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Insights The Homopolar Generator: An Analytical Example - Comments

  1. Nov 27, 2017 #1

    vanhees71

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  2. jcsd
  3. Nov 28, 2017 #2

    Charles Link

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    The method used in this article for the computation of the magnetic field ## B ## for the uniformly magnetized sphere is interesting. That calculation in many E&M textbooks uses the magnetic pole method, and the analogous electrostatic calculation with the dielectric sphere with uniform polarization. The computation is most often performed using Poisson's/LaPlace's equation with the Legendre polynomial solutions. I'm still working my way through the rest of it, but I am finding it very good reading. Thank you @vanhees71 :)
     
  4. Nov 28, 2017 #3

    Charles Link

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    @Greg Bernhardt Suggestion would be to put this in the General Physics section. It is more of a discussion of basic physics principles than it is of an Electrical Engineering gadget.
     
  5. Nov 28, 2017 #4

    Charles Link

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    I performed an additional calculation that might be of interest to show consistency between equations (18) and (29): For a path along the surface of the sphere, (18) shows that the electric field tangent to the surface will be along the lines of longitude. Thereby, I computed ## E_{\theta}=-\frac{M \omega R}{3 c} \, \sin(2 \theta) ## from (18). (Note: ## -\frac{dU_{r=R}}{R \, d \theta}=E_{\theta} ##). ## \\ ## I then computed the component of ## E_{\theta} ## from (29) for the case of a point on the surface ## r=R ## with ## y =0 ## : The components of ## E_x ## and ## E_z ## along ## \hat{a}_{\theta} ## were computed and added together, and were consistent with ## E_{\theta} ## computed from (18). ## \\ ## (29) applies to ## r \geq R ## while (18) applies to ## r \leq R ##, so that this particular calculation only works for ## r=R ##. ## \\ ## For any EMF measurement, this path on the surface (along the lines of longitude) is really the most important in any case, because the probes need to make contact to the sphere. ## \\ ## Editing: One puzzle that arises from this result: perhaps @vanhees71 can provide some insight, is in equation (16), there is no z component of the electric field inside the sphere. This radially inward electric field does have a component along ## \hat{a}_{\theta} ##, (as computed from ## E_{\theta}=- \frac{dU_{r=R}}{R \, d \theta} ## ), and thereby the electric field from equation (29), that does have a non-zero z-component for ## r \geq R ##, and a completely equivalent ## E_{\theta} ## at ## r=R ##, appears to be consistent. The electric field ## E ## is not required to be continuous across the boundary because of the surface charge on the sphere. The parallel components of ## E ## do need to be continuous across the boundary though, and they are, as the calculation linking (18) to (29) shows. Inside the sphere we can have ##E_z=0 ## and outside ##E_z \neq 0 ##. ## \\ ## Additional comment: Very interesting calculations by @vanhees71 computing the EMF for the homopolar generator for a spherical shape. The computations would perhaps be simpler for a cylindrical shape where (18) would be ##U=\frac{M \omega }{2c}(x^2+y^2) ## and the EMF would be across the top face of the cylinder=perhaps not exact because of fringe effects. [Editing: This is rather inexact in the case of a cylinder, because at the endface, the ## B ## field is reduced by a factor of 2 from what it is in the middle]. One additional question I have is, what happens for a thin magnetized disc? The ## B ## field inside the disc is greatly reduced from the ## B ## of a long cylindrical shape, so presumably the EMF that gets generated would be much less than that which is generated from a long cylindrical shape. ## \\ ## Editing: The calculation to compute the surface charge distribution is quite interesting and non-trivial. The Legendre polynomials/multipole expansion which are solutions of the Poisson equation appear to provide the necessary structure to make for the solution. (If I understand it correctly, the surface charge distribution will cause a potential both inside and outside which must consist of multipole expansions, which are the general solution of Poisson's equation in the vacuum). I'm still working through it, but I'm finding it very interesting. Thank you @vanhees71 :)
     
    Last edited: Nov 29, 2017
  6. Nov 29, 2017 #5

    vanhees71

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    Well, I used a sphere instead of a cylinder because of its higher symmetry and being a body of finite extensions. I guess the infinitely long cylinder can be treated similarly, but usually you have trouble because of its infinite extent. I also guess that a thin disk is pretty complicated to treat; maybe one needs numerics to get a full solution.

    The reason that inside the sphere ##E_z=0## is because inside the magnetic field is constant pointing into the ##z## direction, and in the static case ##\vec{E}=-\vec{v}/c \times \vec{B} \perp \vec{e}_z##.
     
