Posted on Categories:Electromagnetism, 物理代写, 电磁学

avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

物理代写|电磁学代写Electromagnetism代考|Vector Potential

We return to Biot-Savart law, and rewrite it as follows (refer also to Fig. 8.14):

$$\mathbf{B}=\frac{\mu_0}{4 \pi} \nabla \times \int_V \frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} d \mathbf{r}^{\prime}$$
where we have used that
$$\nabla\left(\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\right)=-\frac{\mathbf{r}-\mathbf{r}^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3}$$
and
$$\frac{I d \mathbf{s} \times \hat{\mathbf{r}}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^2}=-d \mathbf{r}^{\prime} \frac{\left(\left(\mathbf{r}-\mathbf{r}^{\prime}\right) \times \mathbf{J}\left(\mathbf{r}^{\prime}\right)\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3}$$
where $d V=d \mathbf{r}^{\prime}$ is a small volume element, as indicated in Fig. 8.14. We can now introduce a vector potential of the magnetic field as
$$\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4 \pi} \int_V \frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} d \mathbf{r}^{\prime}$$
and the magnetic field can be written as
$$\mathbf{B}=\nabla \times \mathbf{A}(\mathbf{r})$$

物理代写|电磁学代写Electromagnetism代考|Multipole Expansion

Equation (8.66) gives the vector potential of the magnetic field in terms of the current density $\mathbf{J}\left(\mathbf{r}^{\prime}\right)$ in a localized finite volume $V$. Furthermore, $\mathbf{J}\left(\mathbf{r}^{\prime}\right)$ is zero outside the volume. Suppose that we are interested on finding $A$ outside that volume. For that, similar to scalar potential in electrostatics, we expand the term $1 /\left|\mathbf{r}-\mathbf{r}^{\prime}\right|$ around $\mathbf{r}^{\prime}=0$ using Taylor expansion, as given by Eq. (3.50) (Chap. 3).
Assuming that $|\mathbf{r}| \gg\left|\mathbf{r}^{\prime}\right|$, we can rewrite Eq. (8.66) as follows:

$$\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4 \pi} \int_V \mathbf{J}\left(\mathbf{r}^{\prime}\right)\left(\frac{1}{r}+\frac{\mathbf{r}^{\prime} \cdot \mathbf{r}}{r^3}+\frac{1}{2} \sum_{i, j=1}^3 \frac{3 x_i^{\prime} x_j^{\prime}-\delta_{i j}\left(r^{\prime}\right)^2}{r^5} x_i x_j+\cdots\right) d \mathbf{r}^{\prime}$$
Equation (8.75) can be considered sum of three contributions, namely, $\mathbf{A}_0, \mathbf{A}_1$ and $\mathbf{A}_3$, if we neglect higher order term in the expansion.

The first term, which corresponds to the monopole term in the electrostatic expansion, is
$$\mathbf{A}_0(\mathbf{r})=\frac{\mu_0}{4 \pi r} \int_V \mathbf{J}\left(\mathbf{r}^{\prime}\right) d \mathbf{r}^{\prime}=\frac{\mu_0}{4 \pi r} \int_A\left(\nabla \cdot \mathbf{J}\left(\mathbf{r}^{\prime}\right)\right) d A$$
where the integration is over the surface enclosing the volume $V$ and Stokes’ formula is used. Using the continuity equation of the current density $(\rho=0)$ :
$$\nabla \cdot \mathbf{J}=0$$
we obtain
$$\mathbf{A}_0(\mathbf{r})=0$$

