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Is it possible to change the direction of ions with magnetic field?


jerryyu

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When an atom becomes an positive ion, magnetic field should be able to repel it. But my question is, is this possible? First of all, if magnetic field can change the direction of flowing electrons then for it to change the direction of ion should be possible. But i think the main problem is how far the ions will travel depending on the strength of the magnetic field.

 

P.S. it will be great if anyone know any equations that might help


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P.S. when I say changing the direction of flowing electrons, I mean when electrons are traveling through air.

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Or how the Large Hadron Collider works. The formula you are looking for is the Lorentz force. Note however, that this force due to a magnetic field will always be perpendicular to the current velocity. In effect (and first approximation) you will only change the direction of your ions, not their speed.

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i see, thank you


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Timo,

 

Can you emphasis on why it can't change their speed?


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so are you saying that it doesn't matter how strong the magnetic field is, only the direction will change ?

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Can you emphasis on why it [a magnetic field] can't change their speed?

Intuitively: A force (vector) acting on the ion can be decomposed into two parts: A part parallel to the velocity that pulls the particle to go faster or slower and a perpendicular part that modified the direction. That is of course not a convincing argument' date=' but perhaps gives you an idea what happens.

 

Physically: The force [math'] \vec F = q \vec v \times \vec B [/math] is the change of momentum [math] \vec p = m \vec v [/math] over time: [math] \frac{d\vec p}{dt} = \vec F[/math]. So the change in the square of the momentum is [math] \frac{d(\vec p \cdot \vec p)}{dt} = \vec p \cdot \frac{d\vec p}{dt} + \frac{d\vec p}{dt} \cdot \vec p = 2 \vec p \cdot \frac{d\vec p}{dt} = 2 \vec p \cdot ( q \vec v \times \vec B ) = \frac{2q}{m} \underbrace{\vec p \cdot ( \vec p \times \vec B )}_{=0} = 0[/math]. In other words: [math]\vec p \cdot \vec p = |\vec p|^2 [/math] is the same for all times. Then so is [math]|\vec p|[/math], of course. Then so is [math] |\vec p / m | = |\vec v| [/math], the speed.

 

EDIT: I think Swansont's argument via work is more mainstream than my calculation so it is possibly easier to understand.

Edited by timo
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Yes, but the letters are fairly standard. So if for example you don't know what a vector or a derivative is, then you'd possibly better ignore my calculation. Anyways, the situation about what happens to the speed of an ion traveling through a magnetic field.

- Arrows over objects indicate the object is an R³ vector.

- F is the acting force.

- p is the momentum of the ion.

- q is the electric charge of your ion

- v is the velocity of the ion.

- B is the magnetic field.

- m is the mass of the ion.

- t a time coordinate.

- | ... | is the length of a vector.

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Since an ion has an electric charge it will experience a sideways force if moving across lines of magnetic flux. One small detail - in a cathode ray tube the cathode emits electrons and it is moving electrons that are deflected by a magnetic field. If you are considering positive ions I would expect them to be deflected in the opposite direction to electrons. Also since an ion has a larger mass:charge ratio than an electron I would expect any deflection to be less for any given speed and magnetic field strength.

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Since an ion has an electric charge it will experience a sideways force if moving across lines of magnetic flux. One small detail - in a cathode ray tube the cathode emits electrons and it is moving electrons that are deflected by a magnetic field.

 

they still have the same properties of ions. charge to mass is indeed different but still, you get negative ions as well as positive ions.

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Absolutely agreed. However the poster of this thread specifically mentioned positive ions. Although not that important, I thought I would just correct your statement that crt televisions produce the display by pushing ions about with magnetic fields. They do, as I'm sure you know, actually push a beam of electrons about with magnetic fields. I'm probably just nit picking, if so I apologise.

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  • 3 weeks later...

One more question

 

When the ions is effected by the magnetic field, it will be traveling along the imaginary magnetic field line right? When it travels along the imaginary magnetic field line its speed will not increase or decrease but stay at a constant velocity right? Also, does having a stronger magnetic field means having a bigger magnetic field line?

 

Please correct me if my reasoning is wrong. Thanks.

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One more question

 

When the ions is effected by the magnetic field, it will be traveling along the imaginary magnetic field line right? When it travels along the imaginary magnetic field line its speed will not increase or decrease but stay at a constant velocity right? Also, does having a stronger magnetic field means having a bigger magnetic field line?

