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
$$\textbf{A}=\frac{\mu_o}{4 \pi}\int \frac{\sigma \omega \times...
I have pondered a bit further:
By using Leibniz integral rule
$$\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:
$$\nabla \cdot E=\nabla \cdot...
Sorry but if r-r'=0 within the current integral that would affect
## \nabla (J \cdot \frac{r-r'}{|r-r'|^3})##?
If one uses L'Hopitals on the last element one obtains for the x-element...
How is the term ## a \times \nabla \times b## equal to 0 in
## \nabla (a \cdot b)=a \cdot \nabla b+b \cdot \nabla a + a \times \nabla \times b + b \times \nabla \times a ##
?
Sorry I have updated my last post it was not what I wanted to ask that was in the initial post. Please look at my last post above if interested.
The magnetic vector potential is 0 when ##r'=r## in the denumerator. Is not that the same issue as in:
$$ \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...