- #1
georg gill
- 153
- 6
I am trying to derive that
$$\nabla \times B=\mu_0 J$$
First the derivation starts with the electric field
$$dS=rsin\varphi d\theta r d\varphi $$
$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0} \iint\limits_S \frac{r}{|r|^3} \cdot dS $$
$$\textbf{r}=x\textbf{i}+y\textbf{j}+z\textbf{k}=\textbf{r}=rcos\theta sin \varphi\textbf{i}+rsin\theta sin \varphi\textbf{j}+rcos\varphi\textbf{k}$$$$\frac{\frac{\partial \textbf{r}}{\partial r}} {|\frac{\partial \textbf{r}}{\partial r}|}=cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k}$$$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0 |r|^3} \iint\limits_S rcos\theta sin \varphi\textbf{i}+rsin\theta sin \varphi\textbf{j}+rcos\varphi\textbf{k} \cdot dS$$
$$ =\frac{q}{4 \pi \varepsilon_0 |r|^2}\iint\limits_S cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k} \cdot dS $$$$= \frac{q}{4 \pi \varepsilon_0 |r|^2}\iint\limits_S (cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k}) \cdot (cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k}) rsin\varphi d\theta d\varphi $$
$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0 }\iint\limits_S (cos^2\theta sin^2 \varphi+sin^2\theta sin^2 \varphi+cos^2\varphi) sin\varphi d\theta d\varphi $$
$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0 }\int_0^\pi \int_0^{2\pi} sin\varphi d\theta r d\varphi=\frac{q}{\varepsilon_0} $$
$$\iiint_V \nabla \cdot E dxdydz=\iint\limits_S E \cdot dS = \frac{q}{\varepsilon_0}$$
As for divergence of E:
$$E=\frac{q}{4 \pi \varepsilon_0 r^2 }[\frac{x}{r}\textbf{i}+\frac{y}{r}\textbf{j}+\frac{z}{r}\textbf{k}]$$
after some calculations:
$$\nabla \cdot E=\frac{q}{4 \pi \varepsilon_0}\frac{3r^2-3r^2}{r^5}$$
Which is 0 except in origo
From problem 2.7 and problem 2.8 in Griffiths introduction to electrodynamics we obtain that the E field inside a charged sphere is pointing radially outwards and is given 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:
$$\nabla \cdot E=\nabla \cdot \frac{Qr}{4\pi\varepsilon_0 R^3}[\frac{x}{r}\textbf{i}+\frac{y}{r}\textbf{j}+\frac{z}{r}\textbf{k}]=3\frac{Q}{4\pi\varepsilon_0 R^3}=\frac{\rho}{\varepsilon_0}$$
Introducing the electric potential
$$E=\frac{q}{4 \pi \varepsilon_0 r^2 }[\frac{x}{r}\textbf{i}+\frac{y}{r}\textbf{j}+\frac{z}{r}\textbf{k}]=-\frac{q}{4 \pi \varepsilon_0} \nabla \frac{1}{r}=-\nabla V$$
It can easily be obtained that
$$\frac{\textbf{r}}{r^3}=-\nabla \frac{1}{r}$$
From that we obtain
$$\nabla \cdot E=-\frac{q}{4 \pi \varepsilon_0} \nabla^2 \frac{1}{r} $$
From results above we obtain that for a point charge in origo we have that
$$\nabla^2 \frac{1}{r}=-4 \pi $$
And inside our charged sphere
$$\nabla^2 \frac{1}{r}=-\frac{4 \pi}{V} $$
Above V is the volume of the charged sphere.
