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IZhO 2013, Теория, английски — Задача 2. Square frame

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Всички материали от IZhO 2013

IX International Zhautykov Olympiad/Theoretical Competition · 15 януари 2013 г. · 10 т.

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The time available for the theoretical competition is 4 hours. There are three questions.
Use only the pen provided.
You can use your own calculator for numerical calculations. If you don’t have one, please ask for it from Olympiad organizers.
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Begin each question on a separate sheet of paper.
Fill the boxes at the top of each sheet of paper with your country (Country), your student code (Student Code), the question number (Question Number), the progressive number of each sheet (Page Number), and the total number of Writing sheets used (Total Number of Pages). If you use some blank Writing sheets for notes that you do not wish to be evaluated, put a large X across the entire sheet and do not include it in your numbering.
At the end of the exam, arrange all sheets for each problem in the following order:
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Place the papers inside the envelope and leave everything on your desk. You are not allowed to take any paper out of the room.

Условие

A square frame of side $a=10.0\ sm$ is placed in a uniform magnetic field of $B=1.00\times10^{-1}\ T$ generated by a permanent magnet. Completely neglect edge effects, assuming that the magnetic field between the poles of the magnet is uniform and is equal to zero outside of it. In all parts of the problem assume that at the initial time moment the frame of mass $m=5.00\ g$ is at rest at the edge of the poles of the magnet, see figure.

1. Redraw schematically the figure on the right to identify which of the poles of the magnet is the north $N$ and which is the south $S$ . [0,2 т.]

Part A

Two grey magnet blocks one above the other; between the poles the vertical field lines B point downward through a square frame, and a horizontal arrow F to the right pulls the frame out of the gap.

Let the frame be made of a conductive material with the resistance $R=1.00\times10^{-2}$ ohm and a negligible inductance. At the time moment $t=0$ the frame is subject to the constant force $F=1.00\times10^{-4}\ N$ , which pulls the frame out of the magnetic field.

2. Find an analytical dependence of the frame velocity $v(t)$ on time $t$ assuming that the frame remains at all times between the poles of the magnet. Express your answer in terms of $B,a,m,R$ and $t$ . [1,2 т.]

3. Write down an exact equation to determine the time interval $t_0$ at which the frame moves between the poles of the magnet. Express your answer in terms of $B,a,m,R$ and $t_0$ . Estimate $t_0$ . [0,8 т.]

4. Draw a graph of the frame velocity $v(t)$ on time $t$ in the interval from 0 to 12s. [1,2 т.]

5. Draw a graph of the frame current $I(t)$ on time $t$ in the interval from 0 to 12s. [1,2 т.]

Part B

Let the frame be made of a superconducting material with the inductance of $L=1.00\times10^{-1}\ H$ , the same mass and sizes. At the time moment $t=0$ s the frame is subject to a constant force which pulls the frame out of the magnetic field.

6. Find the minimum force $F_{\min}$ with which the frame can be pulled out of the magnet. Express your answer in terms of $B,a,L$ and calculate its numerical value. [1,8 т.]

7. At the assumptions of section 6 find the minimum time interval $t_0$ of the maximum departure of the frame out of the magnet. Express your answer in terms of $m,B,a,L$ and calculate its numerical value. [0,6 т.]

8. Draw a graph of the frame current $I(t)$ on time $t$ in the interval from 0 to $4t_0$ . [1 т.]

Part C

A square frame with sides $a$ is made of a conductive material with the inductance $L$ , the resistance $R$ and the mass $m$ . At the time moment $t=0$ the frame is given an initial velocity $v_0$ directed exactly as the force in the figure above. It is known that at the time moment when the frame almost leaves the magnet its velocity turns zero.

9. Find an analytical expression for the frame current $I_0$ at the time moment when the frame velocity turns zero. Express your answer in terms of $B,a,m,L,R$ and $v_0$ . [2 т.]

Отговори

Решение

Square frame

In physics, it is agreed that the magnetic field lines begin at the north pole and ends at the South pole. Therefore, the drawing should look like this

The magnet figure redrawn with the upper pole marked N and the lower pole marked S; the field lines B point down from N to S through the frame, the force F points to the right.

