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APhO 2022, Теория — Задача 1. The Stern-Gerlach Experiment

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22nd Asian Physics Olympiad, INDIA 2022 — Theory Q1: The Stern-Gerlach Experiment — Official (English)

General Grading Guidelines

When student’s solutions are correct and s/he also show how solutions were obtained, the student gets full credit. The scheme outlined below is helpful if the student’s answers are partially correct. Attention will be paid to the detailed solution so, if the final answer is correct but it is obtained by incorrect method(s) then no credit will be given. Alternative solutions may exist and will be given due credit.

Partial or full outcomes obtained for later sections in the problem which are incorrect solely because of errors being carried forward from previous sections, but are otherwise reasonable, will not be further penalized. For example a dimensionally wrong answer when carried forward will not get any credit in the subsequent sections. A numerically wrong evaluation when carried forward will get credit in subsequent sections unless the numerical answer is patently wrong (e.g. the value of g is 981 m/sec $^2$ ! )

Incorrect or no labeling of an axis is penalized by -0.1 points

The numerical answer (i) must be correct to +/- 10% AND (ii) must respect significant figures.

It maybe noted that NO micro-marking scheme takes care of all contingencies. A certain amount of discretion rests with and a certain level of judgement is invested in the academic committee.

Условие

The Stern-Gerlach experiment was performed in 1922 and it eventually led to the determination of the magnetic moment of the electron. In this experiment a beam of silver atoms each of mass $m = 1.80 \times 10^{-25}$ kg emerges from an oven kept at temperature T= $1.20 \times 10^{3}$ K (see Figure 1). Assume that on emerging from the oven all the atoms have the same momentum along the direction of the beam ( $z$ -direction). Ignore gravity.

Schematic of the Stern-Gerlach setup: an oven labelled “Ag gas”, “Oven, T = 1200K” emits a beam along the z axis through the gap between magnet poles N (top) and S (bottom); axes x (up) and y (dashed, oblique) meet at O; the beam splits into two and strikes the screen PP' at H, the splitting marked Δx; distances l1 (oven to magnets), l2 (magnet length) and l3 (magnets to screen) are marked below. — Figure 1: Schematic diagram of the Stern-Gerlach setup.

A.1 Speed of the Silver Atoms: The speed $v_z$ of the silver atoms emerging from the oven can be estimated to be $\sqrt{3k_BT/m}$ using the equipartition theorem. Calculate this value. [0,5 т.]

B.1 The Basic Expression: After emerging from the oven the silver atoms move along the $z$ - direction over a distance $l_1 = 0.25$ m. Next, the silver atoms pass between two magnets over a distance $l_2 = 0.5$ m. The magnets produce inhomogeneous magnetic field $B$ in the $x$ -direction with constant gradient $dB/dx$ . Assume that the silver atom has a magnetic moment pointing either in + $x$ -direction or - $x$ -direction, i.e. $\vec{\mu}_s = \pm\mu_s\hat{\imath}$ . After passing through the magnets the silver atoms pass through a further distance $l_3 = 0.25$ m before striking the screen $PP'$ . The distance between the two striking beams on the screen is $\Delta x$ . Derive the expression for the splitting distance $\Delta x$ on the screen. [2 т.]

The Inhomogeneous Magnetic Field:

This part is concerned with the setup to create the inhomogeneous magnetic field ( $dB/dx \neq 0$ ). It consists of a number of sub-parts. Two very long wires parallel to the $z$ -axis carry currents of magnitude $I_0$ and are located at $A_1\,(0,-a,z)$ and $A_2\,(0,a,z)$ (see figure below). The direction of the current passing through $y = -a$ is $-\hat{k}$ and for the one passing through $y = a$ is $\hat{k}$ . The entire system is inside a medium of high relative magnetic permeability $\mu_r$ . We take $\mu$ = $\mu_0\mu_r$ . The wires are insulated and no current leaks into the medium.

x-y plane with origin O: wire A2 at (0, a) carrying current I = I0 k̂ and wire A1 at (0, −a) carrying I = −I0 k̂; points C and D on the x axis joined to A1 and A2; two circles through A1 and A2 centred at C and D, cutting the x axis at R and Q; point P between R and Q with an arrow labelled “Air”; point P0 (xc, yc) on the smaller circle above C (dashed line CP0); point P1(x, y) in the upper right. — Figure 2: The arrangement for the inhomogeneous magnetic field

C.1 Derive the expression for the magnetic field vector at a point $P_1\,(x,y,0)$ in the $x$ - $y$ plane (see Fig. 2). [1,5 т.]

C.2 Consider a circle with its centre C on the $x$ -axis ( $x_c, 0$ ) and with radius equal to the distance $AC$ (see Figure 2). Obtain the direction of the magnetic field at circumferential points $R$ (on the $x$ axis) and at $P_0$ ( $CP_0$ is parallel to the $y$ axis). [0,5 т.]

