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IZhO 2026, Теория — Задача 3. Electron Paramagnetic Resonance

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XXII International Zhautykov Olympiad/Theoretical Competition · 11 януари 2026 г. · 10 т.

THEORETICAL COMPETITION

January 11, 2026

Please read this first:

The time available for the theoretical competition is 4 hours. There are three questions.
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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Use only the front side of Writing sheets. Write only inside the boxed area.
Each task should be solved starting from a new page of the Writing sheets
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Условие

Magnetic Moment

The magnetic moment $\boldsymbol{m}$ of a planar current loop is a vector quantity defined as the product of the electric current $I$ , the area of the loop $S$ , and the unit normal vector $\mathbf{n}$ perpendicular to the plane of the loop:

\boldsymbol{m} = IS\mathbf{n}.

The direction of the magnetic moment vector is determined by the right-hand (corkscrew) rule: if the fingers of the right hand curl in the direction of the current in the loop, the thumb points in the direction of $\boldsymbol{m}$ .

3.1 A circular loop of radius $R$ carries an electric current and has a magnetic moment $\boldsymbol{m}$ . Find the magnetic field induction $\boldsymbol{B}_0$ at the center of the loop.

3.2 The same loop is placed in an external uniform magnetic field of induction $\boldsymbol{B}$ such that the magnetic moment vector makes an angle $\varphi$ with the direction of $\boldsymbol{B}$ . Find the magnitude of the mechanical torque $\boldsymbol{M}$ acting on the loop due to the external magnetic field.

3.3 The loop is slowly rotated in an external uniform magnetic field $\boldsymbol{B}$ so that the direction of its magnetic moment $\boldsymbol{m}$ changes from being aligned with the field ( $\boldsymbol{m}\uparrow\uparrow\boldsymbol{B}$ )) to being opposite to the field ( $\boldsymbol{m}\uparrow\downarrow\boldsymbol{B}$ ). Find the mechanical work $A$ done by the magnetic field during this rotation.

Electron Paramagnetic Resonance

If a charged particle rotates or moves along a closed trajectory, it possesses a mechanical (angular) momentum, and as a result a magnetic moment also arises. In this case, a universal gyromagnetic relation holds, which shows how the magnetic moment and the mechanical moment of the particle are related.

Suppose an electron in an atom moves along a circular orbit such that its orbital angular momentum with respect to the center is $\boldsymbol{L}$ . Such motion can be regarded as equivalent to a circular electric current, which has a magnetic moment $\boldsymbol{m}$ . The magnetic moment and the angular momentum are related by the gyromagnetic relation

\boldsymbol{m} = -g_L \frac{e}{2m_e}\boldsymbol{L},

where $e$ is the elementary charge, $m_e$ denotes the electron mass, and $g_L$ stands for the so-called Landé $g$ -factor.

3.4 Find the Landé factor $g_L$ for the circular orbital motion of an electron.

Electron Paramagnetic Resonance (EPR) is a phenomenon in which a substance containing unpaired electrons absorbs electromagnetic radiation (usually in the microwave range) when placed in a constant magnetic field. This absorption does not occur at arbitrary frequencies but only at a strictly defined one, and is therefore called resonant.

Assume that an atom has a single unpaired electron in its outer shell with zero orbital angular momentum. Such an electron has an intrinsic angular momentum called spin. When a sample of the substance is placed in a constant magnetic field produced by a solenoid with magnetic induction $B$ , the spin can be oriented in two ways relative to the field: along the magnetic field direction with angular momentum projection $+\hbar/2$ , or opposite to it with projection $-\hbar/2$ . These two orientations have different energies, so transitions between them are possible. In what follows, assume that the Landé factor for the spin, $g_s$ , is twice that for the electron's orbital motion. The sample is irradiated with an electromagnetic wave of fixed angular frequency $\omega$ , which can induce transitions of the electron between the two states with different spin projections. Then the magnitude of the external magnetic field induction is slowly varied, while the change in the absorption intensity of the electromagnetic radiation is recorded.

3.5 Find the angular frequency $\omega$ of the external electromagnetic radiation and calculate its numerical value if the maximum absorption occurs at a magnetic field induction $B_0 = 350$ mT.