物理代写|电磁学代写Electromagnetism代考|Vector Potential

$$\mathbf{B}=\frac{\mu_0}{4 \pi} \nabla \times \int_V \frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} d \mathbf{r}^{\prime}$$

$$\nabla\left(\frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\right)=-\frac{\mathbf{r}-\mathbf{r}^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3}$$

$$\frac{I d \mathbf{s} \times \hat{\mathbf{r}}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^2}=-d \mathbf{r}^{\prime} \frac{\left(\left(\mathbf{r}-\mathbf{r}^{\prime}\right) \times \mathbf{J}\left(\mathbf{r}^{\prime}\right)\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3}$$

$$\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4 \pi} \int_V \frac{\mathbf{J}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} d \mathbf{r}^{\prime}$$

$$\mathbf{B}=\nabla \times \mathbf{A}(\mathbf{r})$$

物理代写|电磁学代写Electromagnetism代考|Multipole Expansion

$$\mathbf{A}(\mathbf{r})=\frac{\mu_0}{4 \pi} \int_V \mathbf{J}\left(\mathbf{r}^{\prime}\right)\left(\frac{1}{r}+\frac{\mathbf{r}^{\prime} \cdot \mathbf{r}}{r^3}+\frac{1}{2} \sum_{i, j=1}^3 \frac{3 x_i^{\prime} x_j^{\prime}-\delta_{i j}\left(r^{\prime}\right)^2}{r^5} x_i x_j+\cdots\right) d \mathbf{r}^{\prime}$$

$$\mathbf{A}_0(\mathbf{r})=\frac{\mu_0}{4 \pi r} \int_V \mathbf{J}\left(\mathbf{r}^{\prime}\right) d \mathbf{r}^{\prime}=\frac{\mu_0}{4 \pi r} \int_A\left(\nabla \cdot \mathbf{J}\left(\mathbf{r}^{\prime}\right)\right) d A$$

$$\nabla \cdot \mathbf{J}=0$$

$$\mathbf{A}_0(\mathbf{r})=0$$

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Electromagnetism, 物理代写, 电磁学

avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

物理代写|电磁学代写Electromagnetism代考|Point Charge on the Axis of a Dielectric Cylinder

We want to find the potential caused by a point charge $Q$ at the axis of a dielectric, circular cylinder $\left(\varepsilon_1\right)$. For the remaining space, we assume a different dielectric $\left(\varepsilon_2\right)$ (see Fig. 3.17).
This is by no means a trivial problem, but it can be solved with the results obtained from the previous example. We need to solve Laplace’s equation in region 2, what is certainly possible with an approach similar to (3.185):
$$\varphi_2=\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} \cos (k z) K_0(k r) g_2(k) d k .$$
The factor in front of the integral is just for convenience and could as well be regarded as belonging to $g_2(k)$. In region 1, we have to solve Poisson’s equation for a point charge. This may be done by superposition of the point charge potential according to (3.203) and the general solution of Laplace’s equation according to (3.185), i.e., by
$$\varphi_1=\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} \cos (k z)\left[K_0(k r)+I_0(k r) g_1(k)\right] d k$$
Remember that the general solution of the inhomogeneous (Poisson) equation is obtained from the superposition of the specific solution of the inhomogeneous equation and the general solution of the homogeneous (Laplace) equation. Significant is now, to know the potential of the point charge in the specific form that fits this problem. Another way to view the trial functions of (3.205) and (3.206) is by superposition of the point charge potential and the potential of bound surface charges at the cylinder surface (wall). The following boundary conditions apply for $r=r_0$ :

\begin{aligned} & \left(\frac{\partial \varphi_1}{\partial z}\right){r=r_0}-\left(\frac{\partial \varphi_2}{\partial z}\right){r=r_0}=0 \ & \varepsilon_1\left(\frac{\partial \varphi_1}{\partial r}\right){r=r_0}-\varepsilon_2\left(\frac{\partial \varphi_2}{\partial r}\right){r=r_0}=0 . \end{aligned}
This gives
$$\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} k \sin (k z)\left[K_0\left(k r_0\right)+I_0\left(k r_0\right) g_1(k)-K_0\left(k r_0\right) g_2(k)\right] d k=0$$
and
$$\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} k \cos (k z)\left[-\varepsilon_1 K_1\left(k r_0\right)+\varepsilon_1 I_1\left(k r_0\right) g_1(k)+\varepsilon_2 K_1\left(k r_0\right) g_2(k)\right] d k=0$$
or
\begin{aligned} K_0\left(k r_0\right)+I_0\left(k r_0\right) g_1(k)-K_0\left(k r_0\right) g_2(k) & =0 \ -\varepsilon_1 K_1\left(k r_0\right)+\varepsilon_1 I_1\left(k r_0\right) g_1(k)+\varepsilon_2 K_1\left(k r_0\right) g_2(k) & =0 \end{aligned}