from example 5.11 of Griffiths introduction to electrodynamics 4th edition. And problem 5.29 of Griffiths introduction to electrodynamics third edition one obtains the B field inside a rotating charged sphere. By taking the curl of that one obtains that the curl of the B field inside a rotating charged sphere is ##\nabla \times B=\mu_0 J##
If we instead view the charge as a point charge
The magnetic vector potential gives that B field is the curl of the magnetic vector potential so that:
$$B(r)=\frac{\mu _0}{4\pi}\nabla \times \int \frac{ J(r')}{|r-r'|} dV'$$
$$\nabla \times B(r)=\frac{\mu _0}{4\pi}\nabla \times \nabla \times \int \frac{ J(r')}{|r-r'|} dV'$$
We use the identity:
$$\nabla \times (\nabla \times B)=\nabla(\nabla \cdot B)- \nabla^2 B$$
$$\nabla \times B(r)=\frac{\mu _0}{4\pi}\nabla \nabla \cdot \int \frac{ J(r')}{|r-r'|} dV' - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$
Since it is the current that is the vector we can rewrite the first part when we also use Leibniz integral rule to get the divergence inside the integral:
$$\nabla \times B(r)=\frac{\mu _0}{4\pi}\nabla \int J(r') \cdot \nabla \frac{ 1}{|r-r'|} dV' - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$
Since we assume that the current density is constant
$$\nabla \times B(r)=-\frac{\mu _0}{4\pi}\nabla \int \nabla' \cdot (J(r') \frac{ 1}{|r-r'|}) dV' - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$
Divergence theorem gives that the first part is 0 since we are integrating over the current density volume and it is steady so that no current density goes out of the volume:
$$\nabla \times B(r)= - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$
I have added all these derivations so that if someone would want to answer they could use the derivation that made me approach my problem which is:
Now they use that in the point charge:
$$\nabla^2 \frac{1}{|r-r'|}=-4 \pi $$
$$\nabla \times B(r)= \mu _0 \int J(r') dV'$$
How can they do that since ##\nabla^2 \frac{1}{|r-r'|}=-4 \pi ## is taken from electric forces from the E field and we are looking at the magnetic field?
$$\nabla \times B=\mu_0 J$$
First the derivation starts with the electric field
$$dS=rsin\varphi d\theta r d\varphi $$
$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0} \iint\limits_S \frac{r}{|r|^3} \cdot dS $$
$$\textbf{r}=x\textbf{i}+y\textbf{j}+z\textbf{k}=\textbf{r}=rcos\theta sin \varphi\textbf{i}+rsin\theta sin \varphi\textbf{j}+rcos\varphi\textbf{k}$$$$\frac{\frac{\partial \textbf{r}}{\partial r}} {|\frac{\partial \textbf{r}}{\partial r}|}=cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k}$$$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0 |r|^3} \iint\limits_S rcos\theta sin \varphi\textbf{i}+rsin\theta sin \varphi\textbf{j}+rcos\varphi\textbf{k} \cdot dS$$
$$ =\frac{q}{4 \pi \varepsilon_0 |r|^2}\iint\limits_S cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k} \cdot dS $$$$= \frac{q}{4 \pi \varepsilon_0 |r|^2}\iint\limits_S (cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k}) \cdot (cos\theta sin \varphi\textbf{i}+sin\theta sin \varphi\textbf{j}+cos\varphi\textbf{k}) rsin\varphi d\theta d\varphi $$
$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0 }\iint\limits_S (cos^2\theta sin^2 \varphi+sin^2\theta sin^2 \varphi+cos^2\varphi) sin\varphi d\theta d\varphi $$
$$ \iint\limits_S E \cdot dS = \frac{q}{4 \pi \varepsilon_0 }\int_0^\pi \int_0^{2\pi} sin\varphi d\theta r d\varphi=\frac{q}{\varepsilon_0} $$
