레이블이 전자기학인 게시물을 표시합니다. 모든 게시물 표시
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2018년 4월 15일 일요일

[플라즈마 물리][Plasma Physics]CH1 Introduction - Temperature

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[Youtube Link]

Temperature

The one-dimensional Maxwellian distribution is given by

\[\begin{aligned} f(v)=A e^{-\frac{mv^2}{2K_BT}}\end{aligned}\]

Unlike normal distribution gaussian equation can have a form of

\[\begin{aligned} \label{eq_4} f(x) = \frac{n}{\sqrt{2\pi}\sigma}e^{-\frac{(v-v_{\mu})^2}{2\sigma^2}}\end{aligned}\]

Where \(n\) is the number density. \(fdv\) is the number of particles per [\(m^3\)] with velocity between \(v\) and \(v+dv\), \(\frac{1}{2}mv^2\) is the kinetic energy, and \(K_B\) is the Boltzmann’s constant. The density \(n\), or number of particles per [\(m^3\)], is given by

\[\begin{aligned} n=\int_{-\infty}^{\infty}f(v)dv\end{aligned}\]

so that the constant \(A\) is found to be

\[\begin{aligned} A=n\sqrt{\frac{m}{2\pi K_BT}}\end{aligned}\]

Where \[\begin{aligned} \int_{-\infty}^{\infty}e^{-ax^2}dx = \sqrt{\frac{\pi}{a}} \end{aligned}\] is used.

\(\bullet\) meaning of T = Distribution of the particles

The width of the distribution is characterized by the constant \(T\). By computing the average kinetic energy of particles in the distribution, we can see the exact meaning of \(T\).
\[\begin{aligned} \label{eq_5} E_{av}=\frac{\int_{-\infty}^{\infty}\frac{1}{2}mu^2 f(u) du}{\int_{-\infty}^{\infty} f(u) du}\end{aligned}\]

Defining \(v_{th}=\sqrt{\frac{2K_BT}{m}}\) and \(y=\frac{u}{v_{th}}\), 1-D Maxwellian distribution can be written as

\[\begin{aligned} f(u)=Ae^{-\frac{u^2}{v_{th}^2}}\end{aligned}\]

By substitution average kinetic energy becomes

\[\begin{aligned} E_{av}&=\frac{\frac{1}{2}mAv_{th}^3 \int_{-\infty}^{\infty} e^{-y^2}y^2 dy}{Av_{th}\int_{-\infty}^{\infty} e^{-y^2}dy}\\ &=\frac{\frac{1}{2}mAv_{th}^3 \frac{1}{2}}{A v_{th}}=\frac{1}{4}mv_{th}^2=\frac{1}{2}K_BT\end{aligned}\]

Thus the average kinetic energy is \(\frac{1}{2}K_BT\).
In three dimensions,

\[\begin{aligned} f(u,v,w)=n\left( \frac{m}{2\pi K_BT } \right)^{\frac{3}{2}}e^{-\frac{\frac{1}{2}m\left( u^2 + v^2 + w^2 \right)}{K_BT}}\end{aligned}\]

Using similar calculation we get

\[\begin{aligned} E_{av}=\frac{3}{2}KT\end{aligned}\]

The general result is that \(E_{av}\) equals \(\frac{1}{2}K_BT\) per degree of freedom.

\(\bullet\) Since \(T\) and \(E_{av}\) are so closely related, it is customary in plasma physics to give temperatures in units of energy.

\(\bullet\) To avoid confusion, it is not \(E_{av}\) but the energy corresponding to \(KT\) that is used to denote the temperature.

\(\bullet\) For \(KT=1eV= 1.6\times 10^{-19}[J]\)

\[\begin{aligned} T=\frac{1.6 \times 10^{-19}}{1.38 \times 10^{-23}}=11600\end{aligned}\]

Thus the conversion factor is

\[\begin{aligned} 1eV = 11,600 ^{\circ}K \end{aligned}\]

\(\bullet\) By a \(2eV\) plasma we mean that \(KT=2eV\), or \(E_{av}=3eV\) in three dimensions.

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2018년 4월 14일 토요일

[플라즈마 물리][Plasma Physics]CH4 Waves In Plasma - General ellipse equation angle derivation (tilted ellipse)

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Problem
Derive equation below given from the lecture note.

\[\begin{aligned} \tan(2\theta) = \frac{2E_{0x}E_{0y}}{E_{0x}^2 - E_{oy}^2}\cos \delta\end{aligned}\]

Answer
Figure below shows the equation of an ellipse tilted at an angle of \(\theta\) to the \(E_x\) axis.