 

Please correct me if my reasoning is wrong. Thanks.

 

To alter the path of a beam of electrons (and presumably ions) the electrons have to cut across the magnetic lines of flux, not along them.

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I don't know what he means (can guess it, though). But assume a homogeneous magnetic field (i.e. the field lines are all parallel lines into one direction). If the initial velocity of your ion is parallel to these field lines then the ion will fly parallel to the field lines and be unaffected by the magnetic field. If the initial velocity is perpendicular to the field lines, then the ion will fly in circles perpendicular to the magnetic field because the field affects it via the Lorentz force (the [math]q \vec v \times \vec B[/math] from above which was zero if [math]\vec v[/math] and [math]\vec B[/math] were parallel). If the initial velocity is neither purely parallel nor purely perpendicular then it is a mix of both. The two possibilities become superimposed, causing a screw-formed path of the ion. TonyMcC probably meant "have a velocity that is not parallel to the magnetic field lines" when he said "cut across the magnetic lines of flux".

 

Three quick comments:

- Despite being a construct for visualizing it is uncommon to call the magnetic field lines "imaginary".

- The speed of an ion in a magnetic field does not change, see my previous post. The velocity can change, of course. More precisely, the velocity component perpendicular to the magnetic field lines changes; the component parallel to the field lines remains unchanged.

- If you are interested in understanding it, I suggest you sit down and try to calculate the motion of an ion in a magnetic field - you will be able to answer a lot of possible questions yourself after having done that. Start with some special cases and then see how far you get in generalizing this. This is a topic where a lot of people on sfn are competent enough to help you with the math.

Edited by timo
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Thank you timo. That is exactly what I meant by "cut across". Out of interest I will mention that the devices that I am familiar with deliberately use magnetic fields at right angles to the direction of moving electrons for maximum effect. Examples include cathode ray tubes and cavity magnetrons.

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1) Chose a magnetic field; I recommend [math] \vec B = \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) [/math] (I'll skip units; you can use any units in principle).

2) Put in a particle with charge 1 at the origin with initial velocity [math] \vec v(0) = \vec 0[/math]

3) The acceleration due to the magnetic field is [math] \vec a = \vec F /m[/math]; I posted the force (Lorentz force) above.

4) From knowing the acceleration, the initial velocity and the initial position, try figuring out how the particle moves, i.e. what it's position at some time t (or time t=5 if that is easier for you) is.

5) Now use a different initial velocity [math] \vec v(0) = \left( \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right) [/math] and do the same again.

5) Now use a different initial velocity [math] \vec v(0) = \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right) [/math] and do the same again. You'll want to look up circular motion for this part.

6) Now use a different initial velocity [math] \vec v(0) = \left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right) [/math] and do the same again. You should find that this case can be reduced to the previous cases.

6) Now use a different initial velocity [math] \vec v(0) = \left( \begin{array}{c} v_x \\ v_y \\ v_y \end{array} \right) [/math] and do the same again. You should find that this case is just a generalization of the previous case.

7) Start having ideas yourself (this point can also come sooner, of course).

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  • 1 year later...

To add to the discussion board and along the same lines,....how would free moving ions behave (i.e, in the atmosphere) if there were a pulsating magnetic field emanating from a source within which there was a static electric field,....assuming of course that it could be regulated to be constant ( or not). In other words, if I can share the visualization,....imagine there is a construct made up of three spherical conductive objects that are on the same plane at the corners of an equilateral triangle. On top of this construct, there is a source that generates a magnetic field that can be pulsed at will at variable degrees of intensity,....kinda like a solenoid.

 

How would the free moving atmospheric ions act upon the "construct" if there was a constantly changing magnetic field emanating from this source along which,...from the above thread, the ions can move perpendicularly across? My thinking tis that the "electrostatic" force can attract the ions of opposite charge while the oscillating magnetic field can accelerate the ions toward the electrostatic spheres. The magnetic field, having both north and south poles will channel the ions through the center of the triangle. The idea is to accelerate the groomed ions from the atmosphere through the centre of the solenoid/equilateral-electrostatic electric field so that there is a net force on the object that is greater than gravity.

 

Do you see where I am going with this? Okay,..so call me "out there". I need to go back to school on this one, but I believe there is a way to use the "energy" all around us in a useful and productive manner.

 

The question still stands,....what happens under those circumstances?

 

Thx,...Patman

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