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Условие Отговори Решение

USAPhO 1997, Полуфинал — Задача 5. B1

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Всички материали от USAPhO 1997

United States Physics Team — 1997 Exam 2

Указания

Entia non multiplicanda sunt praeter necessitatem

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Please show all work, and do not hesitate to use explanatory words in addition to equations. Partial credit will be awarded.
Begin each problem on a fresh sheet of paper. In the upper right-hand corner of each page, write your name, problem number, and the page number/total number of pages for this problem (e.g., "A2, 3/4" for "Problem A2, page 3 of four").
You may use a non-programmable calculator, but may not use any tables, books, notes, or collections of formulas.
Work Part A first. You have ninety minutes to complete all four problems.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.

Possibly useful information:

Gravitational field on Earth’s surface: $g = 9.8\ \mathrm{N/kg}$

Gravitational constant: $G = 6.67\times10^{-11}\ \mathrm{N\text{-}m^2 / kg^2}$

Coulomb constant: $k = 1 / 4\pi\epsilon_o = 8.99\times10^{9}\ \mathrm{N\text{-}m^2 / C^2}$

Speed of light in vacuum: $c = 3.00\times10^{8}\ \mathrm{m/s}$

Permeability of empty space: $\mu_o$ , where $\mu_o\epsilon_o = 1/c^2$

Ideal gas constant $R = 8.31\ \mathrm{J / mol\text{-}K}$

Boltzmann’s constant: $k_B = 1.38\times10^{-23}\ \mathrm{J/K}$

Avagadro’s number: $N_A = 6.02\times10^{23}$

1 electron volt = $1\ \mathrm{eV} = 1.60\times10^{-19}\ \mathrm{J}$

Planck’s constant: $h = 6.63\times10^{-34}\ \mathrm{J\text{-}s} = 4.14\times10^{-15}\ \mathrm{eV\text{-}s}$

$(1 - x)^{-1} = 1 + x + x^2 + x^3 + \ldots$ for $|x| < 1$ $\qquad$ $\cos\theta \approx 1 - \theta^2/2$ for $\theta \ll 1$

$(1 + x)^n = 1 + nx + (1/2!)n(n-1)x^2 + \ldots$ for $|x| < 1$ $\qquad$ $\sin\theta \approx \theta$ for $\theta \ll 1$

Условие

Consider an infinitely long, cylindrical cloud of electrons. The cloud has radius $R$ , the number density of electrons has the uniform value $n_o$ , the charge of each electron is $-e$ (where $e > 0$ ), and the mass of each electron is $m$ . The cloud is surrounded by vacuum. There is a constant, uniform magnetic field along the cylinder's axis (the $z$ -axis), so that $\mathbf{B} = B_o\mathbf{k}$ , with $B_o > 0$ . The cloud of electrons is rotating about the $z$ -axis with angular velocity $\omega$ as shown in the figure. The speeds are non-relativistic. [A gas of charged particles is called a "plasma."]

(a) Find the electric field at the radius $r < R$ within the cloud. [6 т.]

An upright cylinder of radius R (radius arrow drawn on the top face) with dotted continuations above and below; the vertical Z-axis with a curved arrow marking the rotation ω; n0 written inside the cylinder.

(b) Find the total force on an electron located at the radius $r$ within the cloud, neglecting any magnetic field induced by the cloud's rotation. (This neglect will be justified in part (f) below.) [6 т.]

(c) Using Newton's second law, show that there are two values of $\omega$ possible for the electrons to steadily orbit the cylinder's axis. Write these values of $\omega$ in terms of two other frequencies relevant to such problems: the cyclotron frequency $\omega_c = eB_o/m$ , and the "plasma frequency" $\omega_p = [\,n_oe^2 / \epsilon_o m\,]^{1/2}$ , where $\epsilon_o$ is the permittivity of empty space. [8 т.]

(d) Find the largest electron number density $(n_o)_{max}$ that can be confined by the applied magnetic field $B_o\mathbf{k}$ . Write your answer in terms of the magnetic field energy density $B^2/2\mu_o$ and the electron's mass energy $mc^2$ . [6 т.]

[This theoretical limit is called the Brillouin limit, and plays a key role in the confinement of plasmas consisting of particles all of like charge.]

(e) At a point located at radius $r$ within the cylindrical cloud, calculate the magnetic field induced by the cloud's rotation. Assume that the field induced on the axis of symmetry is zero. [8 т.]

(f) Determine the force on a charge in the cloud, due to the induced magnetic field of part (e). Show that the magnitude of this force divided by the magnitude of the electric force is $v^2/c^2$ , where $v$ is the speed of the charge and $c$ is the speed of light. [This shows that the effect of the induced field is ignorable when the rotation speeds are non-relativistic.] [6 т.]

