Work req'd to move electron through magnetic field? can't seem to figure it out!!

[Todd Massure]

I need help if anyone can lend me some brain power.
charged particles will travel in circular paths perpendicular to a magnetic field described by the formula T=2pi r/v = 2pi/w = 2pi m/qB
so it will take a certain amount of work to alter the particle from it's course.
I want to figure out how much work it will take to get an electron to move from one point to another that is not on it's regular path.
I am including a picture in a Microsoft word file for discussion.
The red cross at point B is the destination point and I will say that there is a positve chage at that point attracting the electron. The magenta arcs are a rough description of the path that I think the electron would travel and the orange dots represent the perpendicular lines of force.

variables will be:
-positive charge at point B
-magnetic field strength
-distance between starting and destination points
-time elapsed during movement from point A to point B

I appreciate any help
-Todd

[Frank Sanns]

Todd,

Essentially no work is done "bending" the path of a charged particle in a magnetic field. There is synchrotron radiation that is emitted from the angular acceleration but it is very, very small at fusor energies. At particle velocities approaching the speed of light (multiple MeV) it will radiate the power out of the system but it is a non factor at KeV.

My computer is not displaying your B field lines in your document but I assume they are coming out of the plane of the drawing. I assume that the multiple curved paths are due to some kind of pulsing of the magenet field since a permanant magnet does not reverse field in zero distance as the non smooth transition between arcs in the electron path suggests.

Frank S.

[Richard Hull]

I think his mag field is permenant or fixed and the sines are just due to his inability with the program to draw a proper helical path which he more or less denotes at the bottom with the cross section of a circle.

Should be two simultaneous equations. The B-field costs energy if you are using variable strength (settable, adjustable B field electromagnets.) Free, if you use permenant magnets.

RIchard Hull

[Todd Massure]

There were dots on my drawing that didn't show here that were to represent magnetic field lines in the z direction.
I drew the arcs because I was thinking that there would be some rotation of the electron in the field where it would kind of "over rotate" past its mean path and then correct with another arc, I realize now that this is not what would happen if the magnetic field and positive charge are constant.
I came to the conclusion as Frank stated that it really shouldn't require any work for the electron to move through the magnetic field, especially since the magnetic field does not do any work on the particle. Is that pretty much what you all think too?

[Todd Massure]

Actually if the electron has some value of momentum in the + or - x direction I think it would trace out a path like the magenta arc, because that momentum will be conserved.
It still shouldn't take any value of work to move the electron through the field though.

[raneyt]

Todd,

I echo what Frank & Richard stated about work done on the electron.

Some points to remember: if the electron path is parallel to a magnetic field 'line', the electron will trace a helical path around the field line.

If the electron path is perpendicular to a constant & uniform magnetic field (all things being equal), then the electron will describe a path with a curvature proportional to the magnetic field (flux, really).

And yes, if we are talking about using a permanent magnet to produce the field, then no work is done to deflect the electron's path (sort of).

Hope this helps.

Warm Regards,

TIM RANEY

[walter_b_marvin]

getting back to basic physics F~ qV X B where B is the Magnetic Field and V is the Velocity vector. This has to be balanced by the centrifugL force from the particle, which represents in kenetic energy of creation, collision etc., which is mV^2/R. If B is constant the we see that V/R is constant for a fixed magnetic field. This means to get a bigger R You need a bigger |V|, or bigger total kenetic energy. So to increase the orbit, make the initial energy bigger or aqdd energy somehow, to decrease the orbit subtract energy somehow.

[Richard Hull]

Work is done, obviously, by the acceleratory E field.

Richard Hull

[DaveC]

Todd - (sorry to be so late on this interesting question)

The simple physics answer here is in the definition of work (energy) . The energy U is the integral of F*dx, both of which are vectors and the * is the dot product. So the result is a scalar, not a vector.. and... the dot product is

U = FXcos(theta) where F is the force, X is the distance,and theta is the angle between the force and the distance travelled (or direction of motion).

Since a magnetic field causes a force at right angles to the motion of the charged particle, theta is 90 degrees and the work done is zero.

Dave Cooper

[Todd Massure]

Dave, thanks for clearing that up beyond any doubt
-Todd

[Richard Hull]

It is important to remember that zero magnetic field work is done, but if you are accelerating a particle in a mag field with an e-field then work is always done by the e-field along the axis of acceleration. The mag field just makes the particle corkscrew along the axis of travel while accelerating.

One can't move an electron through space or a magnetic field without an acceleratory force and work being done at some point. However, if the particle had a constant velocity V when entering a magnetic field then zero work would be needed to keep it moving inspite of its corkscrewing while traversing the field.

Thus the original question can best be answered by saying that the same amount of work is required to move a particle through a magnetic field as that needed to move it through no magnetic field.

Richard Hull

[esemplectic]

1 July 2005

I must say, I found this thread entertaining. The clearest and most correct answer about work done in a magnetic field was provided by Dave Cooper in his 6/6/2005 post. Bravo Dave for having taken the trouble to learn some elementary physics!

The question, however, seems to be about how to deflect the trajectory of an electron, given a set of initial conditions (position and velocity), so that it intersects a given point B in a homogeneous magnetic field with a stationary charge at B. This has nothing to do with work done during movement in the magnetic field, which Dave has correctly taught us is zero.

The question is fundamentally incomplete, and therefore unanswerable. Various posts have inferred various possible completions of the question and therefore given various answers with various degrees of clarity and correctness.

The information that is missing is the means of perturbing the trajectory: electrostatic? an auxiliary magnet? a kinetic collision? Also is this perturbation applied in the given magnetic field, or prior to entry?

I might infer from the inclusion of a stationary positive charge at B (a completely unphysical scenario) that Todd is supposing this charge will deflect the trajectory through the point B. If that is the case, then Richard Hull's post about work done by an electric field in a magnetic field is correct as far as it goes, and computing that work is simplicity in itself.

Electric fields are conservative, which means the work done on a charge is dependent only upon the initial and final positions, not on the path taken (gravitational fields are also conservative, and a fairly elementary mathematical analysis shows that any inverse square law force field, and only inverse square law force fields are conservative).

Since the charges are of opposite sign, they attract, and the work will be negative, i.e. the electron will be accelerated, by the positive charge. The trouble is, the electron will only travel through the point B for a given static magnetic field, and a given point A for velocities less than some maximum, which can be easily solved for by setting up the equation of motion. This solution is essentially identical to that which gives cutoff conditions in a Hull magnetron (presumably no relation to Richard).

In other cases, the particle will miss B and the work for any point along the trajectory may be computed by integrating F dot s, i.e. the electrostatic force projected onto the unit tangent to the positional trajectory. The value of this integral depends only upon the initial and final positions. This is a simple application of the definition of work: a force acting through a distance.

I could spin my wheels speculating on a myriad of other possible interpretations of Todd's question, but why? The important point here is it is necessary to formulate one's questions completely and in context. Give us a picture of the larger more abstract problem you are examining.

And, oh yes, it is worthwhile learning a little physics, which necessarily involves learning a little calculus, and at least comprehending the utility and application of differential equations. These measures, necessarily, require spending some time off line engaged in solitary, focused, disciplined study.

Bob Henderson