Съдържание

USAPhO 1999, Полуфинал — Задача 6. B2

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 1999

1999 Semi-Final Exam · 40 т.

Условие

Consider the following model of the effect of eddy currents. A rigid wheel, shown in the diagram to the right, consists of ten identical resistors each having resistance $R$ . Five of the resistors are equally spaced around a circumference of radius $r_o$ . The other five resistors form spokes of the wheel with an angle of 72° = 2π/5 radians between adjacent spokes. Assume the resistors have negligible thickness and follow the curvature of the circumference. The wheel rotates about a fixed axis through a circular-sector-wedge-shaped uniform magnetic field $\boldsymbol{B}$ . The sides of the wedge have a length $r_o$ and the angle of the wedge is 72°. The vertex of the wedge coincides with the axis of the rotating wheel. The wheel’s moment of inertia about its center of mass is $I_o$ . At time $t$ , the wheel’s angular velocity is ω(t).

A wheel of radius r_o with five spokes and five rim segments, each containing a resistor (grey rectangles); a curved arrow labelled ω(t) shows the rotation; a shaded 72° circular sector at the bottom, with its vertex at the axis, marks the magnetic field region. — A magnetic field B, directed into the page, exists in the shaded region.

a. Draw a diagram of the wheel which shows the direction and magnitude of the electric current in each of the ten resistors. Express all currents on the diagram as a multiple of the smallest current $I$ . What is $I$ ? [15 т.]

b. Find the total power resistively dissipated by the wheel. [5 т.]

c. Find the angular acceleration of wheel and indicate its direction. [10 т.]

d. Now suppose there are nine equiangular 40° = 2π/9 radian sectors and eighteen identical resistors each having resistance R. Nine resistors are on the spokes and nine are on the circumference between each pair of spokes. The angle of the magnetic field wedge is also reduced to 40°. Repeat Part (a) for this eighteen resistor case. [10 т.]

(contributed by Leaf Turner)

Отговори

Решение

a. Using Faraday's Law

\mathcal{E} = -\frac{d\Phi}{dt} \qquad \text{with} \qquad \Phi = BA.

Let θ be the angle the magnetic field boundary makes with spoke cd shown in the diagram below. Then

A = \left(\frac{\theta}{2\pi}\right)\pi r_o^2 = \left(\tfrac{1}{2}\right)\theta r_o^2

And

\mathcal{E} = -\frac{d}{dt}\left(B\left(\tfrac{1}{2}\right)\theta r_o^2\right) = -\left(\tfrac{1}{2}\right)Br_o^2\omega.

This is the emf induced in the circuit branch cd. By symmetry the currents to the left of cd equal those to the right of cd. $I_5 = 0$ . Since no current branches off at b, $I_4 = I_3 = I$ . The potential drop between point points a and c must be independent of path, so

RI_2 = RI + RI \qquad \text{and} \qquad I_2 = 2I

At point a the total current in must equal the total current out, so

I_1 = 2I + I = 3I.

And finally at point d

I_0 = 3I + 3I = 6I.

For loop acda

\mathcal{E} = -\left(\tfrac{1}{2}\right)Br_o^2\omega = 6IR + 3IR + 2IR = 11IR

I = \frac{1}{22}\frac{Br_o^2\omega}{R}

with the current directions and magnitudes in terms of I shown in the above diagram.

P = \sum I^2R = R\left\{\left(6I\right)^2 + 2\left(3I\right)^2 + 2\left(2I\right)^2 + 4\left(I\right)^2\right\}

P = I^2R\left(36 + 18 + 8 + 4\right) = 66I^2R

P = 66\left(\frac{1}{22}\frac{Br_o^2\omega}{R}\right)^2R = \frac{3}{22}\frac{B^2r_o^4\omega^2}{R}

P = \tau\cdot\omega

τ and ω are in the opposite direction, so

\tau = -\frac{P}{\omega} = -\frac{3}{22}\frac{B^2r_o^4\omega}{R}.

τ is the only torque acting, so

\alpha = \frac{\tau}{I_o} = -\frac{3}{22}\frac{B^2r_o^4\omega}{I_oR}.

Its direction is opposite to ω.

d. The emf determined in Part a. is unchanged. By a treatment similar to Part a., the currents are related as shown in the accompanying diagram. The sequence of current coefficients, 1, 1, 2, 3, 5, 8, 13, 21, are the Fibonacci numbers.

\mathcal{E} = -\left(\tfrac{1}{2}\right)Br_o^2\omega

= 42IR + 21IR + 13IR = 76IR

I = \frac{1}{152}\frac{Br_o^2\omega}{R}

Two wheel diagrams with points a, b, c, b', a', d. Left: currents labelled I0 (spoke cd), I1, I2, I3, I4, I5 with arrows. Right: the same wheel with I5 = 0 and currents I, I, I, I, 2I, 2I, 3I, 3I and 6I in spoke cd.

Wheel with nine spokes and nine rim segments; currents labelled I, I, I, I on the upper rim/spokes, 2I, 2I, 3I, 3I, 5I, 5I, 8I, 8I, 13I, 13I, 21I, 21I and 42I in the spoke inside the field region.

Автори

The 1999 Semi-Final Examination questions were written by the coaches of the United States Physics Team. The coaches are: Academic Director, Dr. Mary Mogge – Professor of Physics at California State Polytechnic University, Pomona, CA; Senior Coach, Dr. Leaf Turner – Physicist in the Theoretical Division of Los Alamos National Laboratory, Los Alamos, NM; Dr. Warren Turner – Physics Teacher at St. Paul's School, Concord, NH. The coaches would also like to thank former academic director, Dr. Larry Kirkpatrick for his many helpful comments.

Оригинал в Архива: Semifinal 1999.pdf · официални решения: Semifinal 1999 Solutions.pdf