# Dragon Notes

UNDER CONSTRUCTION
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# Electrostatics:Solved Problems

$$\bb{Pr1}$$[Magnetization flux density]
A ferromagnetic sphere is uniformly magnetized. Determine the magnetic flux density at the center of the sphere.
[Sol] Key Relations:
$$\ds \bb{E1,2,3}\quad \boxed{\bvec{B}=\frac{\mu_0 I}{4\pi}\oint_{C'}\frac{d\bvec{l'}\times \uvec{R}}{R^2}}$$ $$\hspace{15px}\boxed{\bvec{J}_m=\b{\nabla}\times\bvec{M}}$$$$\hspace{15px}\boxed{\bvec{J}_{ms}=\bvec{M}\times\uvec{n}}$$
The quantity of interest is the B-field at the center of the sphere the information provided describes a field quantity – namely, magnetization. The two
can be related via current – by first finding the total current in the region of interest (the sphere), and then the resultant B-field. $$\vplup$$
$$\vplup$$Let $$b=$$ sphere radius, $$M_0=$$ magnetization intensity:
The total current in a ferromagnetic sphere will be the combination of its surface and volume currents – which are related to the sphere’s
magnetization via [E2,3]:
$$\bvec{J}_m=\b{\nabla}\times (M_0\uvec{z})=\boxed{0=\bvec{J}_m}$$
$$\bvec{J}_{ms}=(M_0\uvec{z})\times\underbrace{\bvec{R}}_{\uvec{n}} =\boxed{\uvec{\phi}[M_0\sin{\theta}]=\bvec{J}_{ms}}$$
Hence, all current is on the surface of the sphere. The surface magnetization density found, $$\bvec{J}_ms$$, forms a cylindrical source symmetry about the
$$z$$-axis; thus, $$\bb{E1}$$ applies. To simplify the computation, divide up the total sphere current into infinitesimal loop current elements, and sum their
contributions to attain the total B-field.
$$\vplup$$Refer to the figure below; the B-field due to a loop of radius $$r_0$$ with (const.) current $$I$$ at a point on its central axis can be determined by via $$\bb{1}$$; first
determine $$d\bvec{B}$$, then integrate over $$C'$$:

$$d\bvec{l'}=[r'd\phi']\uvec{\phi}=\boxed{[r_0d\phi']\uvec{\phi}=d\bvec{l'}}$$
$$\bvec{R}=\bvec{R}_{\t{field}}-\bvec{R}_{\t{source}}=\uvec{r}[0-r']+\uvec{z}[z-0]=\boxed{\uvec{z}[-r_0]+\uvec{z}[z]=\bvec{R}\vphantom{\sqrt{r_0^2+z^2}}} \Rightarrow\boxed{R=\sqrt{r_0^2+z^2}}$$
$$\ds d\bvec{l'}\times\bvec{R}=\uvec{\phi}[r_0 d\phi']\times(\uvec{r}[-r_0]+\uvec{z}[z])=[-r_0^2 d\phi']\overbrace{(\uvec{\phi}\times\uvec{r})}^{=-\uvec{z}}+[r_0 zd\phi'] \overbrace{(\uvec{\phi}\times\uvec{z})}^{=\uvec{r}}\Rightarrow$$
$$\boxed{d\bvec{l'}\times\bvec{R}=\uvec{z}[r_0^2 d\phi']+\uvec{r}[r_0 zd\phi']}$$
$$\ds\Rightarrow \frac{d\bvec{l'}\times\uvec{R}}{R^2}=\frac{d\bvec{l'}\times\bvec{R}}{R^3}=\boxed{\frac{\uvec{z}[r_0^2 d\phi']+\uvec{r}[r_0 zd\phi']}{(r_0^2+z^2)^{3/2}} =\frac{d\bvec{l'}\times\uvec{R}}{R^2}}$$

Putting together, and recognizing that $$\uvec{r}$$ varies with $$\phi'$$ - that is,
$$\uvec{r}=\uvec{r}(\phi')=\uvec{x}[\cos{\phi'}]+\uvec{y}[\sin{\phi'}]\t{ - yields}$$
$$\ds\boxed{d\bvec{B}=\lrbra{\frac{\mu_0 Ir_0}{4\pi (r_0^2+z_0^2)^{3/2}}}^{\swarrow \equiv K}\hspace{-35px}(\uvec{z}[r_0]+ \uvec{x}[z\cos{\phi'}]+\uvec{y}[z\sin{\phi'}])d\phi'}$$