  7. Nov 29, 2017 #6

    Charles Link

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    Upon further study of this, the uniform distribution of charge inside the sphere, along with an electric field inside the sphere of the form ## \vec{E}=A(x \hat{i}+y \hat{j}) ## is quite interesting. Using Gauss' law ## \nabla \cdot \vec{E}=\rho ## results in ## \rho ## being a constant=uniform, but that necessarily means that there must be a surface charge distribution that makes the electric field of this form, because the electric field doesn't point uniformly and radially outward from the center of the sphere, like it would for a uniform distribution of charge. ## \\ ## The Legendre method of determining this surface charge distribution is interesting. Even knowing the final result, I think it would be very difficult to show with a Coulomb's law integral approach that the resulting electric field does in fact have ## \vec{E}=A(x \hat{i}+y \hat{j}) ##.
     
  8. Nov 30, 2017 #7

    Charles Link

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    Verifying equations (10), (11), and (12) is a good exercise in vector calculus. I finally worked my way through them. Thank you @vanhees71 :)
    ## \\ ## One very minor typo: In equation (10) the last ## a ## should be an ## R ##.
     
    Last edited: Nov 30, 2017
  9. Dec 1, 2017 #8

    vanhees71

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    Thanks for the hint. I've corrected it in the Insights article. I'll also correct the pdf version asap.

    I also realized that references to sections do not work as in LaTeX. I've to figure out yet, how to link to sections within an article.
     
  10. Dec 31, 2017 #9

    Charles Link

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    @vanhees71 (Final Editing: Yes, I have it solved !=be sure and see the last paragraph below) The method of solution of ## B ## for the uniformly magnetized sphere from equations (6) to (13) is extremely interesting. I am led to believe there is a similar type of solution that exists for the electric field of a uniformly polarized sphere, using ## E=-\nabla \Phi ##, but I was unable to show the result by trying to compute ## \Phi ## in a similar manner. (A similar symmetry didn't seem to apply because of the ## \cos(\theta) ## term in the expression for ## \Phi ## of the microscopic dipole). Might you be able to show this derivation and/or provide a "link" to it. Up to now, the way I would solve such a problem is by using the Legendre polynomial method. Here is a "link" to a previous post on PF where the OP was basically attempting to do the same calculation: https://www.physicsforums.com/threa...iformly-polarized-sphere.877891/#post-5514160
    ## \\ ## Editing: (see post 7 of this "link" for the Legendre solution for ## \Phi =V ##). For the OP's formula in post 1 of this "link", I get ## \cos(\theta') \sin(\theta') ## in the numerator, but the dot product in the denominator is a rather lengthy expression off sines and cosines of ## \theta ##, ## \theta' ##, and ## \phi ##, and ## \phi' ##. ## \\ ## Additional editing: I was almost able to successfully answer my own question. In the OP's first surface integral, the result is a ## \cos(\theta') ## term for the dot product in the denominator inside the square root sign. The integral then becomes readily solved with an integration by parts, and the same result is obtained for the potential ## V(\vec{r} )## by this method as by the Legendre potentials of post 7 of the "link". I need one final step to justify the form of his first integral expression. (Final edit: And yes, I found the necessary step=see below) ## \\ ## Final editing: I was finally able to justify the terms of the integral contained in the "link", first equation, post 1, by writing the numerator of the integrand as ## \hat{z}'' \cdot \hat{z}' ## and performing the integration over ## d \Omega '' ## over the surface of the sphere. The ## z'' ## axis could be taken along the vector ## r ## without loss of generality, making the dot product term in the denominator equal to ## \cos(\theta'') ##. Meanwhile ## \hat{r''}=\sin(\theta '') \cos(\phi'') \hat{i}''+\sin(\theta'') sin(\phi'') \hat{j}'' +\cos(\theta'') \hat{z}'' ##. The integral over ## d \Omega'' ## vanishes for the ## x ## and ## y ## components because of the ## \phi'' ## integrals with ## cos(\phi'') ## and ## \sin(\phi'') ##. The term that has ## \hat{z}'' ## picks up a ## \sin(\theta'') ## in the ## d \Omega'' ##, and ## \hat{z}'' \cdot \hat{z'}=\cos(\theta) ##. Thereby, the first equation for ## V ## that the OP presents in post 1 of the "link" is indeed the correct one. Finally, the double-prime that I have introduced can be replaced by a single prime. And note: The integral ## \int\limits_{+1}^{-1} \frac{u \, du}{\sqrt{r^2+R^2-2rRu}} ##, where ## u=\cos{\theta} ##, is performed by an integration by parts to get the result.
     
    Last edited: Jan 1, 2018
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