When the frame is removed from the uniform magnetic field, the induced emf can be found from the Faraday law

\varepsilon=-\frac{d\Phi}{dt}=-\frac{d}{dt}(Bax)=Bav(t). \qquad (1)

On the other hand Ohm’s law is written as

\varepsilon=IR, \qquad (2)

Thus, we get the relation between the current and the velocity

I(t)=\frac{Bav(t)}{R}. \qquad (3)

The frame is affected by the force that pulls the frame back into the magnetic field. It is found from Ampere's law

F_A(t)=BaI(t)=\frac{B^2a^2v(t)}{R}. \qquad (4)

Thus, the equation of motion is written as

m\frac{dv(t)}{dt}=F-\frac{B^2a^2v(t)}{R}. \qquad (5)

Solution of equation (5) with the initial condition $v(0)=0$ is

v(t)=\frac{FR}{B^2a^2}\left[1-\exp\left(-\frac{B^2a^2}{mR}t\right)\right]. \qquad (6)

Integrating (6) we get following relation

x(t)=\int_0^t v(t)dt=\frac{FR}{B^2a^2}\left[t+\frac{mR}{B^2a^2}\left(\exp\left(-\frac{B^2a^2}{mR}t\right)-1\right)\right]. \qquad (7)

When the frame leaves the magnetic field

x(t_0)=a, \qquad (8)

Thus, we get an equation for $t_0$

t_0+\frac{mR}{B^2a^2}\left[\exp\left(-\frac{B^2a^2}{mR}t_0\right)-1\right]=\frac{B^2a^3}{FR}. \qquad (9)

Equation (9) is transcendental and cannot be solved analytically. To estimate $t_0$ we can see from equation (6) that for the characteristic time $\tau\sim mR/B^2a^2$ the frame reaches the steady velocity $v_0=FR/B^2a^2$ . We assume that from 0 to $\tau$ the frame moves with the constant acceleration $w=F/m$ , then it moves with the steady velocity $v_0$ . Hence, we get an estimate

t_0\sim\tau+\frac{a-\frac{wt^2}{2}}{v_0}=\frac{B^2a^3}{FR}+\frac{mR}{2B^2a^2}=10.25\ \text{с}. \qquad (10)

Note that numerical solution of equation (9) gives $t_0\approx10.5$ с.

After the $t_0$ the frame continues its motion with the constant acceleration $w$ . At the same time the speed should be a continuous function of time, so the time dependence is written as

v(t)=\begin{cases}\frac{FR}{B^2a^2}\left[1-\exp\left(-\frac{B^2a^2}{mR}t\right)\right], & t<t_0\\ \frac{FR}{B^2a^2}+\frac{F}{m}(t-t_0), & t\ge t_0\end{cases}. \qquad (11)

The corresponding graph is plotted as

Graph of v, m/s (axis label printed as „v,m/a“, 0.00 to 0.04) against t, s (0 to 12): the velocity rises exponentially to 0.01 within about 2 s, stays constant until about 10.3 s, then grows linearly to about 0.045 at 12 s.

While the frame is between the magnetic poles, the current is determined by the equations (3) and (6). After that, the frame current vanishes instantaneously. Thus,

I(t)=\begin{cases}\frac{F}{Ba}\left[1-\exp\left(-\frac{B^2a^2}{mR}t\right)\right], & t<t_0\\ 0, & t\ge t_0\end{cases}. \qquad (12)

The corresponding graph is plotted as

Graph of I, A (0.000 to 0.010) against t, s (0 to 12): the current rises exponentially to 0.010 within about 2 s, stays constant until about 10.3 s and then drops abruptly to zero.

As in the previous part, when the frame is removed from the constant magnetic field, the emf (1) is induced. There is another emf appearing due to the self-induction of the superconductive frame

\varepsilon_L=-L\frac{dI}{dt}. \qquad (13)

Since the resistance of the superconductive frame is zero, Ohm's law for the frame becomes

Bav(t)-L\frac{dI}{dt}=0. \qquad (14)

Taking into account that $I=0$ when $x=0$ we get from equations (13) and (14)

I=\frac{Bax}{L}. \qquad (15)

The corresponding force is given by

F_A=BaI=\frac{B^2a^2x}{L}. \qquad (16)

Thus, the equation of motion is written as

m\frac{d^2x}{dt^2}=F-\frac{B^2a^2}{L}x. \qquad (17)