C.3 Now a small slice of the high magnetic permeability medium between the circles with radii AC and AD is removed and replaced with air at very low pressure (see Fig. 2). One can show from continuity considerations that the magnetic field in this gap region is given by the same expression as if the magnetic medium were not removed. (We shall assume this and you are not required to prove this). Hence state the expression for the magnetic field at the point $(x, 0)$ in the air gap region. [0,5 т.]

D.1 The Force: As mentioned earlier the silver atoms are travelling in the $(x, 0, z)$ plane with their velocities parallel to the z-axis and given by $\vec{v} = v_z\hat{k}$ . Recall also that the magnetic dipole of the silver atom is $\vec{\mu}_s = \pm\mu_s\hat{\imath}$ . Obtain the expression for the magnitude of the force $F_x$ acting on a silver atom along the x-direction in terms of $\mu_s$ , $I_0$ , $a$ $\mu$ and relevant coordinates. [0,5 т.]

E.1 The Magnetic Field and its Gradient: We assume that this same force $F_x$ acts over a small distance $l_2$ along the $z$ axis (Figure 1). Assume also that the silver atoms pass through the mid-point $P$ of $RQ$ (Figure 2). The following experimental values are given:

\frac{\mu}{\mu_0} = 10^4\,; \quad a = 0.60\,cm\,; \quad OC = 0.60cm\,; \quad OD = 0.80\,cm\,; I_0 = 2.00A

Here $\mu_o$ is the magnetic permeability of free space. Obtain the numerical value of the magnitudes of the magnetic field $B_P$ and its gradient $dB_P/dx$ at this mid-point in S.I. units. [2 т.]

F.1 The Magnetic Moment of the Silver Atom: If $v_z = 500\,\mathrm{m{\cdot}s^{-1}}$ and magnetic field is calculates as above, the Stern-Gerlach experiment yields a split of $\Delta x = 0.20$ cm. Obtain the value of the magnetic moment of the silver atom $\mu_s$ in S.I. units. [1,5 т.]

G.1 The spread of the line: The silver atoms may not all have the same speed. Let there be a spread of 20 % in the beam speed. What would be the consequent spread of the dot $\delta x$ on the screen? [0,5 т.]

H.1 The error in the magnetic moment: What is the consequent error bar on the evaluation of the magnetic moment $\delta\mu_s$ ? [0,5 т.]

Отговори

Решение

The Stern-Gerlach Experiment: THE SOLUTION

A.1 Speed of the Silver Atoms: (0.5pt)

We employ the equipartition theorem. Let $\overline{v^2}$ be the mean square speed of the silver atoms in the oven kept at 1200 K. Then

\frac{m\overline{v^2}}{2} = \frac{3k_BT}{2}

where $k_B$ is the Boltzmann constant. This yields the root mean square speed to be $5.255\times10^{2}$ m·s $^{-1}$ .

[0.5]

B.1 The Basic Expression (2pt)

The length $l_1$ is irrelevant and will not be part of the expression.
The magnitude of the acceleration $a$ of the silver atoms in the region defined by $l_2$ is

a = \frac{\mu_s}{m}\frac{dB}{dx}

[0.4]

and it will be either in the + $x$ or - $x$ direction. At the same time it has a constant horizontal velocity $v_z$ . It traverses the region $l_2$ in time $l_2/v_z$ . Thus after traversing the inhomogeneous region the deflection in say the + $x$ direction is

\delta_1 = \frac{1}{2}\frac{\mu_s}{m}\frac{dB}{dx}\frac{l_2^2}{v_z^2}

[0.6]

For the remaining part of the flight the atom will have a constant horizontal speed $v_z$ and a constant vertical speed $v_{x0} = (\mu_s dB/dx)\,(l_2/mv_z)$ . On account of the $v_x$ component the atom will acquire an additional deflection

\delta_2 = l_3 v_{x0}/v_z

This yields

\delta_2 = l_3 l_2\frac{\mu_s}{m v_z^2}\frac{dB}{dx}

[0.4]

The total deflection in the + $x$ direction is $\delta_1 + \delta_2$ . The splitting seen on the screen in this idealized case is twice this amount, e.g. 2 $(\delta_1 + \delta_2)$ . Thus we obtain

\Delta x = 2\frac{\mu_s}{m}\frac{dB}{dx}\frac{l_2}{v_z^2}(l_2/2 + l_3)

-0.3 if factor of 2 is missing.

[0.6]

C.1 The Inhomogeneous Magnetic Field: (1.5pt)

Sketch for C.1: a vertical axis with wire A1 at (0,a) and wire A2 at (0,-a), a horizontal axis arrow from the origin, and the point P at (x,y) in the upper right.