Now atoms of the same type are embedded in an unknown material from which the core of the solenoid is made.

3.6 The magnetic field specified in the previous part was achieved at a current $I_0 = 1.50$ A in the solenoid winding. Determine the new current $I$ in the solenoid winding required for resonant absorption if the magnetic permeability of the unknown material is $\mu = 1.25$ .

Thermodynamic equilibrium

Assume that the core placed in the same magnetic field is in a state of thermodynamic equilibrium at a temperature $T = 50$ K, and the total number of embedded atoms is $N = 1.00 \times 10^{20}$ . Let $n_0$ denote the difference between the number of atoms occupying the lower ( $N_1$ ) and the upper ( $N_2$ ) energy levels.

3.7 Calculate $n_0$ under the given conditions.

When electromagnetic radiation interacts with matter, three processes occur:

Absorption: an atom transitions from a lower energy level $E_1$ to a higher energy level $E_2$ by absorbing a photon. The number of transitions from the lower level per unit time is given by

\frac{dN_1}{dt} = -B_{12}\rho N_1,

where $\rho$ is the energy density of the electromagnetic radiation; 2) Stimulated emission: under the influence of an external photon an atom transitions from a higher energy level $E_2$ to a lower energy level $E_1$ with the emission of another photon. The number of transitions from the upper level per unit time is given by

\frac{dN_2}{dt} = -B_{21}\rho N_2;

Spontaneous transition: a spontaneous transition of an atom from the upper level to the lower level accompanied by the emission of a photon. The number of such transitions from the upper level per unit time is given by

\frac{dN_2}{dt} = -A_{21}N_2.

The constants $B_{12}, B_{21}, A_{21}$ are called Einstein's coefficients. Planck showed that in a state of thermodynamic equilibrium, the energy density of equilibrium electromagnetic radiation is described by the formula:

\rho = \frac{2\hbar\omega^3}{\pi c^3}\frac{1}{\exp\!\left(-\frac{\hbar\omega}{k_B T}\right)-1}.

3.8 Prove that $B_{12} = B_{21}$ .

Presence of an external microwave field source

At the initial moment of time, the system is in thermodynamic equilibrium at the temperature specified above. Then a source of microwave radiation is switched on in such a way that the energy density of the electromagnetic radiation $\rho$ in the sample remains constant in time, and its magnitude is such that spontaneous transitions in the system can be neglected.

3.9 Find the analytical time dependence $n(t)$ of the difference between the numbers of atoms occupying the lower and the upper energy levels, as a function of time $t$ , assuming that $k = B_{12}\rho$ .

3.10 It is known that the difference between the numbers of atoms occupying the lower and the upper energy levels changes by exactly a factor of 2 after a time $\tau = 1.00$ s from the moment the source is switched on. Under these conditions, calculate the power $P$ of the microwave radiation source at the initial moment of time.

In reality, absorption and stimulated emission are not the only processes by which an electron in the upper energy level loses its excess energy. Relaxation processes play an important role, in which the excess energy is transferred to the surrounding matter; it is precisely due to these processes that an equilibrium distribution over energy levels is established.

The relaxation process for level 1 can be described as continuous transitions from level 1 to level 2 and back; the same is true for level 2. The terms describing relaxation for level 1 are written via constants $\alpha_1$ and $\alpha_2$ as

\frac{dN_1}{dt} = -\alpha_1 N_1 + \alpha_2 N_2,

and, consequently, a similar relation can be put down for level 2.

3.11 For the given substance, let $\alpha_1 + \alpha_2 = 0.670 \cdot 10^{-3}\ \mathrm{s^{-1}}$ . Under these conditions, calculate the power of the microwave radiation source in the steady-state regime of spectrum measurement using the electron paramagnetic resonance method.