Consider the transformation between the primed and unprimed coordinates.

\[\begin{aligned} E_x^{\prime} = E_x \cos \theta - E_y \sin \theta\\ E_y^{\prime} = E_x \sin \theta + E_y \cos \theta\end{aligned}\]

\[\begin{aligned} \left( \frac{E_x^{\prime}}{E_{0x}} \right)^2 + \left( \frac{E_y^{\prime}}{E_{0y}} \right)^2 - 2 \left( \frac{E_x^{\prime}}{E_{0x}} \right) \left( \frac{E_y^{\prime}}{E_{0y}} \right) \cos \delta = \sin^2 \delta\end{aligned}\]

\[E_{oy}^2 E_x^{\prime} + E_{0x}^2 E_y^{\prime} - 2 E_{0x}E_{oy} E_{x}^{\prime}E_y^{\prime}\cos \delta = \sin^2 \delta E_{ox}^2 E_{oy}^2\]

By sbustitution and rearranging it,

\[\begin{aligned} &E_x^2 [ E_oy^2 \cos^2 \theta + E_{0x}^2 \sin^2 \theta - 2 \cos \theta \sin \theta E_{0x} E_{0y} \cos \delta ] +\\ &E_y^2 [ E_oy^2 \sin^2 \theta + E_{0x}^2 \cos^2 \theta + 2 \cos \theta \sin \theta E_{0x} E_{0y} \cos \delta ] \\ &E_x E_y [ \boxed{-2 \cos\theta \sin \theta E_{0y}^2 + 2 \cos\theta \sin\theta E_{0x}^2 - 2 (\cos^2 \theta - \sin^2\theta) E_{0x}E_{0y}\cos \delta}] \\ &= E_{0x}^2 E_{0y}^2 \sin^2 \delta\end{aligned}\]

\(E_xE_y\) term must be zero in \(x^{\prime}\) and\(y^{\prime}\)coordinate system.

\[\begin{aligned} \sin(2\theta) (E_{0x}^2 - E_{0y}^2 ) = 2 \cos(2\theta) E_{0x} E_{0y} \cos \delta\end{aligned}\]

\[\begin{aligned} \tan(2\theta) = \frac{2E_{0x}E_{0y}}{E_{0x}^2 - E_{oy}^2}\cos \delta\end{aligned}\]

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[플라즈마 물리][Plasma Physics]CH4 Waves In Plasma - General ellipse equation derivation

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Problem
Derive general elliptical equation for an arbitrary phase difference. Consider the case \(E_{ox} \neq E_{oy}\).

\[\begin{aligned} \left( \frac{E_x}{E_{0x}} \right)^2 + \left( \frac{E_y}{E_{0y}} \right)^2 - 2 \left( \frac{E_x}{E_{0x}} \right) \left( \frac{E_y}{E_{0y}} \right) \cos \delta = \sin^2 \delta\end{aligned}\]

Answer
\[\begin{aligned} E_x(z,t) = E_{0x}\cos(\tau + \delta_x ) \\ E_y(z,t) = E_{0y}\cos(\tau + \delta_y ) \end{aligned}\]

\[\begin{aligned} \frac{E_x(z,t)}{E_{0x}} = \cos( \tau) \cos (\delta_x) - \sin(\tau) \sin (\delta x) \\ \frac{E_y(z,t)}{E_{0y}} = \cos( \tau) \cos (\delta_y) - \sin(\tau) \sin (\delta y) \end{aligned}\]

\[\begin{aligned} \frac{E_x}{E_{0x}}\sin(\delta y) - \frac{E_y}{E_{0y}}\sin (\delta_x) = \cos(\tau) \sin(\delta y -\delta x)\\ \frac{E_x}{E_{0x}}\cos(\delta y) - \frac{E_y}{E_{0y}}\cos (\delta_x) = \sin(\tau) \sin(\delta y -\delta x)\end{aligned}\]

square above two equatiion and add togather leads to the general equation of ellipse.

\[\begin{aligned} \left( \frac{E_x}{E_{0x}} \right)^2 + \left( \frac{E_y}{E_{0y}} \right)^2 - 2 \left( \frac{E_x}{E_{0x}} \right) \left( \frac{E_y}{E_{0y}} \right) \cos \delta = \sin^2 \delta\end{aligned}\]