Let $\vec{A_1P} = \vec{r}_1 = x\hat{i} + (y - a)\hat{j}$ and $\vec{A_2P} = \vec{r}_2 = x\hat{i} + (y + a)\hat{j}$ . This gives for $\vec{B}(x,y)$

\frac{\mu I_0}{2\pi}\left[\frac{\hat{k}\times(x\hat{i} + (y - a)\hat{j}}{r_1^2} - \frac{\hat{k}\times(x\hat{i} + (y + a)\hat{j}}{r_2^2}\right] \qquad (1)

[0.4+0.4]

= \frac{\mu I_0}{2\pi r_1^2 r_2^2}\left[(x\hat{j} - (y - a)\hat{i})(x^2 + (y + a)^2) - (x\hat{j} - (y + a)\hat{i})(x^2 + (y - a)^2)\right]

= \frac{\mu I_0\, a}{\pi r_1^2 r_2^2}\left[2xy\hat{j} + (x^2 - y^2 + a^2)\hat{i}\right] \qquad (2)

[0.7]

Writing the final expression as any correct function of x and y will get full marks. If collecting all the terms component-wise not done, then penalize by -0.1. If an error has been made in simplification then penalize by -0.1.

C.2 Direction at point $R$ : (0.5pt) Field at the point $R$ ( $(x_c + \sqrt{x_c^2 + a^2}, 0)$ is given by substituting $y = 0$ . On simple inspection the $\hat{j}$ component vanishes. Thus $\vec{B}(x, 0) \propto \hat{i}$

[0.2]

Direction at point $P_0$ : Field at $P_0$ ( $(x_c, y_c = (x_c^2 + a^2)^{1/2})$ ) is given, using Eq.(2)

\frac{\mu I_0}{\pi r_1^2 r_2^2}(2x_c(x_c^2 + a^2)^{1/2}\hat{j} + (x_c^2 - x_c^2 - a^2 + a^2)\hat{i})

The $\hat{i}$ component is zero. Thus $\vec{B}(x_c, (x_c^2 + a^2)^{1/2}) \propto \hat{j}$

[0.3]

First Alternative Solution

We can show in general that the field at any point on the circle will be radial (i.e. normal to the circle). We will confine our discussion to the z=0 plane. Consider a point $(x_c, y)$ with radius $\sqrt{x_c^2 + a^2}$ . The equation of a circle with $(x_c, 0)$ as centre and $\sqrt{x_c^2 + a^2}$ as radius is

(x - x_c)^2 + y^2 = x_c^2 + a^2

x^2 - 2\,x\,x_c + y^2 = a^2 \qquad (3)

If at the point $(x_c, y_c)$ the magnetic field is along $\hat{j}$ then the component along $\hat{i}$ is zero. $(x_c, 0)$ is identified with the point C on the figure. The point $y_c$ is then,

x_c^2 - y_c^2 + a^2 = 0

y_c^2 = x_c^2 + a^2 \qquad (4)

Now consider a line joining C $(x_c, 0)$ to any point $P_C(x, y)$ lying on the circle given by eq.(3). The radial vector is $\vec{CP}_C = (x - x_c)\hat{i} + y\hat{j}$ . The magnetic field at $P_C$ is

\propto \vec{B}(x, y, 0) = \left(\frac{\mu I_0}{\pi}\right)\left(2xy\hat{j} + (x^2 - y^2 + a^2)\hat{i}\right)

To show that they are in the same direction, we evaluate the cross product, $\vec{CP}_C\times\vec{B}$ . The cross product is proportional to $\hat{n}$ which is a unit vector along the direction which is normal to both $CP_C$ and $\vec{B}$ and is along $\hat{k}$

\vec{CP}_C\times\vec{B} \propto (2\,x\,y(x - x_c) - y(x^2 - y^2 + a^2))\,\hat{k}

which simplifies to

y(x^2 - 2x\,x_c + y^2 - a^2)\hat{k}

Using eq.(3), this is zero, proving the result.

Second Alternative Solution

To show that the field lines are radial over the circle one may merely show the proportionality of the components of the field and the radius vector. The radius vector is $(x - x_c)\hat{i} + y\hat{j}$ while the magnetic field is proportional to $(x^2 - y^2 + a^2)\hat{i} + 2xy\hat{j}$ . Thus

\frac{y}{2xy} = \frac{1}{2x}

and

\frac{x - x_c}{x^2 - y^2 + a^2} = \frac{1}{2x}

The last step is obtained by observing that the equation of the circle is $(x - x_c)^2 + y^2 = x_c^2 + a^2$ .

