates to specify the state of the system completely. These coordinates can be chosen as $X$, $x$, $\theta_1$, and $\theta_2$, where $\theta_1$ and $\theta_2$ are the angles that each charge makes with respect to a fixed axis perpendicular to $\mathbf{B}$.

 

[haruspex]

I think you need to use vectors for the positions of the particles, or introduce an angle of rotation. The forces arising from the field will not be collinear with their motion.

 

[Hak]

I think you need to use vectors for the positions of the particles, or introduce an angle of rotation. The forces arising from the field will not be collinear with their motion.

I will try to think about it some more. Could you give me some suggestions? I don't think I understand perfectly...

 

[haruspex]

I will try to think about it some more. Could you give me some suggestions? I don't think I understand perfectly...

As the charges are pulled together initially by their charges, which way will the B field push them?

 

[erobz]

I will try to think about it some more. Could you give me some suggestions? I don't think I understand perfectly...

The force acting on the particle defined by $\vec{F} = q \vec{v} \times \vec{B}$ . Your equation adds them as if it is always colinear with the other radial forces, when that force is not defined that way. That is only the case at a particular instant when $\vec{v} \perp \vec{B}$ (initially for instance).

 

[TSny]

I don’t understand the problem statement where it says that the spring has zero rest length and is initially unstretched. Wouldn’t that mean that the two charges are initially at the same location? If so, then the Coulomb attraction would be infinitely strong initially.

 

[haruspex]

That is only the case at a particular instant when $\vec{v} \perp \vec{B}$ (initially for instance).

Did you mean when $\vec v$ has no radial component?

 

[erobz]

Did you mean when $\vec v$ has no radial component?

Maybe I'm confused. They have written that $F = qvB$ in the radial direction for all time, is that correct?

 

[Hak]

Ok, thanks. I'll try to think about it.

 

[haruspex]

Maybe I'm confused. They have written that $F = qvB$ in the radial direction for all time, is that correct?

Who has? I don’t see that in the problem statement in post #1.
If we define the z direction as that of the B field and x as that of the initial extension (as @TSny notes, it cannot be unextended) then the initial movement is along x, making the force from the field along y. Of course, that will lead to a rotation of system and hence to a radial component of the Lorentz force.

 

[haruspex]

Ok, thanks. I'll try to think about it.

Try rewriting your ODEs in post #4 as vector equations. E.g. this doesn’t work because you have a vector on the left but only scalars on the right:

$\mathbf{F} = k_e \frac{q^2}{r^2}$

Alternatively, switch to polar coordinates.

 

[kuruman]

I don’t understand the problem statement where it says that the spring has zero rest length and is initially unstretched. Wouldn’t that mean that the two charges are initially at the same location? If so, then the Coulomb attraction would be infinitely strong initially.

I don't understand this either for exactly the same reasons. The only way it makes sense to me is to say that the spring has relaxed length $L$. Then one can find the equilibrium position $x_0$ (before the masses start moving) using $k (L-x_0)=\dfrac {k_e q^2}{x_0^2}.$ We are told that the spring is initially "inextended" but that's obvious because the spring must, in fact, be compressed.

 

[Hak]

Thank you for your advice and findings. Could you group them more compactly so that I can attempt a reasoning and development? I am slightly confused...

 

[kuruman]

Thank you for your advice and findings. Could you group them more compactly so that I can attempt a reasoning and development? I am slightly confused...

If you are addressing me, I cannot make what I said more compact.

 

[Hak]

If you are addressing me, I cannot make what I said more compact.

I mean everyone, actually. There are many aspects that I cannot group together, it would be useful to arrange them more organically...

 

[TSny]

Thank you for your advice and findings. Could you group them more compactly so that I can attempt a reasoning and development? I am slightly confused...

We're all confused by the statement of the problem. Did you double-check that you typed in the problem statement precisely as given to you (word-for-word) with nothing extra added and nothing left out?

It might be helpful to know where you got the problem. Please specify the source.

 

[Hak]

We're all confused by the statement of the problem. Did you double-check that you typed in the problem statement precisely as given to you (word-for-word) with nothing extra added and nothing left out?

It might be helpful to know where you got the problem. Please specify the source.

This problem comes from an oral examination of a famous school of excellence in Italy, Scuola Normale Superiore. The statement of the problem is as I reported, not a word more nor a word less, translated from Italian. The precise meaning is preserved. A very long written work is required.

 

[TSny]

This problem comes from an oral examination of a famous school of excellence in Italy, Scuola Normale Superiore. The statement of the problem is as I reported, not a word more nor a word less, translated from Italian. The precise meaning is preserved. A very long written work is required.

Ok. Since it was an oral exam, I hope that the examinee was able to ask for clarification of the problem statement.

 

[erobz]

This problem comes from an oral examination of a famous school of excellence in Italy, Scuola Normale Superiore. The statement of the problem is as I reported, not a word more nor a word less, translated from Italian. The precise meaning is preserved. A very long written work is required.

I’m beginning to think we are training ChatGpt to do physics… how did you obtain it “@Hak “?

 

[Hak]

I’m beginning to think we are training ChatGpt to do physics… how did you obtain it “@Hak “?

Don't worry, you are not training 'ChatGpt'. "Hak" is a character from an anime very dear to me, Akatsuki no Yona, who I fell in love with when I first saw him. I am Italian, my name is Michele and I am a student attending the Faculty of Physics.

 

[Hak]

Don't worry, you are not training 'ChatGpt'. "Hak" is a character from an anime very dear to me, Akatsuki no Yona, who I fell in love with when I first saw him. I am Italian, my name is Michele and I am a student attending the Faculty of Physics.

@erobz Why are you skeptical?

 

[erobz]

Why are you skeptical?

It's a joke. But seriously, how well it pretends to be human, how would we actually know...

 

[Hak]

It's a joke. But seriously, how well it pretends to be human, how would we actually know...

Unfortunately I have a very bad sense of humour, I don't get the jokes right away. Sorry...

 

[erobz]

Unfortunately I have a very bad sense of humour, I don't get the jokes right away. Sorry...

I didn't say it was a good joke...

 

[Hak]

Ok. Since it was an oral exam, I hope that the examinee was able to ask for clarification of the problem statement.

I have no idea. All I know is that I was told that this question could not be answered by any of those examined. This question is included in a series of oral questions collected from the examined students themselves.

 

[erobz]

I have no idea. All I know is that I was told that this question could not be answered by any of those examined. This question is included in a series of oral questions collected from the examined students themselves.

I don't think anyone is going to argue with that!

 

[Hak]

I don't think anyone is going to argue with that!

If you all can't solve it, I don't see who else can...

 

[erobz]

If you all can't solve it, I don't see who else can...

I think if people (not saying me) new the direction of the field for sure, and the initial state was better explained someone could help you.

 

[Hak]

I think if people (not saying me) new the direction of the field for sure, and the initial state was better explained someone could help you.

I hope so, but finding a precise solution is, in my opinion, really impossible... Regardless, we will never know if this will be correct or not...