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2018년 4월 12일 목요일

[플라즈마 물리][Plasma Physics]사하 공식 Saha Equation

Saha Equation

\(\bullet\) We live in a small part of the universe where plasmas do not occur naturally; otherwise we would not be alive. The reason for this can be seen from the Saha equation, which tells us the amount of ionization to be expected in a gas in thermal equilibrium. \[\begin{aligned} \label{eq_1} \frac{n_i}{n_n} \simeq 2.4 \times 10^{21} \frac{T^{\frac{3}{2}}}{n_i}e^{\frac{-U_i}{KT}} \end{aligned}\]

where \(n_i\) and \(n_n\) are density(number per \(m^3\)) of ionized atoms and of neutral atoms, respectively. \(T\) is the gas temperature in \(^{\circ}K\), \(K\) is Boltzmann’s constant; \(1.38 \times 10^{-23} [\frac{J}{^{\circ}K}]\), and \(U_i\) is the ionization energy of the gas - that is, the number of joules required to remove the outermost electron from an atom.
In room temperature, \(n_n \simeq 3 \times 10^{25}[m^{-3}]\), \(T \simeq 300 ^{\circ}K\), and \(U_i = 14.5eV\) for nitrogen, where \(1eV = 1.6 \times 10^{-10}[J]\). The fractional ionization \(\frac{n_i}{n_n+n_i}\simeq \frac{n_i}{n_n}\) is rediculously low:

\[\begin{aligned} \frac{n_i}{n_n}\simeq 10^{-122}\end{aligned}\]

Let us define degree of ionization \[\alpha = \frac{n_e}{n_e+n_n}\]. Where \(n_e=n_i\) is the number of electrons or ions per volume [\(cm^3\)], and \(n_n\) is the number of neutrals per volume [\(cm^3\)].

  • \(\alpha << 1\): weakly ionized situation (low temp)

  • \(\alpha = 1\): fully ionized (high temp)

Trend of Saha equation tells us how fractional ionization changes depending on the temperature. As the temperature increases, the ionization increases significantly.

  • \(\frac{n_e}{n_n}\rightarrow 0\) at \(K_BT << U_i\)

  • \(\frac{n_e}{n_n}>> 0\): as \(K_BT >> U_i\)

where again, \(U_i\) is the ionization energy and \(K_BT: T_e \simeq T_i\)

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[전자공학][정리중][Modifying]Capacitor and Inductor

Overview

Passive and Active Elements

On electronic circuits, they contain many components such as transistors or operational amplifiers. However, these components can be classified in to either an active element or a passive element. Passive elements are elements which cannot supply energy by themselves, whereas a battery is an active element which generate energy. Resistors, capacitors and inductors are passive elements which we are going to discuss in this chapter.

Before we begin our discussion, let’s see <Table 1> and briefly think about how Resistors, capacitors and inductors work.

Relation Resistor Capacitor Inductor
Schematic image image image
Voltage \(v_R(t)=i_R(t)R\) \(\frac{1}{C}\int_{t_0}^{t}idt+v(t_0)\) \(v_L(t)= \frac{di_l(L)}{dt}\)
Current \(i_R(t)=\frac{v_R(t)}{R}\) \(i=c\frac{dv_c(t)}{dt}\) \(\frac{1}{L}\int_{t_0}^{t}vdt+i(t_0)\)
Power or Energy \(p=i^2R=\frac{v^2}{R}\) \(w=\frac{1}{2}cv^2\) \(w=\frac{1}{2}li^2\)
Series \(R_{eq}=R_1+R_2+... R_n\) \(C_{eq}=\frac{C_1C_2}{C_1+C_2}\) \(L_{eq}=L_1+L_2+... L_n\)
Parallel \(R_{eq}=\frac{R_1R_2}{R_1+R_2}\) \(C_{eq}=C_1+C_2+... C_n\) \(L_{eq}=\frac{L_1L_2}{L_1+L_2}\)
At DC Voltage Same Open Circuit Short Circuit

Capacitors

Basic Background

Capacitor is composed of two conducting plates separated by a dielectric. When voltage is applied between the plates, +q charges accumulate in one plate while -q charge on the other. Capacitance, which is ability to store an electrical charge, depends on the geometry of the conductors and permittivity of the insulating medium in between the plates. Capacitance of ideal parallel plate capacitor is defined by following equation. \[C=\frac{\epsilon A}{d}\]

Where plate area \(A\) separated by distance \(d\) with a permittivity of \(\epsilon\).