C.3 (0.5pt) Field in the airgap because of the argument presented in the problem continues to be given by Eq.(2). So the field ( $y = 0$ ), is again

\vec{B} = \frac{\mu I_0\, a}{\pi(x^2 + a^2)}\hat{i}

[0.5]

D.1 (0.5pt) The force $F_x$ on a magnetic dipole along the $x$ - direction is

F_x = -\mu_s\frac{\partial B_x}{\partial x} = \frac{\mu_s\mu\, I_0}{\pi}\times\frac{2\,a\,x}{(x^2 + a^2)^2} \qquad (5)

[0.5]

E.1 (2.0pt)

\frac{\mu}{\mu_0} = 10^4; \qquad a = 6.00\times10^{-3}m; \qquad OC = 6.00\times10^{-3}m; \qquad OD = 8.00\times10^{-3}m;

and

I_0 = 2.00\,A

and so at the midpoint P,

y = 0;

x_P = OP = ((1 + \sqrt{2})\times .6 + 1.8)/2 = 1.624\times10^{-2}m

[0.5]

where we have used $OD = .8\times10^{-2}m$ and $DA = 10^{-2}m$ . This gives for $B_x(x_P, 0)$

\frac{\mu}{\mu_0}\frac{\mu_0}{\pi}\frac{I_0\,a}{(x_P^2 + a^2)} = \frac{10^4\times4\times10^{-7}\times2\times.6\times10^{-2}}{(1.624^2 + .6^2)\times10^{-4}}

= 0.16\,T

[1]

We also have

\left(\frac{\partial B_x}{\partial x}\right)_{x_P} = \frac{2\times x_p}{x_p^2 + a^2}\times B_x(x_P, 0) = \frac{2\times1.624\times10^{-2}}{(1.624^2 + .6^2)\times10^{-4}}\times .16 = 17.34\,T\cdot m^{-1}

[0.5]

F.1 The magnetic moment of the silver atom: (1.5pt)

We use

\Delta x = \frac{2\mu_s}{m}\left(\frac{\partial B}{\partial x}\right)_{x_P}\frac{l_2}{v_z^2}\left(\frac{l_2}{2} + l_3\right)

to rewrite

\mu_s = \frac{m\Delta x}{2(\frac{\partial B_x}{\partial x})_{x_P}}\times\frac{1}{[\frac{l_2}{v_z^2}(\frac{l_2}{2} + l_3)]}

[0.5]

= \frac{1.8\times10^{-25}\times2\times10^{-3}}{2\times17.34}\times10^{6} = 1.04\times10^{-23}J\cdot T^{-1}

[1]

G.1 The spread in the line: (0.5pt) The two lines on the screen are separated symmetrically about the centre by $\Delta x$ . So the upper (lower) line is at $\Delta x/2$ from the centre. From Part (2)

\Delta x/2 = \frac{\mu_s}{m}\frac{dB}{dx}\frac{l_2}{v_z^2}(l_2/2 + l_3)

This depends on the beam speed $v_z$ . The spread in this speed leads to a consequent spread in the splitting.

\begin{aligned} \delta(\Delta x/2) &= |\frac{\partial\Delta x/2}{\partial v_z}|\delta v_z \\ &= 2(\Delta x/2)\frac{\delta v_z}{v_z} \\ &= 2(\Delta x/2)\times0.2 \\ &= 0.04cm \end{aligned}

[0.3]

Hence the spread in the line from the centre is 0.1 - 0.04 = 0.06 cm to 0.1 + 0.04 = 0.14 cm.

Credit will also be given if 20% is interpreted as 10% on each side
Answer reported in terms of percentages receive full credit

[0.2]

H.1 Error in the evaluation of the magnetic moment: (0.5pt)

From the previous part we have that the splitting ranges from 0.12 cm to 0.28 cm whereas earlier it was 0.2 cm. The relationship between the splitting and the magnetic moment is linear. So the magnetic moment ranges from (0.12/0.2) to (0.28/0.2) the original value. This yields 0.62 $\times$ 10 $^{-23}$ J·T $^{-1}$ to 1.46 $\times$ 10 $^{-23}$ J·T $^{-1}$ . The total spread is 0.84 $\times$ 10 $^{-23}$ J·T $^{-1}$ about the mean value of 1.04 $\times$ 10 $^{-23}$ J·T $^{-1}$

[0.3]

or in other words

\mu_s = 1.04 \pm 0.42\,\mathrm{J{\cdot}T^{-1}}

[0.2]

Repeat of Figure 2 in the solution: wires at A (current I = I0 k̂) and B (current I = −I0 k̂) on the y axis, points C and D on the x axis, two circles through A and B cutting the x axis at R and Q, point P with an arrow labelled “Air”, point P0 (xc, yc) and P1(x, y).

Бележка под линия

H. S. Mani (former Director, HRI, Prayagraj) and Gautam Datta (DAIICT, Gandhinagar) were the principal authors of this problem. The contributions of the Academic Committee, Academic Development Group, and the International Board are gratefully acknowledged.

Оригинал в Архива: theory-Q1-question.pdf · официални решения: theory-Q1-solution.pdf