Datasets:

lmms-lab

/

OlympiadBench

like 0

1466

Electromagnetism

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\theta^{3} / 6 \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\theta^{2} / 2 \text { for }|\theta| \ll 1 \end{array} $$ Extra Supplementary Reading Materials: Electric Slide Two large parallel plates of area $A$ are placed at $x=0$ and $x=d \ll \sqrt{A}$ in a semiconductor medium. The plate at $x=0$ is grounded, and the plate at $x=d$ is at a fixed potential $-V_{0}$, where $V_{0}>0$. Particles of positive charge $q$ flow between the two plates. You may neglect any dielectric effects of the medium. Context question: a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that $$ v=\mu E $$ where $E$ is the local electric field and $\mu$ is the charge mobility. i. In the steady state, there is a nonzero but time-independent density of charges between the two plates. Let the charge density at position $x$ be $\rho(x)$. Use charge conservation to find a relationship between $\rho(x), v(x)$, and their derivatives. Context answer: $v \frac{\mathrm{d} \rho}{\mathrm{d} x}+\rho \frac{\mathrm{d} v}{\mathrm{~d} x}=0$ Extra Supplementary Reading Materials: a. For large $V_{0}$, the velocity of the positive charges is determined by a strong drag force, so that $$ v=\mu E $$ where $E$ is the local electric field and $\mu$ is the charge mobility. Context question: ii. Let $V(x)$ be the electric potential at $x$. Derive an expression relating $\rho(x), V(x)$, and their derivatives. (Hint: start by using Gauss's law to relate the charge density $\rho(x)$ to the derivative of the electric field $E(x)$.) Context answer: $$ \frac{d^{2} V}{d x^{2}}=-\frac{\rho}{\epsilon_{0}} $$ Context question: iii. Suppose that in the steady state, conditions have been established so that $V(x)$ is proportional to $x^{b}$, where $b$ is an exponent you must find, and the current is nonzero. Derive an expression for the current in terms of $V_{0}$ and the other given parameters. Context answer: \boxed{$I=\frac{9 \epsilon_{0} \mu A V_{0}^{2}}{8 d^{3}}$} Extra Supplementary Reading Materials: b. For small $V_{0}$, the positive charges move by diffusion. The current due to diffusion is given by Fick's Law, $$ I=-A D \frac{\mathrm{d} \rho}{\mathrm{d} x} $$ Here, $D$ is the diffusion constant, which you can assume to be described by the Einstein relation $$ D=\frac{\mu k_{B} T}{q} $$ where $T$ is the temperature of the system. Context question: i. Assume that in the steady state, conditions have been established so that a nonzero, steady current flows, and the electric potential again satisfies $V(x) \propto x^{b^{\prime}}$, where $b^{\prime}$ is another exponent you must find. Derive an expression for the current in terms of $V_{0}$ and the other given parameters. Context answer: \boxed{$I=\frac{6 \mu k_{B} T A \epsilon_{0} V_{0}}{q d^{3}}$}

ii. At roughly what voltage $V_{0}$ does the system transition from this regime to the high voltage regime of the previous part?

Expression

OE_TO_physics_en_COMP

1475

Mechanics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$

a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible. i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision?

Expression

OE_TO_physics_en_COMP

1476

Mechanics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Context question: a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible. i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision? Context answer: \boxed{$u=v(\sin \theta-\mu \cos \theta)$}

ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ?

Expression

OE_TO_physics_en_COMP

1477

Mechanics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Context question: a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible. i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision? Context answer: \boxed{$u=v(\sin \theta-\mu \cos \theta)$} Context question: ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ? Context answer: \boxed{$\mu=\tan \theta$}

b. Now suppose you drop a sphere with mass $m$, radius $R$ and moment of inertia $\beta m R^{2}$ vertically onto the same fixed ramp such that it reaches the ramp with speed $v$. i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case?

Expression

OE_TO_physics_en_COMP

1478

Mechanics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Context question: a. Suppose you drop a block of mass $m$ vertically onto a fixed ramp with angle $\theta$ with coefficient of static and kinetic friction $\mu$. The block is dropped in such a way that it does not rotate after colliding with the ramp. Throughout this problem, assume the time of the collision is negligible. i. Suppose the block's speed just before it hits the ramp is $v$ and the block slides down the ramp immediately after impact. What is the speed of the block right after the collision? Context answer: \boxed{$u=v(\sin \theta-\mu \cos \theta)$} Context question: ii. What is the minimum $\mu$ such that the speed of the block right after the collision is 0 ? Context answer: \boxed{$\mu=\tan \theta$} Context question: b. Now suppose you drop a sphere with mass $m$, radius $R$ and moment of inertia $\beta m R^{2}$ vertically onto the same fixed ramp such that it reaches the ramp with speed $v$. i. Suppose the sphere immediately begins to roll without slipping. What is the new speed of the sphere in this case? Context answer: \boxed{$u=\frac{v \sin \theta}{1+\beta}$}

ii. What is the minimum coefficient of friction such that the sphere rolls without slipping immediately after the collision?