Решение

Magnetic Moment

3.1 The area of a circular coil of radius $R$ is equal to

S = \pi R^2 \qquad (1)

and when current $I$ flows through it, the modulus of its magnetic moment is equal to

m = I\pi R^2. \qquad (2)

The magnetic induction at the center of a circular loop is determined by the expression

B_0 = \frac{\mu_0 I}{2R}, \qquad (3)

from which it follows that

\boldsymbol{B}_0 = \frac{\mu_0\boldsymbol{m}}{2\pi R^3}. \qquad (4)

3.2 If $\boldsymbol{B}$ lies in the plane of the coil, then by analyzing the pairs of Ampere forces acting on the elements of the coil opposite to the magnetic field induction vector, it can be shown that the modulus of the total moment of forces is

M = mB. \qquad (5)

In general, the magnetic moment m forms an angle $\varphi$ with the field. Let us decompose $\boldsymbol{B}$ into components: one in the plane of the coil, and the other perpendicular to it. Obviously, the perpendicular component does not create a torque, causing stretching or compression of the coil, and since the projection of vector $\boldsymbol{B}$ onto the plane of the coil is

B_{\parallel} = B\sin\varphi, \qquad (6)

then the modulus of the resulting torque of Ampere forces is equal to

M = mB\sin\varphi. \qquad (7)

3.3 The elementary work $dA$ for a small rotation of the coil by an angle $d\varphi$ is equal to

dA = Md\varphi. \qquad (8)

When the coil rotates from the state when $\boldsymbol{m}\uparrow\uparrow\boldsymbol{B}$ to the position $\boldsymbol{m}\uparrow\downarrow\boldsymbol{B}$ , the angle changes from 0 to $\pi$ , and the total work is determined by the integral

A = \int_0^{\pi} Md\varphi = 2mB. \qquad (9)

Electron Paramagnetic Resonance

3.4 Let an electron rotate in a circular orbit of radius $R$ with period $T$ , then its motion can be represented as a circular current with a force

I = \frac{e}{T} \qquad (10)

and magnetic moment is obtained as

m = I\pi R^2. \qquad (11)

The angular momentum of an electron moving along a circular trajectory with velocity $v$ is equal to

L = m_e vR, \qquad (12)

from where, taking into account the expression for the period of revolution

T = \frac{2\pi R}{v}, \qquad (13)

we get

g_L = 1. \qquad (14)

In the vector relationship between magnetic and mechanical moments, it is necessary to take into account the sign due to the negative charge of the electron.

3.5 According to paragraph 3.3, to flip the spin, it is necessary to perform work (9), which together with the formula in the condition for the magnetic moment gives

A = g_s B_0 \frac{e\hbar}{2m_e}, \qquad (15)

which is accomplished due to the energy of the photon

E = \hbar\omega, \qquad (16)

that is, the law of conservation of energy is fulfilled as

E = A. \qquad (17)

Thus, using $g_s = 2g_L = 2$ , we get

\omega = \frac{eB_0}{m_e} = 6.15\cdot 10^{10}\ \mathrm{s}^{-1}, \qquad (18)

which is equal to the so-called Larmor frequency of rotation of an electron in a magnetic field.

3.6 In the absence of a core, the magnetic field in the solenoid is proportional to the current, that is,

B_0 \propto I_0, \qquad (19)

and in the presence of the core, the magnetic field is also proportional to the magnetic permeability of the substance

B \propto \mu I. \qquad (20)

The resonance occurs at the same value of the magnetic induction, therefore

I = \frac{I_0}{\mu} = 1.2\ \mathrm{A}. \qquad (21)

Thermodynamic equilibrium

3.7 In a state of thermodynamic equilibrium, the distribution of electrons across energy levels obeys the Boltzmann distribution, therefore

\frac{N_2}{N_1} = \exp\left(-\frac{\hbar\omega}{k_B T}\right), \qquad (22)

on the other hand, the total number of atoms is known and equal

N = N_1 + N_2. \qquad (23)

Thus, from (22) and (23), we obtain under condition $\hbar\omega \ll k_B T$

n_0 = \frac{\hbar\omega}{2k_B T}N = 4.68\cdot 10^{17}. \qquad (24)

3.8 In a state of thermodynamic equilibrium, the rates of upward and downward transitions should be the same, since the number of atoms at the energy levels should not change, in particular for the lower level we have

