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Условие Решение

USAPhO 2001, Полуфинал — B2

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Всички материали от USAPhO 2001

2001 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. It may not be used in graphing mode. Calculators may not be shared. You may not use any tables, books, or collections of formulas. You may use a ruler or straight edge.
Questions with the same point value are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Quantity	Value
Gravitational field at the Earth's surface	$g=9.8$ N/kg
Newton's gravitational constant	$G=6.67\times10^{-11}$ N·m $^2$ /kg $^2$
Coulomb's constant	$k=1/4\pi\epsilon_0=9.0\times10^9$ N·m $^2$ /C $^2$
Biot-Savart constant	$k_m=\mu_0/4\pi=10^{-7}$ T·m/A
Speed of light in a vacuum	$c=3.0\times10^8$ m/s
Boltzmann's constant	$k_B=1.38\times10^{-23}$ J/K
Avogadro's number	$N_A=6.02\times10^{23}$ (mol) $^{-1}$
Ideal gas constant	$R=N_Ak_B=8.31$ J/(mol·K)
Stefan-Boltzmann constant	$\sigma=5.67\times10^{-8}$ J/(s·m $^2$ ·K $^4$ )
Elementary charge	$e=1.6\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.6\times10^{-19}$ J
Planck's constant	$h=6.63\times10^{-34}\text{ J}\cdot\text{s}=4.14\times10^{-15}\text{ eV}\cdot\text{s}$
Electron mass	$m=9.1\times10^{-31}\text{ kg}=0.51\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\simeq1+nx\quad\text{for }\lvert x\rvert\ll1$

Условие

(Cylindrical coordinates are used in this problem and $\hat u_r$ , $\hat u_\theta$ , and $\hat u_z$ , represent unit vectors in the $r$ , $\theta$ and $z$ directions, respectively. The coordinates form a right-handed orthogonal system, e.g., $\hat u_r\times\hat u_\theta=\hat u_z$ .) A particle with charge $q$ and mass $m$ is constrained to move in the xy-plane with velocity that can be expressed as $\vec v=v_r\hat u_r+r\omega\hat u_\theta$ . The particle is initially at rest and a distance $r_o$ from the origin. Two forces act on the particle. One is due to the presence of a uniform magnetic field given by $\vec B=B_o\hat u_z$ . The other force and its corresponding potential energy are:

\vec F=\frac{-2k}{r^3}\hat u_r\qquad\text{and}\qquad U(r)=-\frac k{r^2},

where $k$ is a constant and $r$ is the distance from the origin. The particle executes a trajectory that periodically varies between its initial radius $r_o$ and a minimum radius, $r_{\min}$ . Express your answers in terms of $q$ , $m$ , $v_r$ , $r$ , $\omega$ , $B_o$ , $k$ , $r_o$ , $r_{\min}$ , and the unit vectors.

a. What is the magnetic force acting on the particle? [5 т.]

b. How much work is done by this magnetic force? [5 т.]

c. Use the work-energy theorem to derive an equation relating the variables $v_r$ , $r$ , and $\omega$ to the initial parameters [8 т.]

d. What is the net torque acting on the particle? [7 т.]

e. What is the angular momentum of the particle in terms $q$ , $B_o$ , $r$ , and $r_o$ up to an arbitrary constant vector? [5 т.]

f. What is the particle's angular velocity $\omega$ in terms of $m$ , $q$ , $B_o$ , $r$ , and $r_o$ ? [5 т.]

g. Using your results from (c) and (f), find $r_{\min}$ , the particle's closest approach to the origin. [5 т.]

h. Find the range of $B_o$ for which the particle does not reach the origin. [5 т.]

i. Find $\omega$ for large magnetic fields $B_o$ at each of the limiting radii $r=r_o$ and $r=r_{\min}$ . Write your answer to lowest nonvanishing order in $\dfrac1{B_o}$ . [5 т.]

Решение

a. (5 т.)

The magnetic force is $\vec F=q\vec v\times\vec B$ .

Inserting the expressions for the velocity and magnetic field and taking the cross product,

\vec F=q(v_r\hat u_r+r\omega\hat u_\theta)\times B_o\hat u_z=qB_o(v_r\hat u_r\times\hat u_z+r\omega\hat u_\theta\times\hat u_z)

\vec F=qB_o(r\omega\hat u_r-v_r\hat u_\theta)

b. (5 т.)

At any instant the displacement is in the direction of the velocity while the force is perpendicular to it. Therefore $W=0$ .

dW=\vec F\cdot d\vec l=\vec F\cdot\vec v\,dt=q(\vec v\times\vec B)\cdot\vec v\,dt=0

c. (8 т.)

The magnetic force does no work. The other force has an associated potential energy. Using the work-energy theorem which reduces to conservation of mechanical energy,

U_o+K_o=U+K

Since the particle is initially at rest, its initial kinetic energy is zero. $K_o=0$

Substituting the energy terms

-\frac k{r_o^2}=\frac12mv^2-\frac k{r^2}.

