2
$\begingroup$

I have been thinking about magnetic forces and how they do no work. When looking at the definition of work, it makes sense as the Magnetic forces only change the direction.

My other approach to try and really understand this issue is to consider energy. Suppose we have a charge q with mass m moving with constant speed v in a given direction. Then we pass it through a region of a magnetic field. Without getting too tangled up with the orientation and directions, we know the charge changes direction. So initially we have: $T = T_{translation}=1/2mv^2$ but then with the introduction of the magnetic force there's a change from translational kinetic energy to rotational kinetic energy so $ T = T_{transalation} + T_{rotation} = 1/2mv^2 + 1/2 I\dot{\theta}^2$

So there's a change in the energy but no work done? I don't think I quite understand how this would happen. Sorry if I'm making some dumb assumptions and just can't see it.

$\endgroup$
2
$\begingroup$

You need to be careful in the definitions of $v$ and $\dot{\theta}$. They both hold information about the motion of the particle and need to be defined in such a way that they are always orthogonal. Otherwise you are overcounting the motion.

If you had instead $v_z$ and $\theta$ is the angle around the $\hat{z}$ axis then they are orthogonal to each other. For completeness you would need to include the radius from the $z$-axis, $\rho$. These 3 coordinates fully specify the velocity of a point particle in 3-Dimensions and do not overlap. Now if we start with, $$E_I = \frac{1}{2}mv_z^2 =\frac{1}{2}mv_z^2 + \frac{1}{2}I\dot{\theta}^2,$$ we can write the second term because the initial value of $\dot{\theta} = 0$. Now we turn on a magnetic field. The magnetic field does not change the magnitude of the velocity, it only "shuffles" around the components. This means that you would find part of $v_z$ has now become part of $\dot{\theta}$. We could write the energy after the effect of the magnetic field, $$E_F = \frac{1}{2}mv_z'^2+\frac{1}{2}I\dot{\theta}'^2.$$ Now $\dot{\theta}' \neq 0$ but the value of $v'_z \neq v_z$. If you carefully work out the algebra you would see that the decrease in the velocity in $\hat{z}$ direction is compensated by the increase in the other directions.

To be more explicit you would need to specify a direction for the magnetic field and work out the full 3-Dimensional motion (here I have suppressed motion in $\hat{\rho}$), but you would always see that $|\textbf{v}|$ does not change, and ultimately these are fancy ways of writing, $$E_I = \frac{1}{2}\textbf{v}^2_I = \frac{1}{2}\textbf{v}^2_F = E_F$$

For us to say that the magnetic field did work on the particle we would need to have a change in the energy of the magnetic field, and a corresponding change in the energy of the particle. However in this case the energy of the particle has not changed. Harder to show, but the energy of the magnetic field has not changed either. This means there was no transfer of energy between the two systems and therefore no work was done.

$\endgroup$
  • $\begingroup$ Okay this is great, but that's what I already arrived at. I am not saying the energy comes of out nowhere, I know it is conserved and from the translation kinetic energy we initially get the rotational kinetic energy when the field is turned on. My problem is, since there is a change from kinetic to rotational energy, wouldn't that be considered work done/energy change caused by the magnetic force? $\endgroup$ – Lost In Euclids 5th Postulate Jun 13 at 16:22
  • 1
    $\begingroup$ No Energy is only a single number, work done would be a change to the total energy. In the same way a mass being swung around on a string has constant velocity magnitude (therefore constant energy), but its velocity components are changing between $v_x$ and $v_y$. No work is being done in this case, the energy stays constant and can be described as $E=\frac{1}{2}I\dot{\theta}^2$. Work would be done from energy being added from an external system to this system. Since the energy of the magnetic field does not transfer to the energy of the particle no work is being done. $\endgroup$ – TEH Jun 13 at 16:36
  • 1
    $\begingroup$ I have added this sentiment to the original answer as I think it is as important as my original point. Also another response to your comment, for a point particle there is no separation between rotational and kinetic energy. Rotational energy is simply part of the kinetic energy expressed in weird coordinates. $\endgroup$ – TEH Jun 13 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.