Terminal Characteristics

The relation between charge and capacitance is \[q=Cv\] By looking at this equation, we can know that the plate charge increases as the voltage increases. when we take a derivative for both sides, we can get \[i=\frac{Cdv}{dt}\] where \(i=\frac{dq}{dt}\)

when \(v\) is a constant voltage(which is a DC voltage), then \(i=0\). This means when DC voltage is applied to capacitor, it looks as if it is an open circuit. Meanwhile, abrupt change in it voltage \(v\) is not allowable.

If we integrate equation over time, \[\int_{-\infty}^{t}idt=\int_{-\infty}^{t}C\frac{dv}{dt}dt\] \[v=\frac{1}{C}\int_{-infty}^{t}idt\] \[v=\frac{1}{C}\int_{0}^{t}idt + v(0)\]

\(v(0)\) repr

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[전자기학][Fundamentals of Engineering Electromagnetics]Electrostatic potential energy, Example 3-17 Cheng (P122)


[Fundamentals of Engineering Electromagnetics]
Electrostatic potential energy, Example 3-17 Cheng (P122)


전자공학 학생이 아래 질문을 가지고 질문한 사항에 대해 답변을 하고자 작성하였습니다. 틀렸으면 알려주세요. 제 자신도 공부하기 위함입니다. (I just upload the image from Cheng p122; if there is problem with the copyright please let me know. I am trying to discuss the question that student has)




Problem: Find the energy required to assemble a uniform sphere of charge of radius \(b\) and volume charge density \(\rho_v\)

In terms of total charge

\[\begin{aligned} Q= \frac{4\pi}{3}\rho_v b^3 \end{aligned}\]

Hence we have

\[\begin{aligned} W_e = \frac{3Q^2}{20 \pi \epsilon_0 b}\end{aligned}\]

By looking at the equation below, a lot of people asks why does the half goes away?

\[\begin{aligned} W&= \frac{1}{2}\int_v \rho_v V dv\end{aligned}\]

And the answer is it is not actually going away. Those who might have twice differece might solved problem as below ( used \(V =\frac{q}{4\pi \epsilon_0 r}\))

\[\begin{aligned} W&= \frac{1}{2}\int_v \rho_v V dv \\ &= \frac{1}{2}\int_v \frac{\rho_v^2 \frac{4}{3} \pi r^2}{4 \pi \epsilon_0} r^2 \sin \theta dr d\theta d\phi \\ &=\frac{2}{15}\frac{\rho_v^2 r^5}{\epsilon_0}\\ &=\frac{3}{40}\frac{Q^2}{\epsilon_0 r \pi}\end{aligned}\]

Notice that \(Q = \rho_v \frac{4}{3}\pi r^3\). So there is twice difference. The reason for this is that potential is different inside and outside the effective sphere.
Answer: Think we have sphere with radius R; charge exist in this region. For \(r<R\)

\[\begin{aligned} \oint_s \vec{E} \cdot d\vec{a} &= \frac{\rho}{\epsilon_0}\frac{4}{3}\pi r^3 \\ \vec{E} &= \frac{\rho}{3\epsilon_0} r \\ &=\frac{\rho r}{4\pi \epsilon_0 R^3}\end{aligned}\]

On the last step, we used the fact that charge is inside the \(\frac{4}{3}\pi R^3\). For \(r>R\),

\[\begin{aligned} \vec{E} = \frac{Q}{4\pi \epsilon_0 r^2}\end{aligned}\]

Finally \(E\) field and potential becomes

\[\begin{aligned} \vec{E}(r) = \frac{Q}{4\pi \epsilon_0} \times \begin{cases} \frac{r}{R^3} \quad (r<R) \\ \frac{1}{r^2} \quad (r>R) \end{cases}\end{aligned}\]

\[\begin{aligned} \vec{V}(r) = \frac{Q}{4\pi \epsilon_0} \times \begin{cases} -\frac{1}{2} \frac{r^2}{R^3} + \frac{3}{2R} \\ \frac{1}{r} \end{cases}\end{aligned}\]

Notice that constant term from the potential comes from the Boundary Condition; \(at r=R\), \(V = \frac{Q}{4\pi \epsilon_0 R}\). Finally, let us calculate the energy.

\[\begin{aligned} W &= \frac{1}{2} \int_{0}^{R} \frac{Q}{4\pi \epsilon_0} \rho \left( -\frac{r^2}{2R^3} + \frac{3}{2R} \right) 4\pi r^2 dr \\ &=\frac{1}{2}\int_{0}^{R} \frac{Q \rho}{\epsilon_0}\left( -\frac{r^4}{2R^3} + \frac{3r^2}{2R} \right) dr \\ &= \frac{4}{20}\frac{Q \rho R^2}{\epsilon_0}\\ &= \frac{3}{20}\frac{Q^2}{\pi \epsilon_0 R}\end{aligned}\]

Hence, we get the same answer as David K. Cheng.