Where

v^2=(v_r\hat u_r+r\omega\hat u_\theta)\cdot(v_r\hat u_r+r\omega\hat u_\theta)=v_r^2+r^2\omega^2

Combining the last two equations

-\frac k{r_o^2}=-\frac k{r^2}+\frac12m(v_r^2+r^2\omega^2)\qquad\text{(B2-1)}

d. (7 т.)

The torque about the z-axis is $\vec\tau=\vec r\times\vec F=r\hat u_r\times\vec F$ . The other force $\vec F=-\dfrac{2k}{r^3}\hat u_r$ contributes no torque, since $\hat u_r\times\hat u_r=0$ . Therefore the net torque is due to the magnetic force

\vec\tau=r\hat u_r\times qB_o(r\omega\hat u_r-v_r\hat u_\theta)=-qB_orv_r\hat u_z.

e. (5 т.)

The net torque is equal to the time rate of change of angular momentum. The radial component of the velocity $v_r$ is equal to the time rate of change of $r$ .

\frac{d\vec L}{dt}=\vec\tau=-qB_orv_r\hat u_z=-qB_or\frac{dr}{dt}\hat u_z=-qB_o\frac12\frac{dr^2}{dt}\hat u_z

\frac d{dt}\left(\vec L+qB_o\frac12r^2\hat u_z\right)=0

\vec L+qB_o\frac12r^2\hat u_z=\vec C,

where $\vec C$ is a constant vector.

\vec L=\vec C-qB_o\frac12r^2\hat u_z.

Dropping the vector notation since all motion occurs in the xy-plane,

L=C-qB_o\frac12r^2

f. (5 т.)

For a particle the magnitude of the angular momentum about the z-axis can be written

L=mr^2\omega.

Combining the last two equations

mr^2\omega=C-qB_o\frac12r^2

mr^2\omega+qB_o\frac12r^2=C.

The constant $C$ can be evaluated in terms of the initial conditions

mr^2\omega+qB_o\frac12r^2=mr_o^2\omega_o+qB_o\frac12r_o^2

since the particle is initially at rest $\omega_o=0$ and

mr^2\omega+qB_o\frac12r^2=qB_o\frac12r_o^2

Solving for $\omega$ , we have

mr^2\omega=qB_o\frac12(r_o^2-r^2)

\omega=\frac{qB_o}{2m}\left(\frac{r_o^2}{r^2}-1\right)\qquad\text{(B2-2)}

g. (5 т.)

At $r_{\min}$ , $v_r=0$ , Equation (B2-1) becomes

-\frac k{r_o^2}=-\frac k{r_{\min}^2}+\frac12mr_{\min}^2\omega^2

Bring the other $k$ term to the left and substituting equation (B2-2) at $r_{\min}$

\frac k{r_{\min}^2}-\frac k{r_o^2}=\frac12mr_{\min}^2\left[\frac{qB_o}{2m}\left(\frac{r_o^2}{r_{\min}^2}-1\right)\right]^2

or multiplying by $r_{\min}^2r_o^2$ ,

k(r_o^2-r_{\min}^2)=\frac{r_o^2q^2B_o^2}{8m}(r_o^2-r_{\min}^2)^2.

Canceling the common factor

k=\frac{r_o^2q^2B_o^2}{8m}(r_o^2-r_{\min}^2)

and solving for

r_{\min}^2=r_o^2-\frac{8km}{q^2B_o^2r_o^2}\qquad\text{(B2-3)}

h. (5 т.)

The particle will not reach the origin if $r_{\min}^2>0$ . From the preceding expression, this occurs when

r_o^2>\frac{8km}{q^2B_o^2r_o^2}

or solving for

B_o^2>\frac{8km}{q^2r_o^4}.

Thus

|B_o|>\frac1{r_o^2}\sqrt{\frac{8km}{q^2}}

i. (5 т.)

Evaluating equation (B2-2)

at $r=r_o$

\omega_o=\frac{qB_o}{2m}\left(\frac{r_o^2}{r_o^2}-1\right)=0

at $r=r_{\min}$

\omega=\frac{qB_o}{2m}\left(\frac{r_o^2}{r_{\min}^2}-1\right).

combining with equation (B2-3)

\omega=\frac{qB_o}{2m}\left(\frac{r_o^2}{r_o^2-\frac{8km}{q^2B_o^2r_o^2}}-1\right)=\frac{qB_o}{2m}\left(\frac1{1-\frac{8km}{q^2B_o^2r_o^4}}-1\right).

For large $B_o$ , $\dfrac{8km}{q^2B_o^2r_o^4}$ is small. Using the binomial expansion

\omega=\frac{qB_o}{2m}\left(1+\frac{8km}{q^2B_o^2r_o^4}-1\right)=\frac{4k}{qB_or_o^4}

Оригинал в Архива: Semifinal 2001.pdf · официални решения: Semifinal 2001 Solutions.pdf