Expression

OE_TO_physics_en_COMP

1482

Electromagnetism

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: The electric potential at the center of a cube with uniform charge density $\rho$ and side length $a$ is $$ \Phi \approx \frac{0.1894 \rho a^{2}}{\epsilon_{0}} $$ You do not need to derive this. ${ }^{1}$ For the entirety of this problem, any computed numerical constants should be written to three significant figures.

a. What is the electric potential at a corner of the same cube? Write your answer in terms of $\rho, a, \epsilon_{0}$, and any necessary numerical constants.

Equation,Numerical

,1e-3

OE_TO_physics_en_COMP

1486

Optics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: In this problem, use a particle-like model of photons: they propagate in straight lines and obey the law of reflection, but are subject to the quantum uncertainty principle. You may use small-angle approximations throughout the problem. A photon with wavelength $\lambda$ has traveled from a distant star to a telescope mirror, which has a circular cross-section with radius $R$ and a focal length $f \gg R$. The path of the photon is nearly aligned to the axis of the mirror, but has some slight uncertainty $\Delta \theta$. The photon reflects off the mirror and travels to a detector, where it is absorbed by a particular pixel on a charge-coupled device (CCD). Suppose the telescope mirror is manufactured so that photons coming in parallel to each other are focused to the same pixel on the CCD, regardless of where they hit the mirror. Then all small cross-sectional areas of the mirror are equally likely to include the point of reflection for a photon.

a. Find the standard deviation $\Delta r$ of the distribution for $r$, the distance from the center of the telescope mirror to the point of reflection of the photon.

Expression

OE_TO_physics_en_COMP

1488

Modern Physics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles. a. The force of radiation on a spherical particle of radius $r$ is given by $$ F=P Q \pi r^{2} $$ where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables.

i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.

Expression

OE_TO_physics_en_COMP

1489

Modern Physics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles. a. The force of radiation on a spherical particle of radius $r$ is given by $$ F=P Q \pi r^{2} $$ where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables. Context question: i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun. Context answer: \boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$}

ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$.

Expression

OE_TO_physics_en_COMP

1491

Modern Physics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles. a. The force of radiation on a spherical particle of radius $r$ is given by $$ F=P Q \pi r^{2} $$ where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables. Context question: i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun. Context answer: \boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$} Context question: ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$. Context answer: \boxed{$\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$} Context question: iii. The quality factor is given by one of the following - If $r \ll \lambda, Q \sim(r / \lambda)^{2}$ - If $r \sim \lambda, Q \sim 1$. - If $r \gg \lambda, Q=1$ Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure? Context answer: particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough. Extra Supplementary Reading Materials: b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar system.

i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle $v$ in terms of $M_{\odot}$, distance from sun $R$, and any other fundamental constants.

Expression

OE_TO_physics_en_COMP

1492

Modern Physics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles. a. The force of radiation on a spherical particle of radius $r$ is given by $$ F=P Q \pi r^{2} $$ where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables. Context question: i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun. Context answer: \boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$} Context question: ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$. Context answer: \boxed{$\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$} Context question: iii. The quality factor is given by one of the following - If $r \ll \lambda, Q \sim(r / \lambda)^{2}$ - If $r \sim \lambda, Q \sim 1$. - If $r \gg \lambda, Q=1$ Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure? Context answer: particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough. Extra Supplementary Reading Materials: b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar system. Context question: i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle $v$ in terms of $M_{\odot}$, distance from sun $R$, and any other fundamental constants. Context answer: \boxed{$v=\sqrt{\frac{G M_{\odot}}{R}}$}

ii. Because the particle is moving, the radiation force is not directed directly away from the sun. Find the torque $\tau$ on the particle because of radiation pressure. You may assume that $v \ll c$.