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2018년 4월 11일 수요일

[플라즈마 물리][Plasma Physics]Debye Shielding 디바이 차폐

Example Problem: Debye Shielding
Consider a positive point charge immersed in a plasma as we discussed in the class. Show that the net charge in the Debye shielding cloud exactly cancels the test charge. Assume that the ions are fixed and that \(e\phi << KT_e\). Note that we assumed that ion distribution is similar to that of the electrons in the class.

Answer:

While deriving, we assume potential to be

\[\begin{aligned} \frac{e\phi}{KT_e} << 1 \quad \quad \quad n_i \simeq n_e\\ \lambda_D = \sqrt{\frac{\epsilon KT}{ne^2}}\\ \phi(r) = \frac{e}{4\pi\epsilon_0 r }e^{-\frac{r}{\lambda_D}}\end{aligned}\]

Assume \(\frac{m_i}{m_e} \rightarrow \infty\) so that ions are fixed; \(n_i(r)=n\). Further assume electrons obey the Boltzmann distribution.

\[\begin{aligned} f(u) &= A e^{\frac{\frac{-1}{2}mv^2 + q\phi}{KT_e}}\\ n_e(r) &= ne^{\frac{e\phi(r)}{K_B T}}\end{aligned}\]

\[\begin{aligned} \rho &= -e(n_i - n_e) \\ &=-e \left(n - ne^{\frac{e\phi(r)}{K_BT}} \right)\\ &= ne \left(e^{\frac{e\phi(r)}{K_BT}} - 1 \right)\\\end{aligned}\]

for \(\frac{e\phi}{KT_e} << 1\),

\[\begin{aligned} e^{\frac{e\phi(r)}{K_B T}} = 1 + \frac{1}{2}\frac{e\phi(r)}{K_B T} + ...\end{aligned}\]

\[\begin{aligned} \rho &= ne(1+\frac{e\phi(r)}{K_B T} - 1) \\ &=\frac{ne^2 \phi(r)}{K_B T} \\ &= \frac{\epsilon_0}{\lambda_D^2}\phi(r)\end{aligned}\]

\[\begin{aligned} \int \rho dv &= Q \\ &= \int_V \frac{\epsilon_0}{\lambda_D^2}\frac{e}{4\pi\epsilon_0 r }e^{-\frac{r}{\lambda_D}} r^2 \sin\theta dr d\theta d\phi\\ &= \int \frac{4\pi \epsilon_0}{\lambda_D^2}\frac{e}{4\pi \epsilon_0} r e^{-\frac{r}{\lambda_D}}dr \\ &= \frac{e}{\lambda_D^2} \int_{0}^{\infty} r e^{-\frac{r}{\lambda_D}}dr \\ &= \frac{e}{\lambda_D^2} (\lambda_D^2 e^{-\frac{r}{\lambda_D}}-\lambda_D r e^{-\frac{r}{\lambda_D}}) =0\end{aligned}\]

On the last step, when \(r = \lambda_D\), charge becomes zero. Hence, the Debye sheilding cloud exactly cancels the test charge.

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2018년 4월 10일 화요일

[플라즈마 물리]F.F. Chen Problem 1.5 Solution Debye Shielding

Problem In a strictly steady state situation both the ions and the electrons will follow the Boltzmann relation

\[n_i = n_0 e^{\frac{-q_i \phi }{KT_i}}\]

For the case of an infinite, transparent grid charged to a potential \(\phi\), show that the shielding distance is then given approximately by

\[\lambda_D^{-2} = \frac{ne^2}{\epsilon_0} \left( \frac{1}{KT_e} + \frac{1}{KT_i} \right)\]

Answer

\[\begin{aligned} \nabla \cdot \vec{E} &=\frac{\rho}{\epsilon_0} = \frac{e}{\epsilon_0}(n_i - n_e) \nonumber \\ &= - \frac{e}{\epsilon_0} n_0 \left( e^{-\frac{e\phi}{KT_i}} - e^{\frac{e\phi}{KT_e}} \right) \nonumber \\ &= - \frac{e}{\epsilon_0} n_0 \left( 1 -\frac{e\phi}{KT_i} - 1 - \frac{e\phi}{KT_e} \right) \nonumber \\ \frac{d^2 \phi }{dx^2 }&=\frac{e^2 n_0}{\epsilon_0}\left( \frac{1}{KT_i} + \frac{1}{KT_e} \right) \phi \nonumber\end{aligned}\]