物理代写|电磁学代写Electromagnetism代考|Dirichlet’s Boundary Value Problem and the Fourier-Bessel Series

Consider the cylinder shown in Fig. $3.20$ with radius $r_0$ and height $h$. We want to find the potential inside the charge-free cylinder with the following boundary conditions.
$$\begin{array}{ll} \varphi=\varphi_h(r) & \text { for } z=h \ \varphi=0 & \text { on the remaining surface. } \end{array}$$
This is the cylindrical analogue to the example in Section 3.5.2.1. Again, $m=0$ because of the rotational symmetry. For the radial part $R$, we may choose either $J_0$ or $I_0$, but not $N_0$ or $K_0$ since those let the potential at the axis diverge. The potential has to also vanish for $r=r_0 . J_0$ has zeros for real arguments, $I_0$ does not. The convenient choice is thus $J_0$, For the $z$-dependency, we may choose an approach, for example, based on (3.158) where only the $\sinh$ is an option because of the restriction $\varphi=0$ for $z=0$. We suggest that solving the problem is possible by superposing expressions of the form:
$$J_0(k r) \sinh k z \text {. }$$
Hereby, the condition
$$J_0\left(k r_0\right)=0$$

物理代写|电磁学代写Electromagnetism代考|Point Charge on the Axis of a Dielectric Cylinder

$$\varphi_2=\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} \cos (k z) K_0(k r) g_2(k) d k .$$

$$\varphi_1=\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} \cos (k z)\left[K_0(k r)+I_0(k r) g_1(k)\right] d k$$

$$\left(\frac{\partial \varphi_1}{\partial z}\right) r=r_0-\left(\frac{\partial \varphi_2}{\partial z}\right) r=r_0=0 \quad \varepsilon_1\left(\frac{\partial \varphi_1}{\partial r}\right) r=r_0-\varepsilon_2\left(\frac{\partial \varphi_2}{\partial r}\right) r=r_0=0 .$$

$$\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} k \sin (k z)\left[K_0\left(k r_0\right)+I_0\left(k r_0\right) g_1(k)-K_0\left(k r_0\right) g_2(k)\right] d k=0$$

$$\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} k \cos (k z)\left[-\varepsilon_1 K_1\left(k r_0\right)+\varepsilon_1 I_1\left(k r_0\right) g_1(k)+\varepsilon_2 K_1\left(k r_0\right) g_2(k)\right] d k=0$$

$$K_0\left(k r_0\right)+I_0\left(k r_0\right) g_1(k)-K_0\left(k r_0\right) g_2(k)=0-\varepsilon_1 K_1\left(k r_0\right)+\varepsilon_1 I_1\left(k r_0\right) g_1(k)+\varepsilon_2 K_1\left(k r_0\right) g_2(k) \quad=0$$

物理代写|电磁学代可Electromagnetism代荣|Dirichlet’s Boundary Value Problem and the Fourier-Bessel Series

$$\varphi=\varphi_h(r) \quad \text { for } z=h \varphi=0 \quad \text { on the remaining surface. }$$

$$J_0(k r) \sinh k z .$$

$$J_0\left(k r_0\right)=0$$

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Electromagnetism, 物理代写, 电磁学

avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

物理代写|电磁学代写Electromagnetism代考|Torque on a Current Loop in a Uniform Magnetic Field