Expression (17) is an equation of simple harmonic oscillations with the frequency

\omega=\frac{Ba}{\sqrt{mL}}, \qquad (18)

that are performed near the new equilibrium position with the coordinate

x_0=\frac{FL}{B^2a^2}. \qquad (19)

Obviously, the force $F$ is minimal when

x_0=a/2, \qquad (20)

whence

F_{\min}=\frac{B^2a^3}{2L}=5.00\times10^{-5}\ \text{Н}. \qquad (21)

From previous section 6, the frame reaches the edge of the magnet for a half period of oscillations, thus

t_0=\frac{\pi}{\omega}=\pi\frac{\sqrt{mL}}{Ba}=7.02\ \text{с}. \qquad (22)

Solution of equation (17) with the initial conditions $x(0)=0, x'(0)=0$ is written as

x(t)=\frac{F_{\min}L}{B^2a^2}(1-\cos\omega t)=\frac{a}{2}(1-\cos\omega t). \qquad (23)

According to equation (15) the frame current varies as

I(t)=\frac{Bax(t)}{L}=\frac{Ba^2}{2L}(1-\cos\omega t). \qquad (24)

The corresponding graph is plotted as

Graph of I, A (0.000 to 0.010) against t, s (0 to 28): the current oscillates as 1 − cos between 0 and 0.010 with two full periods, maxima near 7 s and 21 s, minimum near 14 s.

Ohm's law for the frame is written as

Bav-L\frac{dI}{dt}=IR, \qquad (25)

and its equation of motion is as follows

m\frac{dv}{dt}=-BIa. \qquad (26)

Equations (25) and (26) can be rewritten in finite differences as follows

Ba^2-LI_0=qR, \qquad (27)

mv_0=Bqa, \qquad (28)

where $q$ is the charge flown through the circuit.
Solving (27) and (28) together, we obtain

I_0=\frac{B^2a^3-mv_0R}{aBL}. \qquad (29)

Marking scheme

№	Description	points
1	Properly set northern N and southern S poles	0.2	0.2
2	Eq (1)	0.2	1.2
	Eq (2)	0.2
	Eq (3)	0.2
	Eq (4)	0.2
	Eq (5)	0.2
	Solution (6)	0.2
3	Eq (7)	0.2	0.8
	Eq (8)	0.2
	Eq (9)	0.2
	Numerical value of $t_0$	0.2
4	Eq (11)	0.2	1.2
	Graph of $v(t)$ : axis signed and digitized	0.2
	Graph of $v(t)$ : there is a part with a constant velocity	0.2
	Graph of $v(t)$ : there is a part with constant acceleration	0.2
	Graph of $v(t)$ : continuous	0.2
	Graph of $v(t)$ : correct numerical values	0.2
5	Eq (12)	0.2	1.2
	Graph of $I(t)$ : axis signed and digitized	0.2
	Graph of $I(t)$ : there is a part with constant current	0.2
	Graph of $I(t)$ : there is a part with zero current	0.2
	Graph of $I(t)$ : discontinuous	0.2
	Graph of $I(t)$ : correct numerical values	0.2
6	Eq (13)	0.2	1.8
	Eq (14)	0.2
	Relationship (15)	0.2
	Quasi-elastic force (16)	0.2
	Eq of motion (17)	0.2
	Equilibrium point (19)	0.2
	Condition (20)	0.2
	Eq (21) for $F_{\min}$	0.2
	Numerical value $F_{\min}$	0.2
7	Frequency (18)	0.2	0.6
	Eq (22)	0.2
	Numerical value of $t_0$	0.2
8	Eq (23)	0.2	1.0
	Eq (24)	0.2
	Graph of $I(t)$ : axis signed and digitized	0.2
	Graph of $I(t)$ : two periods of current oscillation	0.2
	Graph of $I(t)$ : correct value of amplitude	0.2
9	Ohm’s Law (25)	0.2	2.0
	Eq of motion (26)	0.2
	Eq (27)	0.7
	Eq (28)	0.7
	Expression (29)	0.2
Total			10,0

Оригинал в Архива: IZhO_2013_Theory_eng_fiz.doc · официални решения: IZhO_2013_Theory_eng_sol_fiz.doc