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If $\vec{\mathbf B}=B\vec{\mathbf a}_z$, compute the magnetic flux passing through a hemisphere of radius $R$ centered at the origin and bounded by the plane $z=0$.

Figure


Solution The hemisphere and the circular disc of radius $R$ form a closed surface, as illustrated in the figure; therefore, the flux passing through the hemisphere must be exactly equal to the flux passing through the disc. The flux passing through the disc is

$$\Phi=\int_S\vec{\mathbf B}\cdot\mathrm d\vec{\mathbf s}= \int\limits_0^R\int\limits_0^{2\pi}B\rho\,\mathrm d\rho\,\mathrm d\phi =\pi R^2B$$

The reader is encouraged to verify this result by integrating over the surface of the hemisphere.


According to Maxwell's equations the magnetic flux over a closed surface must be zero, why in this case does not happen?

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It would be great if someone could transcribe the text from the image, and crop the image to just the figure itself. –  David Z May 20 '13 at 5:19
    
@DavidZaslavsky is there some way to get the text next to the image (rather than underneath it) –  Elements in Space Jul 8 '13 at 4:07
    
@UnkleRhaukus no, not on this site. That's okay though, the exact formatting isn't important for these things. –  David Z Jul 8 '13 at 5:43

2 Answers 2

The flux through the closed hemisphere is zero, $$\Phi_{\mathrm{hemi}}+\Phi_{\mathrm{disk}} = 0.$$ This allows us to find the flux through the hemisphere knowing the (more easily calculable) flux through the disk, $$\Phi_{\mathrm{hemi}} = -\Phi_{\mathrm{disk}}.$$

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According to Maxwell's equations the magnetic flux over a closed surface must be zero.

In this case the hemispherical surface in question is not a closed surface, it is an open surface.

If we consider the closed surface (the hemispherical section And the circular base) the total flux passing through will be zero.

Using this information it is clear that the flux leaving thought the hemisphere will be equal in magnitude and opposite in sign to the flux entering through the circular base.

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