Hence,

\[\begin{aligned} \frac{1}{\lambda_D^2} = \frac{e^2 n_0}{\epsilon_0}\left( \frac{1}{KT_i} + \frac{1}{KT_e} \right) \nonumber\end{aligned}\]

\[\lambda_D = \sqrt{\left( \frac{\epsilon_0 K T_e T_i}{ne^2 (T_i + T_e)} \right)}\]

  • \(T_i << T_e \quad \quad \lambda_D \simeq \sqrt{\left( \frac{\epsilon_0 K T_i}{ne^2} \right)}\)

  • \(T_e << T_i \quad \quad \lambda_D \simeq \sqrt{\left( \frac{\epsilon_0 K T_e}{ne^2} \right)}\)

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2018년 4월 9일 월요일

[플라즈마물리]Single Particle Motion - E cross B drift

Problem Consider a particle of charge \(q\) and mass \(m\), initially at rest at (0,0,0), in the presence of a static magntic field \(\vec{B}=B\hat{z}\) and \(\vec{E}=E\hat{y}\).

(a) Taking \(E, B > 0\), sketch the orbit of the particle when \(q>0\).

(b) Derive an exact expression for the orbit \([x(t),y(t),z(t)]\) or the particle. Express your answer in terms of \(E\), \(B\), and \(\omega_c\)

(c) Find the drift velocity after averaging the motion in time. If there were many particles of various charges and masses present, would there be any net current?

(d) Suppose the electric field were replaced by a force \(F\) in the \(y\)-direction. What would be the drift velocity?

Answer (a)

Answer (b)

\(\vec{B}=B\hat{z}\) and \(\vec{E}=E\hat{y}\). From Lorentz equation,

\[m\frac{dv}{dt}=q(\vec{E} + \vec{v}\times \vec{B})\]

\[\begin{aligned} \dot{v_x}&= \quad \quad \quad \quad \frac{q}{m}v_y B \nonumber \\ \dot{v_y}&= \frac{q}{m}\vec{E_y} - \frac{q}{m}v_xB \nonumber \\ \dot{v_z}&= 0\end{aligned}\]

In z-direction \[z(t) = constant = 0\]

\[\begin{aligned} \frac{dv_x}{dt} &= \frac{qB}{m}v_y = \omega_c v_y \nonumber \\ \frac{dv_y}{dt} &= \frac{qE}{m}-\frac{qB}{m}v_x = \frac{qE_y}{m}-\omega_c v_x \nonumber\end{aligned}\]

\[\begin{aligned} \frac{d^2 v_y}{dt^2} &= -\omega_c^2 v_y \nonumber \\ \frac{d^2 v_x}{dt^2} &= -\omega_c^2 \left(v_x-\frac{E}{B} \right) \nonumber\end{aligned}\]

since \(\frac{E}{B}\) is constant, \[\frac{d^2}{dt^2} \left[ v_x-\frac{E}{B} \right] = \frac{d^2 v_x}{dt^2} = -\omega_c^2 \left(v_x-\frac{E}{B} \right)\]

\[\begin{aligned} v_x &= iv_{\perp}e^{i\omega_c t} + \frac{E}{B}\nonumber \\ v_y &= v_{\perp}e^{i\omega_c t} \nonumber \end{aligned}\]

\[\begin{aligned} x(t) &= x(0) + r_L \cos{\omega_c t} +\frac{E}{B}t \nonumber \\ y(t) &= y(0) + r_L \sin{\omega_c t} \nonumber \end{aligned}\]

Answer (c)

\[<v_d> = \frac{\int_{-\infty}^{\infty}v_d f(v) dv}{\int_{-\infty}^{\infty}f(v) dv} = \frac{E}{B}\]

There will be no current because \(\vec{E} \times \vec{B}\) drifts the ions and electrons in the same direction.