Expression

OE_TO_physics_en_COMP

1493

Modern Physics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles. a. The force of radiation on a spherical particle of radius $r$ is given by $$ F=P Q \pi r^{2} $$ where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables. Context question: i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun. Context answer: \boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$} Context question: ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$. Context answer: \boxed{$\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$} Context question: iii. The quality factor is given by one of the following - If $r \ll \lambda, Q \sim(r / \lambda)^{2}$ - If $r \sim \lambda, Q \sim 1$. - If $r \gg \lambda, Q=1$ Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure? Context answer: particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough. Extra Supplementary Reading Materials: b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar system. Context question: i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle $v$ in terms of $M_{\odot}$, distance from sun $R$, and any other fundamental constants. Context answer: \boxed{$v=\sqrt{\frac{G M_{\odot}}{R}}$} Context question: ii. Because the particle is moving, the radiation force is not directed directly away from the sun. Find the torque $\tau$ on the particle because of radiation pressure. You may assume that $v \ll c$. Context answer: \boxed{$\tau=-\frac{v}{c} \frac{L \odot}{4 R c} Q r^{2}$}

iii. Since $\tau=d L / d t$, the angular momentum $L$ of the particle changes with time. As such, develop a differential equation to find $d R / d t$, the rate of change of the radial location of the particle. You may assume the orbit is always quasi circular.

Equation

OE_TO_physics_en_COMP

1494

Modern Physics

$$ \begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} $$ $$ \begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} $$ Extra Supplementary Reading Materials: Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles. a. The force of radiation on a spherical particle of radius $r$ is given by $$ F=P Q \pi r^{2} $$ where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; unless told otherwise, leave your answers in terms of symbolic variables. Context question: i. Given that the total power radiated from the sun is given by $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun. Context answer: \boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$} Context question: ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$. Context answer: \boxed{$\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$} Context question: iii. The quality factor is given by one of the following - If $r \ll \lambda, Q \sim(r / \lambda)^{2}$ - If $r \sim \lambda, Q \sim 1$. - If $r \gg \lambda, Q=1$ Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure? Context answer: particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough. Extra Supplementary Reading Materials: b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar system. Context question: i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle $v$ in terms of $M_{\odot}$, distance from sun $R$, and any other fundamental constants. Context answer: \boxed{$v=\sqrt{\frac{G M_{\odot}}{R}}$} Context question: ii. Because the particle is moving, the radiation force is not directed directly away from the sun. Find the torque $\tau$ on the particle because of radiation pressure. You may assume that $v \ll c$. Context answer: \boxed{$\tau=-\frac{v}{c} \frac{L \odot}{4 R c} Q r^{2}$} Context question: iii. Since $\tau=d L / d t$, the angular momentum $L$ of the particle changes with time. As such, develop a differential equation to find $d R / d t$, the rate of change of the radial location of the particle. You may assume the orbit is always quasi circular. Context answer: \boxed{$-\frac{1}{c^{2}} \frac{L_{\odot}}{R} Q=\frac{8}{3} \pi \rho r \frac{d R}{d t}$}

iv. Develop an expression for the time required to remove particles of size $r \approx 1 \mathrm{~cm}$ and density $\rho \approx 1000 \mathrm{~kg} / \mathrm{m}^{3}$ originally in circular orbits at a distance $R=R_{\text {earth }}$, and use the numbers below to simplify your expression.

s

Numerical

5e13

OE_TO_physics_en_COMP

1502

Thermodynamics

A solid, uniform cylinder of height $h=10 \mathrm{~cm}$ and base area $s=100 \mathrm{~cm}^{2}$ floats in a cylindrical beaker of height $H=20 \mathrm{~cm}$ and inner bottom area $S=102 \mathrm{~cm}^{2}$ filled with a liquid. The ratio between the density of the cylinder and that of the liquid is $\gamma=0.70$. The bottom of the cylinder is above the bottom of the beaker by a few centimeters. The cylinder is oscillating vertically, so that its axis always coincides with that of the beaker. The amplitude of the liquid level oscillations is $A=1 \mathrm{~mm}$.

Find the period of the motion $T$. Neglect the viscosity of the liquid.

s

Numerical

5e-2

OE_TO_physics_en_COMP

1565

Electromagnetism

A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.

1. First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.

Expression

OE_TO_physics_en_COMP

1566

Electromagnetism

A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system. Context question: 1. First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line. Context answer: $\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$

2. The potential due to the line charge could be written as $$ V(r)=f(r)+K, $$ where $K$ is a constant. Determine $f(r)$.

Expression

OE_TO_physics_en_COMP

1572

Modern Physics

Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.

1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).