We showed that a force acts on a current-carrying conductor placed in a magnetic field, given mathematically by Eq. (6.10). Now, consider a current loop placed in a magnetic field, as shown in Fig. 6.8. Equation (6.10) implied that the net magnetic force acting on the loop is zero (see also Eq. (6.14)). However, we will prove in the following that the magnetic field applies a torque on the current loop. For that, consider a rectangular loop carrying a current $I$ in the presence of a uniform magnetic field. First, we assume that the magnetic field is directed parallel to the plane of the loop, as shown in Fig. $6.8$.

The magnetic forces exerted on sides 12 and 34 are both zeroes because these wire segments are parallel to the field $\mathbf{B}$; hence, $\mathbf{L} \times \mathbf{B}=0$ for those two segments. However, the magnetic field is normal to the portions 23 and 34 ; therefore, the magnetic forces on Sects. 23 and 41 are different from zero.

We calculate the magnitude of the forces acting on segments 23 and 41 carrying the current $I$ and length $L=a$ from Eq. (6.8), as
$$F_{23}=F_{41}=\mid I(\mathbf{L} \times \mathbf{B} \mid=I a B$$

However, the direction of $\mathbf{F}{41}$, the force exerted on segment 41, is out of the page, and that of $\mathbf{F}{23}$, the force exerted on segment 23 , is into the page, as indicated in Fig. 6.9. Therefore, those two forces point in opposite directions, and they are not directed along the same line of action; that is, the distance between their lines of action is $b$. Thus, those two forces exert a torque on the loop, such that the loop rotates counterclockwise about the axis passing through the point $O$, as shown in Fig. 6.9. The magnitude of that torque $\tau$ is
$$\tau=F_{23} \frac{b}{2}+F_{41} \frac{b}{2}=(I a B) \frac{b}{2}+(I a B) \frac{b}{2}=I a b B$$
where $b / 2$ is the moment arm about $O$ of each force. Denoting $A=a b$ the area of closed loop, then
$$\tau=I A B$$
We can generalize that result by considering the same current-carrying closed loop, but in a uniform magnetic field making an angle $\theta<90^{\circ}$ with the normal vector to the plane of the loop, as indicated in Fig. 6.10. We also assume that $\mathbf{B}$ is perpendicular to the segment lines 12 and 34 .

物理代写|电磁学代写Electromagnetism代考|Motion of a Charged Particle in a Uniform Magnetic Field

From Eq. (6.1), the force exerted on a charged particle moving in a magnetic field is normal to the velocity of the particle, and hence, perpendicular to its displacement. Therefore, the work done on the particle by the magnetic force is zero because the force is perpendicular to the displacement vector of the particle.

As a particular case, we consider a positively charged particle moving in a uniform magnetic field with an initial velocity vector of the particle perpendicular to the field (see Fig. 6.13). The direction of the magnetic field is pointing into the page. Using the right-hand rule in Eq. (6.1), we find that the direction of the magnetic force is pointing toward a single point at the center of a circle. Therefore, the particle is going to move in a circle in a plane perpendicular to the magnetic field.

The particle moves in this way because the magnetic force $\mathbf{F}B$ is perpendicular to both $\mathbf{v}$ and $\mathbf{B}$ and has a constant magnitude of $q v B$. As the force deflects the particle, the directions of both $\mathbf{v}$ and $\mathbf{F}_B$ change continuously, and $\mathbf{F}_B$ points toward the center of the circle at each position of the particle. Thus, force changes only the direction of $\mathbf{v}$, but it does change its magnitude. The rotation is counterclockwise for that positive charge, and if $q$ is negative, the rotation would be clockwise. Using the second law of Newton for circular motion, we get $$\sum_i F{i r}=m a_r$$
or
$$F_B=q v B=m \frac{v^2}{r}$$
The radius of the circle is
$$r=\frac{m v^2}{q v B}=\frac{m v}{q B}$$

