# The Homopolar Generator: An Analytical Example - Comments

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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 :)

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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.

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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 :)

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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##.

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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}) ##.

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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 ##.

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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.