Expression

OE_TO_physics_en_COMP

1573

Modern Physics

Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$}

2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$.

Expression

OE_TO_physics_en_COMP

1574

Modern Physics

Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$}

3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$.

Expression

OE_TO_physics_en_COMP

1576

Modern Physics

Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题}

5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$.

Expression

OE_TO_physics_en_COMP

1577

Modern Physics

Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem. First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock). By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning. Some mathematics formulas that might be useful - $\sinh x=\frac{e^{x}-e^{-x}}{2}$ - $\cosh x=\frac{e^{x}+e^{-x}}{2}$ - $\tanh x=\frac{\sinh x}{\cosh x}$ - $1+\sinh ^{2} x=\cosh ^{2} x$ - $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$ - $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$ - $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$ Part A. Single Accelerated Particle Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$. Context question: 1. When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame). Context answer: \boxed{$a=\frac{F}{\gamma^{3} m}$} Context question: 2. Calculate the velocity of the particle $\beta(t)=\frac{v(t)}{c}$ at time $t$ (in rest frame), in terms of $F, m, t$ and $c$. Context answer: \boxed{$\beta=\frac{\frac{F t}{m c}}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}$} Context question: 3. Calculate the position of the particle $x(t)$ at time $t$, in term of $F, m, t$ and $c$. Context answer: \boxed{$x=\frac{m c^{2}}{F}\left(\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}-1\right)$} Context question: 4. Show that the proper acceleration of the particle, $a^{\prime} \equiv g=F / m$, is a constant. The proper acceleration is the acceleration of the particle measured in the instantaneous proper frame. Context answer: \boxed{证明题} Context question: 5. Calculate the velocity of the particle $\beta(\tau)$, when the time as experienced by the particle is $\tau$. Express the answer in $g, \tau$, and $c$. Context answer: \boxed{$\beta=\tanh \frac{g \tau}{c}$}

6. ( $\mathbf{0 . 4} \mathbf{~ p t s )}$ Also calculate the time $t$ in the rest frame in terms of $g, \tau$, and $c$.

Expression

OE_TO_physics_en_COMP

1592

Modern Physics

All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the "intrinsic" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term "frequency" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题}

2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$.

Expression

OE_TO_physics_en_COMP

1594

Modern Physics

All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the "intrinsic" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term "frequency" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题}

2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ?

Expression

OE_TO_physics_en_COMP

1596

Modern Physics

All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the "intrinsic" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term "frequency" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} Context question: 2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ? Context answer: \boxed{$\Delta =\gamma B_{0}-\omega$} Context question: 3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is $$ \Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} $$ Context answer: \boxed{证明题}

4. Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )?

Expression

OE_TO_physics_en_COMP

1597

Modern Physics

All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the "intrinsic" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term "frequency" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} Context question: 2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ? Context answer: \boxed{$\Delta =\gamma B_{0}-\omega$} Context question: 3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is $$ \Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} $$ Context answer: \boxed{证明题} Context question: 4. Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )? Context answer: \boxed{$\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$} Extra Supplementary Reading Materials: Part C. Rabi oscillation For an ensemble of $N$ particles under the influence of a large magnetic field, the spin can have two quantum states: "up" and "down". Consequently, the total population of spin up $N_{\uparrow}$ and down $N_{\downarrow}$ obeys the equation $$ N_{\uparrow}+N_{\downarrow}=N $$ The difference of spin up population and spin down population yields the macroscopic magnetization along the $z$ axis: $$ M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} . $$ In a real experiment, two magnetic fields are usually applied, a large bias field $B_{0} \boldsymbol{k}$ and an oscillating field with amplitude $2 b$ perpendicular to the bias field $\left(b \ll B_{0}\right)$. Initially, only the large bias is applied, causing all the particles lie in the spin up states ( $\boldsymbol{\mu}$ is oriented in the $z$-direction at $t=0$ ). Then, the oscillating field is turned on, where its frequency $\omega$ is chosen to be in resonance with the Larmor precession frequency $\omega_{0}$, i.e. $\omega=\omega_{0}$. In other words, the total field after time $t=0$ is given by $$ \boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} . $$

1. In the rotating frame $S^{\prime}$, show that the effective field can be approximated by $$ \boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime}, $$ which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ?

