Recent content by georg gill

about symmetry: if you first calculate the magnetic vector potential for a spherical surface: one starts with the magnetic vector potential formula for a point inside or outside a spherical electrically charged rotating surface $$\textbf{A}=\frac{\mu_o}{4 \pi}\int \frac{\sigma \omega \times...

Do you mean this: If it is post 25 I am having a bit of a problem getting the relevance to post 24. Thanks:rolleyes:

I have pondered a bit further: By using Leibniz integral rule $$\nabla \times B(r)=\frac{\mu _0}{4\pi} \int \nabla \times \frac{ J(r') \times (r-r')}{|r-r|^3}dV'$$ $$ \frac{J(r') \times...

For ##\nabla \times B=J\mu_0## From Griffiths one can obtain that the curl of B inside a rotating sphere is ##\nabla \times B=J\mu_0## but that is for an arbitrary large sphere. So it seems that the current outside r-r' is not taken into account for that that value of ##\nabla \times B## since...

I guess for gauss law I would unerstand that you can calculate the E-field inside a charged sphere as $$E=\frac{Q\textbf{r}}{4\pi\varepsilon_0 R^3}$$ by using unit vector and obtaining the divergence inside the charged sphere one obtains: $$\nabla \cdot E=\nabla \cdot...

Sorry but if r-r'=0 within the current integral that would affect ## \nabla (J \cdot \frac{r-r'}{|r-r'|^3})##? If one uses l'hospital's on the last element one obtains for the x-element...

Yes. Think I got it. 0 due to dot product with the vector J that does not go out of the surface. Thanks!

How is the term ## a \times \nabla \times b## equal to 0 in ## \nabla (a \cdot b)=a \cdot \nabla b+b \cdot \nabla a + a \times \nabla \times b + b \times \nabla \times a ## ?

Sorry but how is ## \nabla (J \cdot \frac{r-r'}{|r-r'|^3})=(J(r') \cdot \nabla) \frac{r-r'}{|r-r'|^3} ##? $$(J(r') \cdot \nabla) \frac{r-r'}{|r-r'|^3}=(J_x\frac{\partial}{\partial x}+J_y\frac{\partial}{\partial y}+J_z \frac{\partial}{\partial...

$$\nabla \times B(r)=\frac{\mu _0}{4\pi} \int \nabla \times J(r') \times \frac{ (r-r')}{|r-r|^3}dV'$$ using the vector identity: $$\nabla \times (A \times B) = (B \cdot \nabla)A - B(\nabla \cdot A) - (A \cdot \nabla )B + A(\nabla \cdot B)$$ ##A=J## and ##B=\frac{r-r'}{|r-r'|^3}## since...

Sorry I have updated my last post it was not what I wanted to ask that was in the initial post. Please look at my last post above if interested. The magnetic vector potential is 0 when ##r'=r## in the denumerator. Is not that the same issue as in: $$ \frac{\mu _0}{4\pi} \nabla^2 \int...

I have tried to go through a proof for this: $$\nabla \cdot \vec{A}=\frac{\mu _0}{4\pi}\nabla \cdot \int \frac{ J(r')}{|r-r'|} dV'$$ again we use $$ \nabla \cdot \frac{ 1}{|r-r'|}=-\nabla' \cdot \frac{ 1}{|r-r'|}$$ And by similar reasoning as in my first post in the 5th last equation...

I am trying to derive that $$\nabla \times B=\mu_0 J$$ First the derivation starts with the electric field $$dS=rsin\varphi d\theta r d\varphi $$ $$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0} \iint\limits_S \frac{r}{|r|^3} \cdot dS $$...

The problem is that I want a proof for why integral and curl can be intechanged because they operate on differnet variables. I have been given the argumentation before. But I need a proof. For example if you start with the integral Thanks for the proof!

I have tried to update the prime notation. Should I move it to homework? The problem is that I want a proof for why integral and curl can be intechanged because they operate on different variables. I have been given the argumentation before. But I need a proof. For example if you start with the...