Съдържание

Условие a. Magnetic Moments. (20 т.)b. Diamagnetism. (30 т.)Отговори Решение

USAPhO 2007, Полуфинал — Задача 6. B2

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Всички материали от USAPhO 2007

AAPT AIP United States Physics Team 2007 — 2007 Semi-Final Exam · 50 т.

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 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.
Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie

A1 – 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s, or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Questions with the same point value are not necessarily of the same difficulty.
Do not discuss the contents of this exam with anyone until after March 27th.
Good luck!

Possibly Useful Information - (Use for both part A and for part B)

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Coulomb’s constant	$k = 1/4\pi\varepsilon_o = 8.99 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}$
Biot-Savart constant	$k_m = \mu_o/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}$
Speed of light in a vacuum	$c = 3.00 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_B = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_A = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_A k_B = 8.31\ \mathrm{J/(mol\cdot K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
Elementary charge	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV\cdot s}$
Electron mass	$m = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ ; $\cos\theta \approx 1 - \frac{1}{2}\theta^2$

Условие

A model of the magnetic properties of materials is based upon small magnetic moments generated by each atom in the material. One source of this magnetic moment is the magnetic field generated by the electron in its orbit around the nucleus. For simplicity, we will assume that each atom consists of a single electron of charge $–e$ and mass $m_e$ , a single proton of charge $+e$ and mass $m_p \gg m_e$ , and that the electron orbits in a circular orbit of radius $R$ about the proton.

a. Magnetic Moments. (20 т.)

Assume that the electron orbits in the $x$ - $y$ plane.

i. Calculate the net electrostatic force on the electron from the proton. Express your answer in terms of any or all of the following parameters: $e$ , $m_e$ , $m_p$ , $R$ , and the permittivity of free space, $\varepsilon_0$ , where

\varepsilon_0 = \frac{1}{4\pi k}.

( $k$ is the Coulomb’s Law constant). [3 т.]

ii. Determine the angular velocity $\omega_0$ of the electron around the proton in terms of any or all of the following parameters: $e$ , $m_e$ , $R$ , and $\varepsilon_0$ . [5 т.]

iii. Derive an expression for the magnitude of the magnetic field $B_e$ due to the orbital motion of the electron at a distance $z \gg R$ from the $x$ - $y$ plane along the axis of orbital rotation of the electron. Express your answer in terms of any or all of the following parameters: $e$ , $m_e$ , $R$ , $\omega_0$ , $z$ , and the permeability of free space $\mu_0$ . [8 т.]

iv. A small bar magnet has a magnetic field far from the magnet given by

B = \frac{\mu_0}{2\pi}\frac{m}{z^3},

where $z$ is the distance from the magnet on the axis connecting the north and south poles, $m$ is the magnetic dipole moment, and $\mu_0$ is the permeability of free space. Assuming that an electron orbiting a proton acts like a small bar magnet, find the dipole moment $m$ for an electron orbiting an atom in terms of any or all of the following parameters: $e$ , $m_e$ , $R$ , and $\omega_0$ . [4 т.]

b. Diamagnetism. (30 т.)

We model a diamagnetic substance to have all atoms oriented so that the electron orbits are in the $x$ - $y$ plane, exactly half of which are clockwise and half counterclockwise when viewed from the positive $z$ axis looking toward the origin. Some substances are predominantly diamagnetic.

i. Calculate the total magnetic moment of a diamagnetic substance with $N$ atoms. Write your answer in terms of any or all of the following parameters: $e, m_e, R, N$ , and $\mu_0$ . [3 т.]

ii. An external magnetic field $\vec{B}_0 = B_0\hat{z}$ is applied to the substance. Assume that the introduction of the external field doesn’t change the fact that the electron moves in a circular orbit of radius $R$ . Determine $\Delta\omega$ , the change in angular velocity of the electron, for both the clockwise and counterclockwise orbits. Throughout this entire problem you can assume that $\Delta\omega \ll \omega_0$ . Write your answer in terms of $e, m_e$ , and $B_0$ only. [6 т.]

iii. Assume that the external field is turned on at a constant rate in a time interval $\Delta t$ . That is to say, when $t = 0$ the external field is zero and when $t = \Delta t$ the external field is $\vec{B}_0$ . Determine the induced emf $\mathcal{E}$ experienced by the electron. Write your answer in terms of any or all of the following parameters: $e, m_e, R, N$ , $B_0$ , $\omega_0$ , and $\mu_0$ . [6 т.]

iv. Verify that the change in the kinetic energy of the electron satisfies $\Delta K = e\,\mathcal{E}$ . This justifies our assumption in (ii) that $R$ does not change. [6 т.]

v. Determine the change in the total magnetic moment $\Delta m$ for the $N$ atoms when the external field is applied, writing your answer in terms of $e, m_e, R, N, \mu_0$ and $B_0$ . [6 т.]

vi. Suppose that the uniform magnetic field used in the previous parts of this problem is replaced with a bar magnet. Would the diamagnetic substance be attracted or repelled by the bar magnet? How does your answer show this? [3 т.]