The path of integration is along the source: $$C':\underbrace{r'=r_0}_{=\t{const}},0\leq\phi'\leq2\pi,\underbrace{z'=0}_{=\t{const}}$$. Integrating,
\ds\begin{align}\bvec{B}&=\oint_{C'}d\bvec{B}=K\left[\vphantom{\intlim{0}{2\pi}}\right. \uvec{z}r_0\underbrace{\intlim{0}{2\pi}d\phi'}_{=2\pi}+\uvec{x}z\underbrace{\intlim{0}{2\pi}\cos{\phi'}d\phi'}_{=0:\ [\sin{\phi'}] _0^{2\pi}=0}+\uvec{y}z\underbrace{\intlim{0}{2\pi}\sin{\phi'}d\phi'}_{=0:\ [-\cos{\phi'}]_0^{2\pi}=0}\left.\vphantom{\intlim{0}{2\pi}}\right]\\ &= \uvec{z}[2\pi r_0K]=\uvec{z}\left[(2\pi r_0)\Frac{\mu_0 Ir_0}{4\pi(r_0^2+z^2)^{3/2}}\right]\Rightarrow\end{align}
$$\boxed{\bvec{B}_{\t{loop}}=\uvec{z}\left[\frac{1}{2}\mu_0 r_0^2I\right]\frac{1}{(r_0^2+z^2)^{3/2}}}\ \ \bb{1^*}$$
Cylindrical source symmetry could have also been used to justify a claim that field cylindrical $$\uvec{r}$$-components cancel to zero.
$$\vplup$$For the sphere, the total B-field at the origin can be determined via the vector-sum of differential field contributions from differential current source
loops distributed over the sphere’s surface:
The surface integral of $$d\bvec{B}_{\t{loop}}$$ contributions will yield the total magnetic flux intensity, $$\bvec{B}$$, at the center of the sphere. To express the line current
element $$dI_m$$ in terms of the surface current density $$\bvec{J}_{\t{ms}}$$, apply a virtual impulse construct – as no ‘real’ surface is oriented orthogonal to the
spherical shell of interest:
The R-limits of integration aren’t taken to be $$(-\infty,\infty)$$, as that would include the contribution of a separate loop, positioned symmetrically opposite
about the origin to the one of interest. Defining $$I_m$$ to be the total current resultant from integrating $$\bvec{J}_m$$ over the surface of the sphere, we get:
$$\ds I_m=\iint_S\bvec{J}_{\t{m}}\cdot d\bvec{S}$$
A virtual impulse construct can be applied here by expressing the volume current density $$\bvec{J}_{\t{m}}$$ in terms of the surface current density $$\bvec{J}_{\t{ms}}$$ and an
impulse function: $$\bvec{J}_{\t{m}}=\uvec{\phi}[J_{\t{ms}}\delta (R-b)]$$. Then,
\begin{align}I_m &= \intlim{0}{\pi}\lrbra{\intlim{R=0}{R\rightarrow\infty}[RJ_{\t{ms}}\delta (R-b)]dR}d\theta \overbrace{[\uvec{\phi}\cdot\uvec{\phi}]}^{=1}\\ &= \intlim{0}{\pi}bJ_{\t{ms}}d\theta\Rightarrow\boxed{dI_m=J_{\t{ms}}bd\theta}\end{align}
Applying the differential form of $$\bvec{B}_{\t{loop}}$$ in $$\bb{1^*}$$, and taking source coordinates $$(\theta\rightarrow\theta')$$, we get
$$\ds d\bvec{B}=\uvec{z}\lrbra{\frac{\mu_0}{2}}\frac{\overbrace{r_0^2}^{=R\sin{\theta'}}}{\underbrace{(r_0^2+z^2)^{3/2}}_{=R^3}}dI_m= \uvec{z}\lrbra{\frac{\mu_0}{2}}\frac{\t{sin}^2(\theta')}{\underbrace{R}_{=b,=\t{const}}}\overbrace{J_{\t{ms}}}^{=M_0\sin{\theta'}}bd\theta'\Rightarrow$$
$$\ds\boxed{d\bvec{B}=\underbrace{\uvec{z}\lrbra{\frac{\mu_0 M_0}{2}}}_{\equiv K'}\t{sin}^3(\theta')d\theta'}$$
Summing the differential loop B-field contributions over the surface of the sphere, we get
$$\ds\bvec{B}=K\intlim{\theta'=0}{\theta'=\pi}\t{sin}^3(\theta')d\theta' =K\intlim{\theta'=0}{\theta'=\pi}\underbrace{\sin{\theta'} (1-\t{cos}^2(\theta'))d\theta'}_{u=\cos{\theta'},\ du=-\sin{\theta'}d\theta' \\ =\int (u^2-1)du=u^3/3-u} = \frac{K}{3}\overbrace{[\t{cos}^3(\theta')-3 \cos{\theta'}]_0^\pi}^{=4}\Rightarrow$$$\bxred{\bvec{B}=\uvec{z}\lrbra{\frac{2}{3}\mu_0 M_0}}$