Statistics-lab™可以为您提供washington.edu PHYS322 electrodynamics电动力学的代写代考和辅导服务！

PHYS322 electrodynamics课程简介

This course provides a deeper look into the theory of electricity and magnetism.

We shall study:

– magnetostatics

– magnetic fields in matter

– Maxwell’s equations

– electromagnetic waves

PREREQUISITES 

Upon successful completion of this course you will be able to:

• Understand the implications of Ampère’s and Biot-Savart’s laws
• Understand the nature of the vector potential
• Understand the behaviour of magnetic fields in medium
• Be perfectly familiar with Maxwell’s equations
• Know the nature of momentum and angular moment in Electrodynamics
• Understand the details of propagation of electromagnetic waves

PHYS322 electrodynamics HELP（EXAM HELP， ONLINE TUTOR）

a. The stress tensor for an electrostatic field in vacuum is given by: \sigma_{ij}=-\epsilon_0\left(E_iE_j-\frac{1}{2}\delta_{ij}E^2\right)σij​=−ϵ0​(Ei​Ej​−21​δij​E2) where $\sigma_{ij}$ is the stress tensor, $E_i$ is the electric field, $\delta_{ij}$ is the Kronecker delta symbol, and $\epsilon_0$ is the permittivity of free space.

For an infinite parallel-plate capacitor with the lower plate carrying the charge density $-\sigma$ and the upper plate carrying the charge density $+\sigma$, the electric field is constant in the region between the plates and zero outside the plates. Therefore, the stress tensor reduces to:

\sigma_{ij}=-\epsilon_0\left(E_iE_j-\frac{1}{2}\delta_{ij}E^2\right)=-\epsilon_0\frac{1}{2}\delta_{ij}E^2σij​=−ϵ0​(Ei​Ej​−21​δij​E2)=−ϵ0​21​δij​E2

Since the electric field is pointing in the $z$-direction, the only non-zero elements of the stress tensor are:

\sigma_{xx}=\sigma_{yy}=-\frac{1}{2}\epsilon_0E^2σxx​=σyy​=−21​ϵ0​E2 \sigma_{zz}=\epsilon_0E^2σzz​=ϵ0​E2

The off-diagonal elements are zero since there is no shear stress.

b. The force per unit area on the top plate is given by the integral of the normal component of the stress tensor over the surface of the plate:

\mathbf{F}=\int_S \mathbf{\sigma} \cdot \mathbf{\hat{n}} dSF=∫S​σ⋅n^dS

where $\mathbf{\hat{n}}$ is the unit normal vector to the surface $S$.

For the top plate at $z=+d/2$, the unit normal vector is $\mathbf{\hat{n}}=-\mathbf{\hat{z}}$, and the surface element is $dS=dxdy$. Therefore, the force per unit area on the top plate is:

\frac{F}{A}=\frac{\int_{-L/2}^{L/2}\int_{-W/2}^{W/2}\sigma_{zz}dxdy}{LW}=\frac{\sigma_{zz}WL}{LW}=\sigma_{zz}=\epsilon_0E^2AF​=LW∫−L/2L/2​∫−W/2W/2​σzz​dxdy​=LWσzz​WL​=σzz​=ϵ0​E2

where $L$ and $W$ are the dimensions of the plates. Thus, the force per unit area on the top plate is equal to the normal stress on the plate, which is $\epsilon_0E^2$.

To find the vector potential $\mathbf{A}$ at the center of the loop, we can use the Biot-Savart law, which relates the magnetic field $\mathbf{B}$ produced by a current to the vector potential $\mathbf{A}$:

\mathbf{B}(\mathbf{r})=\frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{r’})\times(\mathbf{r}-\mathbf{r’})}{|\mathbf{r}-\mathbf{r’}|^3}d^3r’B(r)=4πμ0​​∫∣r−r′∣3J(r′)×(r−r′)​d3r′

where $\mathbf{J}$ is the current density, $\mu_0$ is the permeability of free space, and the integral is taken over the volume of the wire.

For a circular loop of radius $R$ carrying a current $I(t)=kt$, the current density is given by $\mathbf{J}=\frac{I}{\pi R^2}\mathbf{\hat{\phi}}$, where $\mathbf{\hat{\phi}}$ is the unit vector in the azimuthal direction. Thus, the magnetic field at the center of the loop is:

\mathbf{B}(\mathbf{0})=\frac{\mu_0 k}{4\pi R^2}\int_0^{2\pi}\frac{\mathbf{\hat{\phi}}\times(\mathbf{\hat{z}}R)}{R^2}Rd\phi=\frac{\mu_0 k}{2R}\mathbf{\hat{z}}B(0)=4πR2μ0​k​∫02π​R2ϕ^​×(z^R)​Rdϕ=2Rμ0​k​z^

where we have used the fact that $\mathbf{r}-\mathbf{r’}=R\mathbf{\hat{\phi}}$ and $|\mathbf{r}-\mathbf{r’}|=R$ at the center of the loop. Note that the magnetic field is directed in the $z$-direction.

The retarded vector potential $\mathbf{A}$ is related to the magnetic field by:

\mathbf{B}=\nabla\times\mathbf{A}B=∇×A

Since the magnetic field has only a $z$-component, the vector potential must have only an $x$- and $y$-component. Using the expression for the magnetic field, we can solve for the $x$- and $y$-components of the vector potential by taking the curl:

\nabla\times\mathbf{B}=\nabla\times(\nabla\times\mathbf{A})=\nabla(\nabla\cdot\mathbf{A})-\nabla^2\mathbf{A}=\mu_0\frac{\partial\mathbf{B}}{\partial t}=0∇×B=∇×(∇×A)=∇(∇⋅A)−∇2A=μ0​∂t∂B​=0

Since $\nabla\cdot\mathbf{A}=0$ (the vector potential is divergence-free), we have:

\nabla^2\mathbf{A}=-\mu_0\frac{\partial\mathbf{B}}{\partial t}=0∇2A=−μ0​∂t∂B​=0

Thus, the vector potential must satisfy Laplace’s equation. Since the magnetic field is only in the $z$-direction, we can choose the vector potential to have the form:

\mathbf{A}(x,y,0)=\frac{\mu_0 k}{4\pi}\frac{1}{\sqrt{x^2+y^2}}(-y\mathbf{\hat{x}}+x\mathbf{\hat{y}})A(x,y,0)=4πμ0​k​x2+y2​1​(−yx^+xy^​)

where we have used the fact that the current is increasing linearly with time to justify the use of the retarded time

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

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