Отговори

Решение

a. Magnetic Moments

i. From Coulomb’s Law,

F = \frac{e^2}{4\pi\epsilon_0 R^2} \qquad (B2-1)

ii. For circular motion,

F = \frac{m_e v^2}{R} = m_e R\omega_0^2, \qquad (B2-2)

The force is provided by the Coulomb force, so

m_e R\omega_0^2 = \frac{e^2}{4\pi\epsilon_0 R^2}, \qquad (B2-3)

\omega_0 = \sqrt{\frac{e^2}{4\pi\epsilon_0 m_e R^3}} \qquad (B2-4)

iii. From the law of Biot and Savart,

\vec{B}_e = \frac{\mu_0 i}{4\pi}\oint\frac{d\vec{s} \times \vec{r}}{r^3}, \qquad (B2-5)

B_e = \frac{\mu_0 i}{4\pi}2\pi R\frac{R}{(z^2 + r^2)^{3/2}}, \qquad (B2-6)

\approx \frac{\mu_0 i R^2}{2z^3}. \qquad (B2-7)

For the current, $i$ , we can write

i = \frac{q}{t} = \frac{e\omega_0}{2\pi}. \qquad (B2-8)

Then

B_e = \frac{\mu_0 e\omega_0 R^2}{4\pi z^3}. \qquad (B2-9)

iv. By substitution,

m = \frac{e\omega_0 R}{2}. \qquad (B2-10)

b. Diamagnetism

i. If half go one way and half go the other, $M = 0$ .

ii. Additional force from magnetism,

F_B = qvB_0 = eR\omega B_0 \qquad (B2-11)

modifies previous central force problem to give

m_e R\omega^2 = \frac{e^2}{4\pi\epsilon_0 R^2} \pm eR\omega_0 B_0, \qquad (B2-12)

where the positive sign corresponds to anticlockwise motion, the negative to clockwise motion.

A little math,

m_e R(\omega^2 - \omega_0^2) = \pm eR\omega B_0, \qquad (B2-13)

m_e(\omega - \omega_0)(\omega + \omega_0) = \pm e\omega B_0, \qquad (B2-14)

m_e(\Delta\omega)(2\omega_0) = \pm e\omega_0 B_0, \qquad (B2-15)

where in the last line we have used the approximation $\omega \approx \omega_0$ . Then

\Delta\omega = \pm\frac{eB_0}{2m_e}. \qquad (B2-16)

iii. The emf is given by

\mathcal{E} = n\frac{\Delta\Phi}{\Delta t} = \frac{\Delta n}{\Delta t}\Phi, \qquad (B2-17)

but $\Delta n/\Delta t$ is a measure of the number of turns made by the electron in a time interval $\Delta t$ , so

\frac{\Delta n}{\Delta t} = \frac{\omega_0 R}{2\pi R} = \frac{\omega_0}{2\pi}. \qquad (B2-18)

Then

\mathcal{E} = \frac{\omega_0}{2\pi}B_0\pi R^2 = \frac{1}{2}\omega_0 b_0 R^2. \qquad (B2-19)

iv. The change in kinetic energy is given by

\Delta K = \Delta\left(\frac{1}{2}m_e\omega^2R^2\right), \qquad (B2-20)

= m_e R^2\omega\ \Delta\omega, \qquad (B2-21)

\approx m_e R^2\omega_0\ \Delta\omega, \qquad (B2-22)

= m_e\omega_0 R^2\left(\pm\frac{eB_0}{2m_e}\right), \qquad (B2-23)

= e\mathcal{E}. \qquad (B2-24)

v. $\Delta M = N\delta m$ , where $N$ is the number of atoms, and $\Delta m$ the change in magnetic moment in each. The change is

\Delta m = \Delta\left(\frac{e\omega_0 R}{2}\right). \qquad (B2-25)

= \frac{eR}{2}\Delta\omega, \qquad (B2-26)

= \frac{e^2R^2B_0}{4m_e}, \qquad (B2-27)

\Delta M = N\frac{e^2R^2B_0}{4m_e}. \qquad (B2-28)

vi. Repelled, by Lenz’s law.

Оригинал в Архива: olympiad_2007_semi-final.pdf · официални решения: olympiad_2007_semi-final_soln.pdf