Съдържание

Условие Отговори Решение

USAPhO 2003, Полуфинал — Задача 3. A3

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Всички материали от USAPhO 2003

AAPT / AIP United States Physics Team 2003 — 2003 Semi-Final Exam

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

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	$k = 1/4\pi\varepsilon_0 = 9.0 \times 10^{9}\ \mathrm{N{\cdot}m^2/C^2}$
Biot-Savart constant	$k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T{\cdot}m/A}$
Speed of light in a vacuum	$c = 3.0 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_\mathrm{A} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_\mathrm{A}k_\mathrm{B} = 8.31\ \mathrm{J/(mol{\cdot}K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s{\cdot}m^2{\cdot}K^4)}$
Elementary charge	$e = 1.6 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.6 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J{\cdot}s} = 4.14 \times 10^{-15}\ \mathrm{eV{\cdot}s}$
Electron mass	$m = 9.1 \times 10^{-31}\ \mathrm{kg} = 0.51\mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ ; $\cos\theta \approx 1 - \tfrac{1}{2}\theta^2$

Условие

Cross-section of a shaded slab of thickness L (double arrow at left) extending indefinitely left and right (dashed edges); axes z up and y to the right from a point in the middle marked with a dot, labelled 'x out of page'. — Figure a

The same slab of thickness L with axes z and y and 'x out of page'; above it a square loop of side a (both sides labelled a) carrying current I (arrow pointing down on its left), at a distance b above the top surface of the slab (double arrow b). — Figure b

A conducting slab has infinite extent in the $x$ and $y$ directions and thickness $L$ in the $z$ direction as shown in Figure a above. The slab is centered at $z = 0$ and carries a uniform current density $\vec{J} = J\hat{i}$ where $\hat{i}$ , $\hat{j}$ , and $\hat{k}$ are unit vectors in the $x$ , $y$ , and $z$ directions, respectively.

a. Find the magnetic field $\vec{B}$ at all points. [10 т.]

b. A square wire loop of side $a$ is placed a distance $b$ above the slab (as shown in Figure b above). The loop has unit normal vector $\hat{n} = \cos\theta\hat{i} + \sin\theta\hat{j}$ and applied current $I$ . (Figure b depicts the special case $\theta = 0$ .) What is the net force and the net torque on the loop as a function of $\theta$ ? [10 т.]

c. The applied current $I$ is now removed from the loop described above and the current density in the slab $\vec{J} = J\hat{i}$ is reduced to zero over time $T$ . The wire used to construct the loop has resistance per unit length $S$ . How much charge flows through each cross section of the loop wire due to the reduction in current density? [5 т.]

Отговори

Решение

a. The slab has infinite extent in the $x$ and $y$ direction. Therefore the magnetic field $\vec{B}$ can depend at most on $z$ . By the right hand rule and symmetry $\vec{B}$ is to the left for positive $z$ and to the right for negative $z$ .

\vec{B} = -B(z)\hat{j} \qquad z > 0

\vec{B} = +B(z)\hat{j} \qquad z < 0

$\vec{B}$ can be found from Ampere’s law. To find $\vec{B}$ outside the slab, $z > L/2$ , use the loop shown in Figure a. The loop segments parallel to the $z$ direction are perpendicular to $\vec{B}$ and will not contribute.

\oint \vec{B}\cdot d\vec{l} = \mu_o I_{enc}

Bw + Bw = \mu_o(LwJ)

B = \tfrac{1}{2}\mu_o LJ \qquad \text{for } \lvert z\rvert > L/2.

To find $\vec{B}$ inside the slab, $z < L/2$ , use the loop shown in Figure b. Once again, the loop segments parallel to the $z$ direction are perpendicular to $\vec{B}$ and will not contribute.

\oint \vec{B}\cdot d\vec{l} = \mu_o I_{enc}

Bw + Bw = \mu_o(2zwJ)

B = \mu_o Jz \qquad \text{for } \lvert z\rvert < L/2.

Including the direction,

\vec{B} = \tfrac{1}{2}\mu_o JL\hat{j} \qquad -L/2 > z,

\vec{B} = -\mu_o Jz\hat{j} \qquad +L/2 > z > -L/2,

\vec{B} = -\tfrac{1}{2}\mu_o JL\hat{j} \qquad z > L/2

b. In the region of the loop, $\vec{B}$ is uniform, $\vec{B} = -\tfrac{1}{2}\mu_o JL\hat{j}$ . The force on the top segment is equal and opposite to the force on the bottom segment. The force on the left segment it equal and opposite to the force on the right segment. The net force on the loop is zero.

\vec{F}_{net} = 0.

The net torque is given by

\vec{\tau}_{net} = \vec{\mu}\times\vec{B}

where $\vec{\mu}$ , the magnetic dipole moment is

\vec{\mu} = IA\hat{n} = Ia^2\left(\cos\theta\hat{i} + \sin\theta\hat{j}\right).

\vec{\tau}_{net} = Ia^2\left(\cos\theta\hat{i} + \sin\theta\hat{j}\right)\times\left(-\tfrac{1}{2}\mu_o JL\hat{j}\right)

\vec{\tau}_{net} = -\tfrac{1}{2}\mu_o JLIa^2\left(\cos\theta\hat{i}\times\hat{j} + \sin\theta\hat{j}\times\hat{j}\right)

\vec{\tau}_{net} = -\tfrac{1}{2}\mu_o JLIa^2\cos\theta\hat{k}

c. $\vec{B}$ is uniform so the flux through the loop is

\Phi = \vec{B}\cdot A\hat{n} = -\tfrac{1}{2}\mu_o JL\hat{j}\cdot a^2\left(\cos\theta\hat{i} + \sin\theta\hat{j}\right)

\Phi = -\tfrac{1}{2}\mu_o JLa^2\sin\theta

Using Faraday’s Law, the emf

\mathcal{E} = -\frac{\Delta\Phi}{\Delta t} = \tfrac{1}{2}\mu_o La^2\sin\theta\frac{\Delta J}{\Delta t}.

The emf is also

\mathcal{E} = IR = \frac{\Delta Q}{\Delta t}R

where

R = (4a)S.

\frac{\Delta Q}{\Delta t} = \frac{1}{2R}\mu_o La^2\sin\theta\frac{\Delta J}{\Delta t} = \frac{1}{8aS}\mu_o La^2\sin\theta\frac{\Delta J}{\Delta t}.

Since $J$ has been reduced to zero over time $T$ , the charge $Q$ flowing in time $T$ is

Q = \frac{\mu_o aLJ}{8S}\sin\theta

Slab of thickness L with a rectangular Amperian loop of width w and height 2z straddling the whole slab; arrows along the top and bottom segments; field B arrow pointing left above the slab and right below it. — Figure a

Slab of thickness L with a rectangular Amperian loop of width w and height 2z lying inside the slab; field arrows B pointing left in the upper half and right in the lower half. — Figure b

Оригинал в Архива: Semifinal 2003.pdf · официални решения: Semifinal 2003 Solutions.pdf