Съдържание

Условие 2A. MAGNETIC FOCUSING SOLENOID (4 points)2B. MAGNETIC FOCUSING (FRINGING FIELD) (6 points)ANSWER FORM ANSWER FORM Решение

APhO 2005, Теория — Задача 2. Question 2 MAGNETIC FOCUSING

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от APhO 2005

6th Asian Physics Olympiad — theoretical competition · 10 т.

Внимание: Бележка към темата

Complete3native whole questions/9requests,14Q+30final-key pages, full derivations, both original swing alternatives, all native rubrics and four blank forms. Printed reversed subpart labels, conflicting swing counts, mirror allocations, field rounding, unnamed-point text and aberrant/overlaid native graphics remain source-attributed and literal.

Native cover and instructions — Wednesday, April 27th, 2005

Please read this first: 1. The time available is 5 hours for the theoretical competition. 2. Use only the pen provided. 3. Use only the front side of the paper. 4. Begin each part of the problem on a separate sheet. 5. For each question, in addition to the blank sheets where you may write, there is an answer form where you must summarize the results you have obtained. Numerical results should be written with as many digits as are appropriate to the given data. 6. Write on the blank sheets of paper whatever you consider is required for the solution of the question. Please use as little text as possible; express yourself primarily in equations, numbers, figures, and plots. 7. Fill in the boxes at the top of each sheet of paper used by writing your Country No and Country Code, your student number (Student No), the number of the question (Question No), the progressive number of each sheet (Page No), and the total number of blank sheets used for each question (Total No of pages). Write the question number and the section letter of the part you are answering at the top of each sheet. If you use some blank sheets of paper for notes that you do not wish to be marked, put a large X across the entire sheet and do not include it in your numbering. 8. At the end of the exam, arrange all sheets for each problem in the following order; ; answer form ; used sheets in order ; the sheets you do not wish to be marked ; unused sheets and the printed question Place the papers inside the envelope and leave everything on your desk. You are not allowed to take any sheets of paper out of the room.

Условие

There exist many devices that utilize fine beams of charged particles. The cathode ray tube used in oscilloscopes, in television receivers or in electron microscopes. In these devices the particle beam is focused and deflected in much the same manner as a light beam is in an optical instrument.

Beams of particles can be focused by electric fields or by magnetic fields. In problem 2A and 2B we are going to see how the beam can be focused by a magnetic field.

2A. MAGNETIC FOCUSING SOLENOID (4 points)

Figure 2.1 shows an electron gun situated inside (near the middle) a long solenoid. The electrons emerging from the hole on the anode have a small transverse velocity component. The electron will follow a helical path. After one complete turn, the electron will return to the axis connecting the hole and point F. By adjusting the magnetic field $B$ inside the solenoid correctly, all the electrons will converge at the same point F after one complete turn. Use the following data:

The voltage difference that accelerates the electrons $V=10\,\mathrm{kV}$
The distance between the anode and the focus point F, $L=0.50\,\mathrm m$
The mass of an electron $m=9.11\times10^{-31}\,\mathrm{kg}$
The charge of an electron $e=1.60\times10^{-19}\,\mathrm C$
$\mu_0=4\pi\times10^{-7}\,\mathrm{H/m}$
Treat the problem non-relativistically

Figure 2.1. Electron gun and focusing solenoid.

a Calculate $B$ so that the electron returns to the axis at point F after one complete turn. [3 т.]

b Find the current in the solenoid if the latter has 500 turns per meter. [1 т.]

2B. MAGNETIC FOCUSING (FRINGING FIELD) (6 points)

Two pole magnets positioned on horizontal planes are separated by a certain distance such that the magnetic field between them be $B$ in vertical direction (see Figure 2.2). The poles faces are rectangular with length $l$ and width $w$ . Consider the fringe field near the edges of the poles (fringe field is field particularly associated to the edge effects). Suppose the extent of the fringe field is $b$ (see Fig. 2.3). The fringe field has two components $B_x\mathbf i$ and $B_z\mathbf k$ . For simplicity assume that $|B_x|=B|z|/b$ where $z=0$ is the mid plane of the gap, explicitly:

when the particle enters the fringe field $B_x=+Bz/b$ ,

when the particle enters the fringe field after traveling through the magnet, $B_x=-Bz/b$ .

Fig. 2.2. Overall view (note that θ is very small).