Answer (d)

\[\begin{aligned} \vec{v_f}&= \frac{1}{q} \frac{\vec{F} \times \vec{B}}{B^2} \nonumber \\ &=\frac{FE}{qB}\hat{y}\end{aligned}\]

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2018년 4월 8일 일요일

[대학원-전자기학]Vector Calculus Problem 2

Problem 2: Prove \(-\nabla \frac{1}{|\vec{r}-\vec{r}\prime|}=\frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}\)

Let \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\]

\[\nabla \frac{1}{|\vec{r}-\vec{r}\prime|} = \frac{\partial}{\partial x}\hat{x} + \frac{\partial}{\partial y}\hat{y} + \frac{\partial}{\partial z}\hat{z}\frac{1}{|\vec{r}-\vec{r}\prime|}\]

Let us consider only \(\hat{x}\) components. \[\begin{aligned} \frac{\partial}{\partial x}\hat{x}\frac{1}{\zeta^{\frac{1}{2}}} &= \frac{1}{2}\zeta^{-\frac{3}{2}}\zeta\prime \nonumber \\ &=\frac{x_1 - x_2}{\zeta^{\frac{3}{2}}}\hat{x} \nonumber \\ &=\frac{x_1 - x_2}{|\vec{r}-\vec{r}\prime|^3}\hat{x} \nonumber\end{aligned}\]

Same procedure occur for the other two component. When we add all, final result become \[-\nabla \frac{1}{|\vec{r}-\vec{r}\prime|}=\frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}\]

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[대학원-전자기학]Vector Calculus Problem 1

Problem 1: Prove \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)

Answer:
\[\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}= \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{x_1-x_2}{|\vec{\eta}^3|} & \frac{y_1-y_2}{|\vec{\eta}^3|} & \frac{z_1-z_2}{|\vec{\eta}^3|} \end{vmatrix}\] Where \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\vec{\eta} = \vec{r}-\vec{r}\prime\], \[|\vec{\eta}|=\sqrt{(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\] \[\begin{aligned} \nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3} &= (\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\eta}^3|}))\vec{x} \nonumber \\ & -(\frac{\partial}{\partial x}(\frac{z_1-z_2}{|\vec{\eta}^3|}) - \frac{\partial}{\partial z}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{y} \nonumber \\ & +(\frac{\partial}{\partial x}(\frac{y_1-y_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial y}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{z} \nonumber\end{aligned}\]

Since \[\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(z_1 - z_2)(y_1 - y_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]

\[\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(y_1 - y_2)(z_1 - z_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]

They cancel out each others. Same procedure occur for the other two components. Hence, \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)

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2018년 4월 6일 금요일

[플라즈마 물리]Normalization Constant & Average Kinetic Energy

Problem The 3D Maxwellian distribution is given by
\[f(\vec{v}) = A_3 e^{\left( -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} \right)}\]
with \[n = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\vec{v})dv_x dv_y dv_z\]
(a) Show that the normalization constant is given by \[A_3 = n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}}\]
(b) Show that the average kinetic energy is \[E_{av} = \frac{3}{2}KT\]
Answer (a)

Notice that
\[\begin{aligned} \label{eq_gauss} \int_{-\infty}^{\infty} e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}\end{aligned}\]
\[\begin{aligned} n &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\vec{v})dv_x dv_y dv_z \nonumber \\ &= A_3 \int_{-\infty}^{\infty} e^{ -\frac{m(v_x^2)}{2KT}}dv_x \int_{-\infty}^{\infty} e^{ -\frac{m(v_y^2)}{2KT}}dv_y \int_{-\infty}^{\infty} e^{ -\frac{m(v_z^2)}{2KT}}dv_z \nonumber \\ &= A_3 \left( \sqrt{\frac{ 2 K T \pi}{m}} \right)^{3}\end{aligned}\]
Hence,
\[A_3 = n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}}\]
Answer (b)

Average energy can be calculated as
\[\begin{aligned} E_{av} &= \frac{\int_{-\infty}^{\infty} \frac{1}{2}mv^2 f(\vec{v})dv}{\int_{-\infty}^{\infty}f(\vec{v})dv} \nonumber \\ &=\frac{m}{2n}\int_{-\infty}^{\infty} (v_x^2 + v_y^2 + v_z^2) A_3 e^{ -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} }dv\end{aligned}\]
The denominator is \(n\) as we calculated in previous problem. Notice that, \[\int x^2 e^{-ax^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}\]
If we calculate the first term of equation,
\[\begin{aligned} E_{av_x} &= \frac{m}{2n} \int_{-\infty}^{\infty} A_3(v_x^2 ) e^{ -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} }dv_x \nonumber \\ &=\frac{m}{2n}n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}} \frac{1}{2} \left( \frac{2 KT}{m} \right)^{\frac{3}{2}}\sqrt{\pi}\frac{2KT \pi}{m} \nonumber \\ &=\frac{1}{2}KT\end{aligned}\]
Hence, if we calculate \(v_y\) and \(v_z\) components, final result becomes \[E_{av} = \frac{3}{2}KT\]
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2018년 4월 5일 목요일