Expression

OE_TO_physics_en_COMP

1599

Modern Physics

All matters in the universe have fundamental properties called spin, besides their mass and charge. Spin is an intrinsic form of angular momentum carried by particles. Despite the fact that quantum mechanics is needed for a full treatment of spin, we can still study the physics of spin using the usual classical formalism. In this problem, we are investigating the influence of magnetic field on spin using its classical analogue. The classical torque equation of spin is given by $$ \boldsymbol{\tau}=\frac{d \boldsymbol{L}}{d t}=\boldsymbol{\mu} \times \boldsymbol{B} $$ In this case, the angular momentum $\boldsymbol{L}$ represents the "intrinsic" spin of the particles, $\boldsymbol{\mu}$ is the magnetic moment of the particles, and $\boldsymbol{B}$ is magnetic field. The spin of a particle is associated with a magnetic moment via the equation $$ \boldsymbol{\mu}=-\gamma \boldsymbol{L} $$ where $\gamma$ is the gyromagnetic ratio. In this problem, the term "frequency" means angular frequency (rad/s), which is a scalar quantity. All bold letters represent vectors; otherwise they represent scalars. Part A. Larmor precession Context question: 1. Prove that the magnitude of magnetic moment $\mu$ is always constant under the influence of a magnetic field $\boldsymbol{B}$. For a special case of stationary (constant) magnetic field, also show that the angle between $\boldsymbol{\mu}$ and $\boldsymbol{B}$ is constant. (Hint: You can use properties of vector products.) Context answer: \boxed{证明题} Context question: 2. A uniform magnetic field $\boldsymbol{B}$ exists and it makes an angle $\phi$ with a particle's magnetic moment $\boldsymbol{\mu}$. Due to the torque by the magnetic field, the magnetic moment $\boldsymbol{\mu}$ rotates around the field $\boldsymbol{B}$, which is also known as Larmor precession. Determine the Larmor precession frequency $\omega_{0}$ of the magnetic moment with respect to $\boldsymbol{B}=B_{0} \boldsymbol{k}$. Context answer: \boxed{$\omega_{0}=\gamma B_{0}$} Extra Supplementary Reading Materials: Part B. Rotating frame In this section, we choose a rotating frame $S^{\prime}$ as our frame of reference. The rotating frame $S^{\prime}=\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ rotates with an angular velocity $\omega \boldsymbol{k}$ as seen by an observer in the laboratory frame $S=(x, y, z)$, where the axes $x^{\prime}, y^{\prime}, z^{\prime}$ intersect with $x, y, z$ at time $t=0$. Any vector $\boldsymbol{A}=A_{x} \boldsymbol{i}+A_{y} \boldsymbol{j}+A_{z} \boldsymbol{k}$ in a lab frame can be written as $\boldsymbol{A}=A_{x}{ }^{\prime} \boldsymbol{i}^{\prime}+A_{y}{ }^{\prime} \boldsymbol{j}^{\prime}+A_{z}{ }^{\prime} \boldsymbol{k}^{\prime}$ in the rotating frame $S^{\prime}$. The time derivative of the vector becomes $$ \frac{d \boldsymbol{A}}{d t}=\left(\frac{d A_{x}{ }^{\prime}}{d t} \boldsymbol{i}^{\prime}+\frac{d A_{y}{ }^{\prime}}{d t} \boldsymbol{j}^{\prime}+\frac{d A_{z}{ }^{\prime}}{d t} \boldsymbol{k}^{\prime}\right)+\left(A_{x}{ }^{\prime} \frac{d \boldsymbol{i}^{\prime}}{d t}+A_{y}{ }^{\prime} \frac{d \boldsymbol{j}^{\prime}}{d t}+A_{z}{ }^{\prime} \frac{d \boldsymbol{k}^{\prime}}{d t}\right) $$ $$ \left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}=\left(\frac{d \boldsymbol{A}}{d t}\right)_{r o t}+(\omega \mathbf{k} \times \boldsymbol{A}) $$ where $\left(\frac{d \boldsymbol{A}}{d t}\right)_{l a b}$ is the time derivative of vector $\boldsymbol{A}$ seen by an observer in the lab frame, and $\left(\frac{d A}{d t}\right)_{\text {rot }}$ is the time derivative seen by an observer in the rotating frame. For all the following problems in this part, the answers are referred to the rotating frame $S^{\prime}$. Context question: 1. Show that the time evolution of the magnetic moment follows the equation $$ \left(\frac{d \boldsymbol{\mu}}{d t}\right)_{r o t}=-\gamma \boldsymbol{\mu} \times \boldsymbol{B}_{e f f} $$ where $\boldsymbol{B}_{\text {eff }}=\boldsymbol{B}-\frac{\omega}{\gamma} \boldsymbol{k}^{\prime}$ is the effective magnetic field. Context answer: \boxed{证明题} Context question: 2. For $\boldsymbol{B}=B_{0} \boldsymbol{k}$, what is the new precession frequency $\Delta$ in terms of $\omega_{0}$ and $\omega$ ? Context answer: \boxed{$\Delta =\gamma B_{0}-\omega$} Context question: 3. Now, let us consider the case of a time-varying magnetic field. Besides a constant magnetic field, we also apply a rotating magnetic field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, so $\boldsymbol{B}=B_{0} \boldsymbol{k}+\boldsymbol{b}(t)$. Show that the new Larmor precession frequency of the magnetic moment is $$ \Omega=\gamma \sqrt{\left(B_{0}-\frac{\omega}{\gamma}\right)^{2}+b^{2}} $$ Context answer: \boxed{证明题} Context question: 4. Instead of applying the field $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}+\sin \omega t \boldsymbol{j})$, now we apply $\boldsymbol{b}(t)=b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$, which rotates in the opposite direction and hence $\boldsymbol{B}=B_{0} \boldsymbol{k}+b(\cos \omega t \boldsymbol{i}-\sin \omega t \boldsymbol{j})$. What is the effective magnetic field $\boldsymbol{B}_{\text {eff }}$ for this case (in terms of the unit vectors $\boldsymbol{i}^{\prime}, \boldsymbol{j}^{\prime}, \boldsymbol{k}^{\prime}$ )? What is its time average, $\overline{\boldsymbol{B}_{\text {eff }}}$ (recall that $\overline{\cos 2 \pi t / T}=\overline{\sin 2 \pi t / T}=0$ )? Context answer: \boxed{$\mathbf{B}_{\mathrm{eff}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}+b\left(\cos 2 \omega t \mathbf{i}^{\prime}-\sin 2 \omega t \mathbf{j}^{\prime}\right)$ , $\overline{\mathbf{B}_{\mathrm{eff}}}=\left(B_{0}-\frac{\omega}{\gamma}\right) \mathbf{k}^{\prime}$} Extra Supplementary Reading Materials: Part C. Rabi oscillation For an ensemble of $N$ particles under the influence of a large magnetic field, the spin can have two quantum states: "up" and "down". Consequently, the total population of spin up $N_{\uparrow}$ and down $N_{\downarrow}$ obeys the equation $$ N_{\uparrow}+N_{\downarrow}=N $$ The difference of spin up population and spin down population yields the macroscopic magnetization along the $z$ axis: $$ M=\left(N_{\uparrow}-N_{\downarrow}\right) \mu=N \mu_{z} . $$ In a real experiment, two magnetic fields are usually applied, a large bias field $B_{0} \boldsymbol{k}$ and an oscillating field with amplitude $2 b$ perpendicular to the bias field $\left(b \ll B_{0}\right)$. Initially, only the large bias is applied, causing all the particles lie in the spin up states ( $\boldsymbol{\mu}$ is oriented in the $z$-direction at $t=0$ ). Then, the oscillating field is turned on, where its frequency $\omega$ is chosen to be in resonance with the Larmor precession frequency $\omega_{0}$, i.e. $\omega=\omega_{0}$. In other words, the total field after time $t=0$ is given by $$ \boldsymbol{B}(t)=B_{0} \boldsymbol{k}+2 b \cos \omega_{0} t \boldsymbol{i} . $$ Context question: 1. In the rotating frame $S^{\prime}$, show that the effective field can be approximated by $$ \boldsymbol{B}_{\text {eff }} \approx b \boldsymbol{i}^{\prime}, $$ which is commonly known as rotating wave approximation. What is the precession frequency $\Omega$ in frame $S^{\prime}$ ? Context answer: \boxed{$\Omega=\gamma b$} Context question: 2. Determine the angle $\alpha$ that $\boldsymbol{\mu}$ makes with $\boldsymbol{B}_{\text {eff }}$. Also, prove that the magnetization varies with time as $$ M(t)=N \mu(\cos \Omega t) . $$ Context answer: \boxed{证明题}

3. Under the application of magnetic field described above, determine the fractional population of each spin up $P_{\uparrow}=N_{\uparrow} / N$ and spin down $P_{\downarrow}=N_{\downarrow} / N$ as a function of time. Plot $P_{\uparrow}(t)$ and $P_{\downarrow}(t)$ on the same graph vs. time $t$. The alternating spin up and spin down population as a function of time is called Rabi oscillation.

Expression

OE_TO_physics_en_COMP