A parallel narrow beam of particles, each of mass $m$ and positive charge $q$ enters the magnet (near the center) with a high velocity $v$ parallel to the horizontal plane. The vertical size of the beam is comparable to the distance between the magnet poles. A certain beam enters the magnet at an angle $\theta$ from the center line of the magnet and leaves the magnet at an angle $-\theta$ (see Figure 2.4. Assume $\theta$ is very small). Assume that the angle $\theta$ with which the particle enters the fringe field is the same as the angle $\theta$ when it enters the uniform field.

Figure 2.4. Top view; native field icon preserved.

2B The beam will be focused due to the fringe field. Calculate the approximate focal length if we define the focal length as illustrated in Figure 2.5 (assume $b\ll l$ and assume that the $z$ -component of the deflection in the uniform magnetic field $B$ is very small).

[6 т.]

ANSWER FORM

Header fields: Country no; Country code; Student No.; Question No.; Page No.; Total No. of pages.

2A a): $B$ (formula) = ___; $B$ = ___ mT.

2A b): $I$ (formula) = ___; $I$ = ___ ampere.

2B: Focus length (formula) = ___.

Решение

2A. MAGNETIC FOCUSING SOLENOID

The original Questions and Solutions packet repeats the question with this diagram variant.

SOLUTION

a) In magnetic field, the particle will be deflected and follow a helical path.

Lorentz Force in a magnetic field $B$ ,

\frac{mv_\perp^2}{R}=ev_\perp B.\qquad (1)

Where $v_\perp$ is the transverse velocity of the electron, $R$ is the radius of the path.

Since $v_\perp=\omega R$ ( $\omega=2\pi/T$ is the particle angular velocity and $T$ is the period), then,

m\frac{2\pi}{T}=eB.\qquad (2)

To be focused, the period of electron $T$ must be equal to $L/v_\parallel$ , where $v_\parallel$ is the parallel component of the velocity.

We also know,

eV=\frac12m(v_\perp^2+v_\parallel^2)\approx\frac12mv_\parallel^2.\qquad (3)

All the information above leads to

B=\frac{2^{3/2}\pi\sqrt{mV/e}}{L}.\qquad (4)

Numerically $B=4.24\,\mathrm{mT}$ .

b) The magnetic field of the Solenoid:

B=\mu_0in.\qquad (5)

i=\frac B{n\mu_0}.\qquad (6)

Numerically $i=6.75\,\mathrm A$ .

[Marking Scheme] THEORETICAL Question 2A — Magnetic Focusing Solenoid

Part	Marks	Printed item
a. (3.0)	0.3	Equation $mv_\perp^2/R=ev_\perp B$
a.	0.1	Equation $v_\perp=\omega R$
a.	0.1	Equation $\omega=2\pi/T$
a.	0.3	Equation $m(2\pi/T)=eB$
a.	0.2	$T=L/v_\parallel$
a.	0.5	Energy $eV=\frac12m(v_\perp^2+v_\parallel^2)\approx\frac12mv_\parallel^2$
a.	1.0	$B=2^{3/2}\pi\sqrt{mV/e}/L$
a.	0.5	$B=4.23\,\mathrm{mT}$
b. (1.0)	0.5	$B=\mu_0in$
b.	0.3	$i=B/(n\mu_0)$
b.	0.2	$i=6.75\,\mathrm A$

Note: Propagation errors will not be considered.

2B. MAGNETIC FOCUSING (FRINGING FIELD)

The key repeats the complete question with these native diagram variants, followed by the full derivation.

Fig. 2.2. Overall view from the original solution packet.

Figure 2.3. Fringe field from the original solution packet.

SOLUTION

The magnetic force due to the fringe field on charge $q$ with velocity $v$ is

\vec F=q\vec v\times\vec B.\qquad (1)

The $z$ -component of the force obtained from the cross product is

F_z=q(v_xB_y-v_yB_x)=-qv_yB_x=-qv\sin\theta\frac{Bz}{b}.\qquad (2)

The vertical momentum gained by the particle after entering the fringe field

\Delta P_z=\int F_z\,dt=-\frac{qvBz\sin\theta}{b}\Delta t=-\frac{qvBz\sin\theta}{b}\frac b{v\cos\theta}=-qBz\tan\theta.\qquad (3)

The particle undergoes a circular motion in the constant magnetic field $B$ region

\frac{mv^2}{R}=qvB.\qquad (4)

v=\frac{qBR}{m}=\frac{qBl}{2m\sin\theta}.\qquad (5)

Therefore,

\sin\theta=\frac{qBl}{2mv}.\qquad (6)

After the particle exits the fringe field at the other end, it will gain the same momentum.

