# Magnetic Force Question

1. Jun 5, 2012

### schaefera

So, I have an apparent paradox:

Consider a particle moving with constant velocity, v, along the x-axis.

It now enters a constant magnetic field with is parallel to the y-axis. So there is a force, F=q(vxB) which acts on it, and points in the positive z-direction at first, but then as the charge is accelerated, the force continues to rotate causing the point to moving along a circle in the planes parallel to the yz-plane, but it still moves at a constant speed in the x direction because no force acts on it in either the +x or -x direction.

So, the charge has been accelerated (and it now has velocity components moving in SHM (circles) in the y and z directions) meaning that its KE must have increased, since the x-component of velocity is unchanged... what could have done this work to cause the change in KE if the force can do no work since it is always perpendicular to the displacement?

We know this because F is perpendicular to v, and v points in the direction of an infinitesimal displacement....

Seems like a paradox to me!

2. Jun 5, 2012

### Dickfore

Kinetic energy depends on the speed of the particle, which remains unchanged when the acceleration is perpendicular to the velocity vector. Thus, your conclusion that the kinetic energy is changed is wrong.

3. Jun 5, 2012

### schaefera

But if it started out not moving in the y or z directions, and then suddenly it IS moving in those directions, it speed must have increased (it has accelerated!)

Acceleration perpendicular to velocity might not change the speed once it's been moving, but to start the particle moving in a circle it certainly does.

4. Jun 5, 2012

### Dickfore

No, it does not. Also, if the initial velocity was in the x direction, and the magnetic field is in the y-direction, then the force, as well as the initial acceleration is in the z direction. Thus, the particle would move in the xz plane, perpendicular to the magnetic field.

5. Jun 5, 2012

### schaefera

But it must-- consider it more quantitatively.

Initial velocity: <a, 0, 0>
Velocity JUST after entering field: <a, 0, c>
And clearly after a little bit of time there is a velocity component pointing in towards the middle of the circle...

So the velocity vector has increased in length, and speed must have increased.

6. Jun 5, 2012

### Dickfore

This is wrong. The velocity JUST after entering the field must be the same as initial velocity, since velocity cannot change discontinuously by a finite acceleration.

7. Jun 5, 2012

### schaefera

It's not changing discontinuously, obviously, it's changing according to how the force influences it! Initially after entering the field, it feels a force q*a*B (if a is the initial x-component of velocity and B is the magnitude of the field).

So acceleration= (qaB)/m where m is its mass. So this acceleration will have caused an infinitesimal change in velocity which, for brevity's sake, I called c.

8. Jun 5, 2012

### schaefera

But even if we only look at it qualitatively, for some time the particle moves in the x-direction at constant speed. Suddenly, the magnetic field turns on and it's moving (for an instant) in the x- and z-directions, and then for after that it is moving in some combination of the x and z directions in a circle of a plane parallel to the xz-plane.

9. Jun 5, 2012

### schaefera

Wait, so maybe it's like this: it's moving in the x-direction, and when the magnetic field turns on it suddenly doubles back on itself in the shape of a circle and moves in that circle forever after? So you basically have "stopped it" in its tracks and made it move in a circle rather than forward like it had been.

Does this generalize if it had been moving in a different direction to start with (would it be making circles in some plane that isn't parallel to any of the coordinate planes)?

10. Jun 5, 2012

### schaefera

Finally, if we know that the trajectory must satisfy both the magnetic force law and Newton's second law, is there any general way of solving the (vector) differential equation

F=q(vxB)=ma, where v, a, and B are all vectors?

So the open questions at this point are: (1) Does the magnetic field "stop" the forward motion of the particle and convert it into circular motion, doubling back on its original path for a little bit? (2) Is this a general result even in slanted planes? (3) Is there a general way to solve a vector differential equation like q(vxB)=ma?

11. Jun 6, 2012

### Couchyam

(3) Fortunately, the differential equation you mentioned is linear (the sum of any two solutions is also a solution). In general, linear differential equations are very easy to solve. Guess an exponential solution (turn the differential equation into an algebraic equation), and then try to find all the solutions. It may be simpler to consider solutions unique up to changing reference frame, and you will need some linear algebra.
(2) This isn't a very well posed question. Obviously changing reference frames will cause the plane you were considering before to become "slanted". If you mean to ask, what happens when the magnetic field is not perpendicular to the motion, then you will answer this question by answering (3) above. The short answer is this: the component of the particle's velocity parallel to the magnetic field is constant.
(1) instantaneously the motion of the particle will be circular. In magnetostatics it will not double back tangent to its initial path, although it may intersect, because returning tangent to its initial path would require a time-dependent field.