$$\iiint_V \nabla \cdot E dxdydz=\iint\limits_S E \cdot dS = \frac{q}{\varepsilon_0}$$
As for divergence of E:
$$E=\frac{q}{4 \pi \varepsilon_0 r^2 }[\frac{x}{r}\textbf{i}+\frac{y}{r}\textbf{j}+\frac{z}{r}\textbf{k}]$$
after some calculations:
$$\nabla \cdot E=\frac{q}{4 \pi \varepsilon_0}\frac{3r^2-3r^2}{r^5}$$
Which is 0 except in origo
From problem 2.7 and problem 2.8 in Griffiths introduction to electrodynamics we obtain that the E field inside a charged sphere is pointing radially outwards and is given 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:
$$\nabla \cdot E=\nabla \cdot \frac{Qr}{4\pi\varepsilon_0 R^3}[\frac{x}{r}\textbf{i}+\frac{y}{r}\textbf{j}+\frac{z}{r}\textbf{k}]=3\frac{Q}{4\pi\varepsilon_0 R^3}=\frac{\rho}{\varepsilon_0}$$
Introducing the electric potential
$$E=\frac{q}{4 \pi \varepsilon_0 r^2 }[\frac{x}{r}\textbf{i}+\frac{y}{r}\textbf{j}+\frac{z}{r}\textbf{k}]=-\frac{q}{4 \pi \varepsilon_0} \nabla \frac{1}{r}=-\nabla V$$
It can easily be obtained that
$$\frac{\textbf{r}}{r^3}=-\nabla \frac{1}{r}$$
From that we obtain
$$\nabla \cdot E=-\frac{q}{4 \pi \varepsilon_0} \nabla^2 \frac{1}{r} $$
From results above we obtain that for a point charge in origo we have that
$$\nabla^2 \frac{1}{r}=-4 \pi $$
And inside our charged sphere
$$\nabla^2 \frac{1}{r}=-\frac{4 \pi}{V} $$
Above V is the volume of the charged sphere.
from example 5.11 of Griffiths introduction to electrodynamics 4th edition. And problem 5.29 of Griffiths introduction to electrodynamics third edition one obtains the B field inside a rotating charged sphere. By taking the curl of that one obtains that the curl of the B field inside a rotating charged sphere is ##\nabla \times B=\mu_0 J##
If we instead view the charge as a point charge
The magnetic vector potential gives that B field is the curl of the magnetic vector potential so that:
$$B(r)=\frac{\mu _0}{4\pi}\nabla \times \int \frac{ J(r')}{|r-r'|} dV'$$
$$\nabla \times B(r)=\frac{\mu _0}{4\pi}\nabla \times \nabla \times \int \frac{ J(r')}{|r-r'|} dV'$$
We use the identity:
$$\nabla \times (\nabla \times B)=\nabla(\nabla \cdot B)- \nabla^2 B$$
$$\nabla \times B(r)=\frac{\mu _0}{4\pi}\nabla \nabla \cdot \int \frac{ J(r')}{|r-r'|} dV' - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$
Since it is the current that is the vector we can rewrite the first part when we also use Leibniz integral rule to get the divergence inside the integral:
$$\nabla \times B(r)=\frac{\mu _0}{4\pi}\nabla \int J(r') \cdot \nabla \frac{ 1}{|r-r'|} dV' - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$
Since we assume that the current density is constant
$$\nabla \times B(r)=-\frac{\mu _0}{4\pi}\nabla \int \nabla' \cdot (J(r') \frac{ 1}{|r-r'|}) dV' - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$
Divergence theorem gives that the first part is 0 since we are integrating over the current density volume and it is steady so that no current density goes out of the volume:
$$\nabla \times B(r)= - \frac{\mu _0}{4\pi} \nabla^2 \int \frac{ J(r')}{|r-r'|} dV'$$
I have added all these derivations so that if someone would want to answer they could use the derivation that made me approach my problem which is:
Now they use that in the point charge:
$$\nabla^2 \frac{1}{|r-r'|}=-4 \pi $$
$$\nabla \times B(r)= \mu _0 \int J(r') dV'$$
How can they do that since ##\nabla^2 \frac{1}{|r-r'|}=-4 \pi ## is taken from electric forces from the E field and we are looking at the magnetic field?