The total vertical momentum gained by the particle is

\begin{aligned}(\Delta P_z)_{\mathrm{total}}&=2\Delta P_z=-2qBz\tan\theta\\&\approx-2qBz\left(\frac{qBl}{2mv}\right)=-\frac{q^2B^2zl}{mv}.\end{aligned}\qquad (7)

Note that for small $\theta$ , we can approximate $\tan\theta\approx\sin\theta$ .

Meanwhile, the momentum along the horizontal plane (xy-plane) is

p=mv.\qquad (8)

From the geometry in figure 4, we can get the focal length by the following relation,

\frac{|\Delta P_z|}{p}=\frac{|Z|}{f}.\qquad (9)

f=\frac{m^2v^2}{q^2B^2l}.\qquad (10)

[Marking Scheme] THEORETICAL Question 2B — Magnetic Focusing (Fringing Field) (6.0)

Marks	Printed item
0.25	Lorentz force $\vec F=q\vec v\times\vec B$
0.25	$F_z=q(v_xB_y-v_yB_x)$
0.25	$F_z=-qv_yB_x=-qv\sin\theta Bz/b$
0.5	Change of momentum $\Delta P_z=\int F_z\,dt$
0.75	$\Delta P_z=-qBz\tan\theta$
0.5	$mv^2/R=qvB$
0.25	$\sin\theta=qBl/(2mv)$
0.5	$(\Delta P_z)_{\mathrm{total}}=2\Delta P_z$
0.25	$(\Delta P_z)_{\mathrm{total}}=-q^2B^2zl/(mv)$
0.5	$p=mv$
1.0	$\vert \Delta P_z\vert _{\mathrm{total}}/p=\vert Z\vert /f$
1.0	$f=m^2v^2/(q^2B^2l)$

Note: No propagation error will be considered here.

Reader note: original key authority and question/key order

The retained th_solutions.pdf is headed 6th Asian Physics Olympiad, Pekanbaru–Indonesia2005, THEORETICAL COMPETITION, Questions and Solutions, FINAL VERSION and contains matching full worked solutions and marking schemes. The question cover carries an APhO.ORG digital-signature attestation. This transcription uses the retained event-branded final key, preserving the original bytes and printed defects. Native key pages2–6 label Gas Volume as a(3.0) and Small Oscillation as b(2.0); the question and the repeated stem label frequency a(2), volume b(3). The key’s letters/order and the question’s letters/order are both retained.

Reader note: two printed swing alternatives and literal formulas

The key prints Solution1 ending n=10/3 and ALTERNATE SOLUTION ending n=5. Its marking table prints both as paired alternatives. Both complete derivations survive without choosing or silently reconciling the two outcomes. Solution1 Eq2 lacks an equals sign between L_B and the following mr_u² angular-velocity term; Eq4 has a minus potential-energy term. The rubric also prints its own literal energy expressions. These are retained rather than substituted with an inferred correction.

Reader note: literal magnetic-field result and reference

The solenoid calculation prints B=4.24mT while its rubric prints4.23mT. Both values survive. The fringe-field derivation literally refers to figure4; the diagram captions remain2.4 and2.5, and its Eq9 uses ΔP_z while the rubric prints a total subscript. These source forms remain literal.

Reader note: vector glyph correspondence and malformed native graphics

Native question pages10–11 show aberrant vector accents, and Figure2.4 contains an aberrant field icon. Personally viewed MuPDF and retained Poppler previews agree on those forms, so they are not represented as a detected loss peculiar to one renderer. The matching repeated prose on native key pages21–22 prints actual vector arrows; typed vector notation follows that retained source correspondence. All question diagrams and distinct key diagram versions preserve actual original pixels, including the malformed field icon and overlaid key Figures3.1,3.2,3.4. No graphic is redrawn or repaired.

Reader note: literal mirror allocations and unnamed-point gap

The question and repeated stem give3A2.5points,3B2points,3C5.5points; the original marking tables giveA3.0,B2.0,C5.0. Native point allocations remain unchanged. Question3C and its key literally say atoms at point without naming the point; this original gap remains in both transcribed passages. The printed frequency equality0.5 and50% conclusion remain literal rather than replaced by a recomputed value. no further companion is listed in the exact retained Theory archive folder.

Оригинал в Архива: th_problems.pdf · официални решения: th_solutions.pdf