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I have invented this little problem to help me understand the magnetic force better.

Imagine 3 positive charges all on a line. The first charge is separated by distance D from the second charge, and the second charge by distance D from the third charge. So

               +  ---------------- + ------------------ +
                         D                  D

The 3 charges are moving downwards at a constant velocity. The first and third charge have x axis positions which are permanently fixed - no force can change their minds. The middle charge is in perfect, peaceful free fall. For a long time the 3 charges have been moving in unison, downwards - the negative j-hat direction. The middle charge is in equilibrium between the electric repulsive forces due to the other two charges. It experiences no vertical forces, as it has constant vertical velocity.

Suddenly the 3 charges enter a B field. The magnetic field points into the page, so there is a magnetic force pointing towards the right on the middle charge. The other two charges don't matter.

What happens to the middle charge?

I am asking for the classical electromagnetism answer to the question, and in particular I am wondering:

By experiment it is so that there is the stated force on the magnetic field. But this force cannot do work. Therefore it should not be able to displace the charge in the direction of the 3rd, rightmost charge.

How can classical physics explain what's next? What exactly is happening here in terms of work? What principles underlie the hypothesis that the middle charge's vertical velocity falls, if that is the salient hypothesis?

I assume that there are no external forces acting on the system.

Enjoy.

PS. obviously these charges have inertia - all equal (and gravitational fields sum to zero with electric at the middle charge).

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    $\begingroup$ Can you elaborate as to why you think the magnetic force can't change the direction of a particle's motion? $\endgroup$
    – J. Murray
    Commented Oct 20, 2020 at 0:39
  • $\begingroup$ @J.Murray The Coulombic force depends on horizontal position in this case. So moving in the rightwards direction would mean fighting the Coulombic force. At the instant they enter the B field, the middle charge purely experiences a horizontal force. If you can see what source of energy could supply the work for the magnetic force to displace the middle charge in the rightwards direction, then please explain. $\endgroup$
    – juicy_adrian_berger
    Commented Oct 20, 2020 at 9:24
  • $\begingroup$ The climb up the electric potential is done at the expense of removing kinetic energy from the middle charge. As soon as it is displaced, its speed will be reduced by a net Coulomb force. $\endgroup$
    – ProfRob
    Commented Oct 20, 2020 at 10:06

4 Answers 4

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The rate of doing work is $\vec{F}\cdot \vec{v}$.

Since the magnetic component of the Lorentz force is $q\vec{v}\times \vec{B}$ then this force is always perpendicular to the velocity and does no work.

In the absence of electric field from the other charges, the middle charge would execute a circular path at constant speed and kinetic energy. No work would be done.

Because of the other charges, the charge will initially veer to the right without changing speed, but will then experience a further accelerating force towards the left (let's call that $-x$) and upwards ($+y$) due to the net Coulomb force from the charges that will have slightly larger negative y displacement compared to the central charge. The Coulomb force does do work and the middle particle will be initially slowed and fall behind the outer two charges and execute a complex trajectory that will depend on the dimensions of the problem, the initial velocity and strength of the electric and magnetic fields.

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  • $\begingroup$ Thanks for the response. I'm trying to understand how the magnetic component could end up slowing down the middle charge. If the component causes a dx displacement to the right (let's say negligible work is done), then the velocity changes by dv towards the right, so the next instant of the magnetic component will have a component in the positive y direction, so it slows down (vertically) - all because we allowed that a dx change towards the rightmost charge involves 0 work. Is that right? $\endgroup$
    – juicy_adrian_berger
    Commented Oct 20, 2020 at 9:43
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    $\begingroup$ @AdriaanBerger It doesn't. It changes the direction of the velocity. Once the direction has changed then that means the $v_y$ component becomes smaller. Once this results in a displacement of the middle charge away from the centre there will be a Coulomb force towards -x and towards +y because the middle charge is also displaced somewhat above the other two charges because its $v_y$ is smaller. $\endgroup$
    – ProfRob
    Commented Oct 20, 2020 at 10:01
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At the moment the system enters the region with the magnetic field it experiences a time dependent magnetic field, which, because this is the same thing as a rotation of the electric field, can do work. This displaces the charge to the left until the magnetic field becomes static. The charges stays in its new position. Perhaps transient oscillations may destroy my simple scenario ...

No work is done by a static magnetic field.

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What happens to the middle charge?

It will get deflected to the right as well if the magnetic force is greater than the electric force trapping it by the two charges as you stated. That is, if the net electric force manages to keep the middle charge restricted to move in the y-axis, and the magnetic force is greater than this force, there will be motion in the x-axis. The Lorentz force states

$\vec F = q( \vec v \times \vec B)$

By experiment it is so that there is the stated force on the magnetic field. But this force cannot do work. Therefore it should not be able to displace the charge in the direction of the 3rd, rightmost charge

If a magnetic field is applied this field will definitely apply a force on the middle charge (and the other two) if it is moving. It most definitely can cause a displacement (once again if the net magnetic component of Lorentz force is greater than that caused by the electric force of the other two, there will be a horizontal displacement).

How can classical physics explain what's next? What exactly is happening here in terms of work? What principles underlie the hypothesis that the middle charge's vertical velocity falls, if that is the salient hypothesis?

Your assumption is incorrect and classical physics explains this perfectly. The reason why no work is done is because the Lorentz force acts in a direction perpendicular to the velocity of the charges.

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Suddenly the 3 charges enter a B field. The magnetic field points into the page, so there is a magnetic force pointing towards the right on the middle charge.

Where isn‘t a magnetic force pointing towards the right. The deflection is the result of the interaction of the electrons magnetic dipole with the external magnetic field.

By experiment it is so that there is the stated force on the magnetic field. But this force cannot do work. Therefore it should not be able to displace the charge.

A hovering charge will not be displaced, only a moving - non-parallel to the external magnetic field - will start a spiral path. And, yes, the external magnetic field is not weakened over time. As you correctly mentioned, it does not add energy to the charge to deflect it.

How can classical physics explain what's next? What exactly is happening here in terms of work?

Sometimes we forget that the electron has not only a charge but also a magnetic dipole. Only this dipole can interact with the external magnetic field. If you now consider that accelerated electrons (and any deflection from a straight line is such a deflection) emit electromagnetic radiation, it is clear why the electron is deflected sideways (the emission of photons always causes a recoil at the emitter).

The above mentioned moving charge in relation to the external magnetic field is the final key to understanding the spiral path of the electron. Remember that an electron is not exhausted by the emission of photons. What exhausts itself is the kinetic energy of the moving electron.

In the usual arrangement - without the acceleration of the electron towards the earth - the electron rotates spirally inwards and comes to a standstill, giving off all its kinetic energy to the EM radiation. In your arrangement the speed of the electron increases due to the acceleration towards the earth, and the spiral track has a larger radius at the beginning. By the way, with really high speeds and large magnetic fields, you end up with a free electron laser with its microwave, terahertz, visible, ultraviolet or even X-ray EM radiation.

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