物理代写|电磁学代写Electromagnetism代考|Torque on a Current Loop in a Uniform Magnetic Field

$$F_{23}=F_{41}=\mid I(\mathbf{L} \times \mathbf{B} \mid=I a B$$

$$\tau=F_{23} \frac{b}{2}+F_{41} \frac{b}{2}=(I a B) \frac{b}{2}+(I a B) \frac{b}{2}=I a b B$$

$$\tau=I A B$$

物理代写|电磁学代写Electromagnetism代考|Motion of a Charged Particle in a Uniform Magnetic Field

$$\sum_i F i r=m a_r$$

$$F_B=q v B=m \frac{v^2}{r}$$

$$r=\frac{m v^2}{q v B}=\frac{m v}{q B}$$

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Posted on Categories:Electromagnetism, 物理代写, 电磁学

avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

物理代写|电磁学代写Electromagnetism代考|Resonance Scattering

For resonance scattering,
$$\omega \simeq \omega_{0}$$
and
\begin{aligned} f(\omega) &=\frac{\omega^{4}}{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+\left(\frac{2 e^{2} \omega^{3}}{3 m c^{3}}\right)^{2}} \ &=\frac{\omega_{0}^{4}}{\left(\omega_{0}-\omega\right)^{2}\left(\omega_{0}+\omega\right)^{2}+\left(\frac{2 e^{2} \omega^{3}}{3 m c^{3}}\right)^{2}} \ & \simeq \frac{\omega_{0}^{2}}{4\left(\omega_{0}-\omega\right)^{2}+\left(\frac{2 e^{2} \omega_{0}^{2}}{3 m c^{3}}\right)^{2}}=\frac{\omega_{0}^{2} / 4}{\left(\omega_{0}-\omega\right)^{2}+\frac{1}{4} \gamma^{2}} \end{aligned}
where
$$\gamma=\frac{2 e^{2} \omega_{0}^{2}}{3 m c^{3}}=\frac{2}{3}\left(\frac{e^{2}}{m c^{2}}\right) \frac{\omega}{c} \omega_{0}=\frac{4 \pi}{3} \frac{r_{e}}{\lambda} \omega_{0}$$
Then
\begin{aligned} d \sigma &=r_{e}^{2} \sin ^{2} \theta d \Omega f(\omega)=r_{e}^{2} \sin ^{2} \theta d \Omega \frac{\omega_{0}^{2} / 4}{\left(\omega_{0}-\omega\right)^{2}+\frac{1}{4} \gamma^{2}} \ &=\frac{1}{4} r_{e}^{2} \sin ^{2} \theta d \Omega \frac{\omega_{0}^{2} /(\gamma / 2)^{2}}{1+\left[\left(\omega-\omega_{0}\right) /(\gamma / 2)\right]^{2}} \end{aligned}

物理代写|电磁学代写Electromagnetism代考|Multipole Expansion

A current or a charge distribution is periodic if it is of the following form:
\begin{aligned} &\mathbf{j}(\mathbf{x}, t)=\mathbf{j}{0}(\mathbf{x}) e^{-i \omega t} \ &\rho(\mathbf{x}, t)=\rho{0}(\mathbf{x}) e^{-i \omega t} \end{aligned}
This case is distinct from the case of a charge vibrating harmonically with frequency $\omega$ :
$$\mathbf{x}{p}(t)=\mathbf{x}{0} \cos \omega t$$
In this last case,
\begin{aligned} \mathbf{j}(\mathbf{x}, t) &=e \dot{\mathbf{x}}{p} \delta\left(\mathbf{x}-\mathbf{x}{p}(t)\right) \ &=-e \omega \mathbf{x}{0} \sin \omega t \delta\left(\mathbf{x}-\mathbf{x}{0} \cos \omega t\right) \ \rho(\mathbf{x}, t) &=e \delta\left(\mathbf{x}-\mathbf{x}{p}(t)\right)=e \delta\left(\mathbf{x}-\mathbf{x}{0} \cos \omega t\right) \end{aligned}