[대학원-전자기학]문제풀이 델타함수와 발산

Problem Prove that \[\nabla^2 (\frac{1}{r}) = -4\pi \delta(\vec{x})\] where \(r=|\vec{x}|\)
Answer

\[\begin{aligned} \nabla^2 \frac{1}{r} = \nabla \cdot (\nabla \frac{1}{r}) = \nabla \cdot (-\frac{1}{r^2}\hat{r}) = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \left( \frac{-1}{r^2}\right) \right) \nonumber\end{aligned}\]
Hence, the intergral of \(\nabla^2 \frac{1}{r} = 0\)
\[\int_v \nabla^2 \frac{1}{r} dv = 0\]
However, when we use the divergence theorem, we get different result.
\[\begin{aligned} \int_v \nabla^2 \frac{1}{r} dv &= \oint_s -\frac{1}{r^2}\hat{r} d\vec{a} \nonumber \\ &=\oint_s - \frac{1}{r^2} r^2\sin\theta d\theta d\phi \nonumber \\ &= -4\pi \nonumber\end{aligned}\]
This problem arise in the origin. Hence, we introduce delta function to solve this problem.
\[\nabla^2 (\frac{1}{r}) = -4\pi \delta(\vec{x})\]
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2018년 4월 4일 수요일

[대학원-전자기학]문제풀이

Problem Prove the symmetry property of the Green’s function satisfying the Dirichlet boundary condition \[G(\vec{x},\vec{x}^{\prime}) = G(\vec{x}^{\prime}, \vec{x})\]

Refer to Jackson Problem (1.14)

Answer

Green function \(G(\vec{x},\vec{x}^{\prime})\) satisfies Dirichlet boundary conditions; bound region \(\Omega\) with boundary \(d\Omega\), \(G(\vec{x},\vec{x}^{\prime})=0 \quad \forall x^{\prime} \in \partial \Omega\)

\[\int_V (\phi \nabla^2 \psi - \psi \nabla^2 \phi)d^3x = \oint_s \left[ \phi \frac{\partial \psi}{\partial n} - \psi \frac{\partial \phi }{\partial n}\right]da\]

Substitute \(\phi = G(\vec{x},\vec{y})\) and \(\psi = G(\vec{x}^{\prime},\vec{y})\).

\[\begin{aligned} \int_{\Omega} \left[ G(\vec{x},\vec{y})\nabla^2 G(\vec{x}^{\prime},\vec{y}) - G(\vec{x}^{\prime},\vec{y}) \nabla^2 G(\vec{x},\vec{y}) \right]d^3y =\nonumber \\ \int_{\partial \Omega}\left[ G(\vec{x}, \vec{y})\frac{\partial }{\partial n}G(\vec{x}^{\prime},\vec{y}) - G(\vec{x}^{\prime},\vec{y}) \frac{\partial}{\partial n} G(\vec{x},\vec{y}) \right]da \nonumber\end{aligned}\]

\(G(\vec{x}, \vec{x}^{\prime})\) satisfies Dirichlet B.C. Hence \(RHS=0\); \(\Phi\) is known on the surface and \(F\) can be chosen to make \(G_D(\vec{x},\vec{x}^{\prime}=0)\) Since \[\nabla^2 G(\vec{x},\vec{y}) = -4\pi \delta^3(\vec{x}-\vec{y})\]

\[\begin{aligned} \int_{\Omega} \left[ G(\vec{x},\vec{y})\nabla^2 G(\vec{x}^{\prime},\vec{y}) - G(\vec{x}^{\prime},\vec{y}) \nabla^2 G(\vec{x},\vec{y}) \right]d^3y = \nonumber \\ \int_{\Omega}\left[ -4\pi G(\vec{x},\vec{y}) \delta^3(\vec{x}^{\prime} - \vec{y})+G(x^{\prime},\vec{y}) 4\pi \delta^3(\vec{x}-\vec{y}) \right]d^3y \nonumber \\\end{aligned}\]

Finally

\[G(\vec{x},\vec{x}^{\prime}) = G(\vec{x}^{\prime}, \vec{x})\]

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