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Discharging a magnetized sphere
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[QUOTE="ELB27, post: 4964278, member: 516388"] First of all, it was very silly of me to omit the constant of integration - thank you for noticing. The correct expression reads ##K = -\frac{dQ}{dt}\frac{\cos\theta + c}{4\pi R\sin\theta}## where ##c## is a constant and the minus sign comes from the fact that ##\frac{dQ}{dt}<0## but ##K## is a magnitude. This not only has a correct physical interpretation if ##c≥1## but also gives the correct answer regardless of the exact value of ##c##. Secondly, I would like to follow your approach for consistency. Call ##dq## the charge during time ##dt## lost between the south pole and the arbitrary latitude ##\theta## while the total lost charge be ##dQ##. Since we assume that the density will remain uniform throughout, ##\frac{dq}{dQ} = \frac{\sigma A}{\sigma A_{total}} = \frac{A}{A_{total}}## where ##A## is the area of the spherical surface from the latitude to the south pole and ##A_{total}## is the total surface area of the sphere. Using a simple surface integral: ##A = \int_0^{2\pi}\int_{\theta}^{\pi}R^2\sin\theta d\theta d\phi = 2\pi R^2(1+\cos\theta)##. The total area is, of course, ##A_{total} = 4\pi R^2##. Substituting, we get ##dq = \frac{1+\cos\theta}{2}dQ##. Let ##I## be the total current flowing from the south pole to the latitude. By definition, ##I = \frac{dq}{dt} = -\frac{dQ}{dt}\frac{1+\cos\theta}{2}##, the minus sign, again, comes from the fact that ##\frac{dQ}{dt}<0## but ##I## is a magnitude. Now, by definition, ##K = \frac{dI}{dl_{⊥}} = -\frac{dQ}{dt}\frac{1+\cos\theta}{2}\frac{1}{2\pi R\sin\theta} = -\frac{dQ}{dt}\frac{1+\cos\theta}{4\pi R\sin\theta}## where ##dl_{⊥}## is an infinitesimal width of the surface current perpendicular to the current direction. Evidently, ##c=1## in my original approach. I have just two more questions. First, is there any simple boundary condition I can use to determine ##c## right away? Secondly, I never understood the formal definition ##K = \frac{dI}{dl_{⊥}}##. It says to take a derivative, but I [I]always[/I] end up [I]dividing[/I] by the width instead of differentiating. Can you shed some light on this? Thank you very much for your answer! I highly appreciate it and your approach was quite clever. EDIT: Concerning my first question, would a zero current at the south pole be an appropriate condition? (I mean physically, mathematically it all turns out correctly anyway). [/QUOTE]
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