Отговори

Решение

(a) Use Gauss' law:

\text{flux of } \vec{E} = \frac{1}{\epsilon_o}(\text{charge enclosed})

E\,2\pi rh = \frac{1}{\epsilon_o}(n_oe)(\pi r^2h)

E = \left(\frac{n_oe}{2\epsilon_o}\right)r

\vec{E} = \frac{n_oe}{2\epsilon_o}\,r\,(-\hat{r}) \qquad \text{radially } \underline{\text{in}}

[gaussian surface: cylinder, $r \le R$ , height $h$ ; no flux through end caps ( $\vec{E}\cdot\hat{n} = 0$ there)]

(b)

\begin{aligned} \vec{F} &= q(\vec{E} + \vec{v}\times\vec{B}) \\ &= -e\left[-\frac{n_oer}{2\epsilon_o}\hat{r} + r\omega B_o\hat{r}\right] \end{aligned}

[ $\vec{B} = B_o\hat{k}$ , $\vec{v} = r\omega\hat{\theta}$ , $\hat{\theta}\times\hat{k} = \hat{r}$ ]

\vec{F} = \left(\frac{n_oe^2r}{2\epsilon_o} - r\omega B_oe\right)\hat{r}

(c)

\vec{F} = m\vec{a}

\left(\frac{n_oe^2r}{2\epsilon_o} - r\omega B_oe\right)\hat{r} = m\frac{(r\omega)^2}{r}(-\hat{r})

\Rightarrow\ \frac{n_oe^2}{2\epsilon_o} - \omega B_oe + m\omega^2 = 0

\omega^2 - \left(\frac{Be}{m}\right)\omega + \frac{n_oe^2}{2\epsilon_om} = 0

\omega^2 - \omega_c\omega + \tfrac{1}{2}\omega_p^2 = 0

[ $\omega_c \equiv \dfrac{B_oe}{m}$ , $\omega_p^2 \equiv \dfrac{n_oe^2}{\epsilon_om}$ ]

\omega = \frac{\omega_c}{2} \pm \frac{1}{2}\sqrt{\omega_c^2 - 2\omega_p^2}

\omega = \frac{\omega_c}{2}\left(1 \pm \sqrt{1 - 2\left(\frac{\omega_p}{\omega_c}\right)^2}\right)

(d) Since $\omega$ is real, this requires

1 - 2\left(\frac{\omega_p}{\omega_c}\right)^2 \ge 0

\omega_p^2 \le \frac{1}{2}\omega_c^2

\frac{n_oe^2}{\epsilon_om} \le \frac{1}{2}\frac{e^2B_o^2}{m^2}

n_o \le \frac{B_o^2\epsilon_o}{2m} \qquad \text{so} \qquad (n_o)_{max} = \frac{B_o^2\epsilon_o}{2m}

(e)

(n_o)_{max} = \frac{B_o^2\epsilon_o}{2m} \qquad \text{use } \mu_o\epsilon_o = \frac{1}{c^2}

= \frac{B_o^2}{2m(\mu_oc^2)}

= \frac{B_o^2/2\mu_o}{mc^2} = \frac{\text{magnetic field energy density}}{\text{electron mass energy}}

(e) Apply Ampere's law,

\oint_C \vec{B}\cdot d\vec{l} = \mu_oI_S \qquad \text{where } C = \text{boundary of } S

B'h = \mu_o\int_S j\,da

where

\begin{aligned} j &= \text{current density} \\ &= (n_oe)v \\ &= n_oer'\omega \end{aligned}

(or $\Sigma\, r\Delta r$ )

B'h = \mu_o(n_oe\omega)\int_0^r r'(dr'\,h)

B'h = \mu_on_oe\omega h\int_0^r r'\,dr'

B' = \mu_on_oe\omega\frac{r^2}{2}

[ $\vec{B}'$ = induced field $= B'\hat{k}$ ; given $\vec{B}' = 0$ on axis; note $\vec{B}'\cdot d\vec{l} = 0$ horizontally]

(f) Force due to induced magnetic field is $\vec{F}'$ where

\begin{aligned} \vec{F}' &= -e\,\vec{v}\times\vec{B}' = -e(r\omega\hat{\theta})\times\left(\mu_on_oe\omega\frac{r^2}{2}\hat{k}\right) \\ &= \frac{\mu_on_oe^2\omega^2r^3}{2}(-\hat{r}) \end{aligned}

[ $\hat{\theta}\times\hat{k} = \hat{r}$ ; $v = r\omega$ ]

|\vec{F}'| = \tfrac{1}{2}\mu_on_oe^2v^2r

\therefore\ \frac{|\vec{F}'|}{|\vec{F}_{el.}|} = \frac{\tfrac{1}{2}\mu_on_oe^2v^2r}{\tfrac{1}{2}\dfrac{n_oe^2r}{\epsilon_o}} = \mu_o\epsilon_ov^2 = \frac{v^2}{c^2}

(the denominator is $|e\vec{E}|$ where $|\vec{E}|$ is from part (a))

Hand sketch for (a): cross-section of the cloud, a circle of radius R with a dashed concentric Gaussian circle of radius r; notes: gaussian surface: cylinder, r ≤ R, height h; no flux through end caps (E·n = 0 there).

Hand sketch for (e): the rotating cylinder (rotation ω about the axis) with a rectangular Amperian loop C bounding surface S in a plane through the axis, height h, from radius r to outside; a hatched strip da = h dr at radius r' of width dr; current density j circulating; induced field B' upward; notes: B' = induced field = B' k; given B' = 0 on axis; note B'·dl = 0 horizontally.

Оригинал в Архива: Semifinal 1997.pdf · официални решения: Semifinal 1997 Solutions.pdf