and the current an charge do not depend on time as $e^{-i \omega t}$. Eventually, we can make the following expansion:
\begin{aligned} &\mathbf{j}(\mathbf{x}, t)=\sum_{n=1}^{\infty} \mathbf{j}{n}(\mathbf{x}) e^{-i n \omega t} \ &\rho(\mathbf{x}, t)=\sum{n=1}^{\infty} \rho_{n}(\mathbf{x}) e^{-i n \omega t} \end{aligned}
A harmonic oscillator can give dipole, quadrupole, .., radiation. In the present chapter we examine these elementary types of radiation separately.

物理代写|电磁学代写Electromagnetism代考|Resonance Scattering

$$f(\omega)=\frac{\omega^{4}}{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+\left(\frac{2 e^{2} \omega^{3}}{3 m c^{3}}\right)^{2}} \quad=\frac{\omega_{0}^{4}}{\left(\omega_{0}-\omega\right)^{2}\left(\omega_{0}+\omega\right)^{2}+\left(\frac{2 e^{2} \omega^{3}}{3 m c^{3}}\right)^{2}} \simeq \frac{\omega_{0}^{2}}{4\left(\omega_{0}-\omega\right)^{2}+\left(\frac{2 e^{2} \omega 0_{0}^{2}}{3 m c^{3}}\right)^{2}}=\frac{\omega_{0}^{2} / 4}{\left(\omega_{0}-\omega\right)^{2}+\frac{1}{4} \gamma^{2}}$$

$$\gamma=\frac{2 e^{2} \omega_{0}^{2}}{3 m c^{3}}=\frac{2}{3}\left(\frac{e^{2}}{m c^{2}}\right) \frac{\omega}{c} \omega_{0}=\frac{4 \pi}{3} \frac{r_{e}}{\lambda} \omega_{0}$$

$$d \sigma=r_{e}^{2} \sin ^{2} \theta d \Omega f(\omega)=r_{e}^{2} \sin ^{2} \theta d \Omega \frac{\omega_{0}^{2} / 4}{\left(\omega_{0}-\omega\right)^{2}+\frac{1}{4} \gamma^{2}} \quad=\frac{1}{4} r_{e}^{2} \sin ^{2} \theta d \Omega \frac{\omega_{0}^{2} /(\gamma / 2)^{2}}{1+\left[\left(\omega-\omega_{0}\right) /(\gamma / 2)\right]^{2}}$$

物理代写|电磁学代写Electromagnetism代考|Multipole Expansion

$$\mathbf{j}(\mathbf{x}, t)=\mathbf{j} 0(\mathbf{x}) e^{-i \omega t} \quad \rho(\mathbf{x}, t)=\rho 0(\mathbf{x}) e^{-i \omega t}$$

$$\mathbf{x} p(t)=\mathbf{x} 0 \cos \omega t$$

$$\mathbf{j}(\mathbf{x}, t)=e \dot{\mathbf{x}} p \delta(\mathbf{x}-\mathbf{x} p(t)) \quad=-e \omega \mathbf{x} 0 \sin \omega t \delta(\mathbf{x}-\mathbf{x} 0 \cos \omega t) \rho(\mathbf{x}, t)=e \delta(\mathbf{x}-\mathbf{x} p(t))=e \delta(\mathbf{x}-\mathbf{x} 0 \cos \omega t)$$

$$\mathbf{j}(\mathbf{x}, t)=\sum_{n=1}^{\infty} \mathbf{j} n(\mathbf{x}) e^{-i n \omega t} \quad \rho(\mathbf{x}, t)=\sum n=1^{\infty} \rho_{n}(\mathbf{x}) e^{-i n \omega t}$$

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。