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# 物理代写|电磁学代写Electromagnetism代考|Cylindrical Boundary Value Problems

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## 物理代写|电磁学代写Electromagnetism代考|Separation of Variables

The separation of variables method is also important in solving magnetostatic problems. It will suffice to discus a few examples in cylindrical coordinates.

In Sect. 5.1, we have restricted ourselves to Cartesian coordinates, which enabled us to use the known solution of the scalar Poisson equation to solve Poisson’s vector equation (5.12). It is also possible to use curvilinear coordinates, for example, cylindrical coordinates. Then one needs to solve the set of equations given in (5.14). For simplicity reasons, we restrict ourselves to rotationally symmetric fields, and assume that there are only azimuthal currents.
$$\mathbf{g}=\left\langle 0, g_{\varphi}(r, z), 0\right\rangle .$$
Based on Sect. 5.2.4, one concludes that $\mathbf{A}$ also has only an azimuthal component
$$\mathbf{A}=\left\langle 0, \mathbf{A}{\varphi}(r, z), 0\right\rangle .$$ According to (5.14) it is for $\mathbf{A}{\varphi}$ :
$$\nabla^2 \mathbf{A}{\varphi}(r, z)-\frac{\mathbf{A}{\varphi}(r, z)}{r^2}=-\mu_0 g_{\varphi}(r, z) .$$
In particular for the current-free space
$$\nabla^2 \mathbf{A}{\varphi}-\frac{\mathbf{A}{\varphi}}{r^2}=0$$
or written more explicitly using (3.33):
$$\frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial}{\partial r} \mathbf{A}{\varphi}(r, z)+\frac{\partial^2}{\partial z^2} \mathbf{A}{\varphi}(r, z)-\frac{\mathbf{A}_{\varphi}(r, z)}{r^2}=0 .$$

## 物理代写|电磁学代写Electromagnetism代考|Structure of Rotationally Symmetric Magnetic Fields

The function $r A_{\varphi}(r, z)$ is constant along field lines. From
$$\mathbf{B}=\nabla \times \mathbf{A}$$
follows with (5.145) that
$$\mathbf{B}=\left\langle-\frac{\partial}{\partial z} A_{\varphi}, 0, \frac{1}{r} \frac{\partial}{\partial r}\left(r A_{\varphi}\right)\right\rangle .$$
Therefore, for rotational symmetry $\partial / \partial \varphi=0$, it is:
\begin{aligned} \mathbf{B} \bullet \nabla\left(r A_{\varphi}\right) & =B_r \frac{\partial}{\partial r}\left(r A_{\varphi}\right)+B_z \frac{\partial}{\partial z}\left(r A_{\varphi}\right) \ & =-\frac{\partial}{\partial z} A_{\varphi} \frac{\partial}{\partial r}\left(r A_{\varphi}\right)+\frac{1}{r} \frac{\partial}{\partial r}\left(r A_{\varphi}\right) \frac{\partial}{\partial z}\left(r A_{\varphi}\right)=0 . \end{aligned}
This means that $r A_{\varphi \varphi}(r, z)$ is the flux function of the rotationally symmetric field. The lines $r A_{\varphi}(r, z)=$ const. are the field lines laying in the $r-z$ plane $(\varphi=$ const. $)$

The field used initially based on $\mathbf{g}$ as of eq. (5.144) is not the most general rotationally symmetric field, which is rather
$$\mathbf{g}=\left\langle g_r(r, z), g_{\varphi}(r, z), g_z(r, z)\right\rangle$$
which yields

$$\mathbf{B}=\left\langle B_r(r, z), B_{\varphi}(r, z), B_z(r, z)\right\rangle .$$
Both vector fields are source-free:
\begin{aligned} & \nabla \bullet g=\frac{1}{r} \frac{\partial}{\partial r}\left(r g_r\right)+\frac{\partial}{\partial z}\left(g_z\right)=0 \ & \nabla \bullet \mathbf{B}=\frac{1}{r} \frac{\partial}{\partial r} r B_r+\frac{\partial}{\partial z}\left(B_z\right)=0 . \end{aligned}

## 物理代写|电磁学代写Electromagnetism代考|Separation of Variables

$$\mathbf{g}=\left\langle 0, g_{\varphi}(r, z), 0\right\rangle$$

$$\mathbf{A}=\langle 0, \mathbf{A} \varphi(r, z), 0\rangle .$$

$$\nabla^2 \mathbf{A} \varphi(r, z)-\frac{\mathbf{A} \varphi(r, z)}{r^2}=-\mu_0 g_{\varphi}(r, z) .$$

$$\nabla^2 \mathbf{A} \varphi-\frac{\mathbf{A} \varphi}{r^2}=0$$

$$\frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial}{\partial r} \mathbf{A} \varphi(r, z)+\frac{\partial^2}{\partial z^2} \mathbf{A} \varphi(r, z)-\frac{\mathbf{A}_{\varphi}(r, z)}{r^2}=0 .$$

## 物理代写|电磁学代写Electromagnetism代考|Structure of Rotationally Symmetric Magnetic Fields

$$\mathbf{B}=\nabla \times \mathbf{A}$$

$$\mathbf{B}=\left\langle-\frac{\partial}{\partial z} A_{\varphi}, 0, \frac{1}{r} \frac{\partial}{\partial r}\left(r A_{\varphi}\right)\right\rangle$$

$$\mathbf{B} \bullet \nabla\left(r A_{\varphi}\right)=B_r \frac{\partial}{\partial r}\left(r A_{\varphi}\right)+B_z \frac{\partial}{\partial z}\left(r A_{\varphi}\right) \quad=-\frac{\partial}{\partial z} A_{\varphi} \frac{\partial}{\partial r}\left(r A_{\varphi}\right)+\frac{1}{r} \frac{\partial}{\partial r}\left(r A_{\varphi}\right) \frac{\partial}{\partial z}\left(r A_{\varphi}\right)=0$$

$$\mathbf{g}=\left\langle g_r(r, z), g_{\varphi}(r, z), g_z(r, z)\right\rangle$$

$$\mathbf{B}=\left\langle B_r(r, z), B_{\varphi}(r, z), B_z(r, z)\right\rangle$$

$$\nabla \bullet g=\frac{1}{r} \frac{\partial}{\partial r}\left(r g_r\right)+\frac{\partial}{\partial z}\left(g_z\right)=0 \quad \nabla \bullet \mathbf{B}=\frac{1}{r} \frac{\partial}{\partial r} r B_r+\frac{\partial}{\partial z}\left(B_z\right)=0$$

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