\frac{dN_1}{dt} = -B_{12}\rho N_1 + A_{21}N_2 + B_{21}\rho N_2 = 0, \qquad (25)

from which we obtain for the energy density of electromagnetic radiation

\rho = \frac{A_{21}}{B_{21}\left(\frac{N_1B_{12}}{N_2B_{21}} - 1\right)}. \qquad (26)

Taking into account relation (22) and comparing with Planck's formula, we conclude that

B_{12} = B_{21}. \qquad (27)

Presence of an external microwave field source

3.9 Taking into account the equality of the Einstein coefficients and neglecting spontaneous transitions, equation (25) can be rewritten as

\frac{dN_1}{dt} = -kN_1 + kN_2, \qquad (28)

which, taking into account (23), gives rise to

\frac{dn}{dt} = -2kn. \qquad (29)

Using the initial condition $n(0) = n_0$ , we obtain the solution of equation (29) in the form

n(t) = n_0\exp(-2kt). \qquad (30)

It is interesting to note that under the influence of an external source of an alternating field, the difference in the number of atoms at the two levels drops to zero over time.

3.10 From formula (30) it follows that

k = \frac{\ln 2}{2\tau}. \qquad (31)

Since during the transition of each atom from the lower level to the upper one quantum of energy of the alternating field $\hbar\omega$ is absorbed, and during each reverse transition the same energy is released, then the expression describing the absorption of energy $E$ of the alternating field in the core has the form

\frac{dE}{dt} = kn\hbar\omega, \qquad (32)

and this means that at the initial moment of time the power of the source is equal to

P = \frac{dE}{dt}(0) = \frac{n_0\hbar\omega\ln 2}{2\tau} = 1.05\ \mathrm{mW}. \qquad (33)

3.11 Let us consider the case of the absence of an external field, then the equation given in the condition for level 1 is rewritten as

\frac{dn}{dt} = N(\alpha_2 - \alpha_1) - n(\alpha_1 + \alpha_2). \qquad (34)

Since equation (34) must include the condition of thermodynamic equilibrium, then when the derivative on the left becomes zero, the equilibrium concentration must be equal to $n_0$ , from which it follows that

N = n_0\frac{(\alpha_1 + \alpha_2)}{(\alpha_2 - \alpha_1)}. \qquad (35)

Substituting (35) into (34) and adding the term from (29), we obtain an equation for taking into account the presence of an alternating field

\frac{dn}{dt} = -2kn - (n - n_0)(\alpha_1 + \alpha_2). \qquad (36)

Since it is required to determine the power of the source in a stationary mode, the difference in the number of atoms at the lower and upper levels must be constant, that is, $dn/dt = 0$ , from which it follows

n = \frac{n_0}{1 + \frac{2k}{(\alpha_1 + \alpha_2)}}. \qquad (37)

The power of the source is still determined by equation (42) and taking into account that $2k \gg (\alpha_1 + \alpha_2)$ , we finally obtain

P = \frac{1}{2}n_0\hbar\omega(\alpha_1 + \alpha_2) = 1.01\ \mathrm{\mu W}. \qquad (38)

Note that, unlike 3.10, the power of the source does not depend on the energy density of the electromagnetic field; this is the so-called saturation mode.

	Content	Points
3.1	Formula (1): $S = \pi R^2$	0.2	0.8
	Formula (2): $m = I\pi R^2$	0.2
	Formula (3): $B_0 = \frac{\mu_0 I}{2R}$	0.2
	Formula (4): $\boldsymbol{B}_0 = \frac{\mu_0\boldsymbol{m}}{2\pi R^3}$	0.2
3.2	Formula (5): $M = mB$	0.2	0.6
	Formula (6): $B_{\parallel} = B\sin\varphi$	0.2
	Formula (7): $M = mB\sin\varphi$	0.2
3.3	Formula (8): $dA = Md\varphi$	0.2	0.4
	Formula (9): $A = 2mB$	0.2
3.4	Formula (10): $I = \frac{e}{T}$	0.2	1.0
	Formula (11): $m = I\pi R^2$	0.2
	Formula (12): $L = m_e vR$	0.2
	Formula (13): $T = \frac{2\pi R}{v}$	0.2
	Formula (14): $g_L = 2$	0.2
3.5	Formula (15): $A = g_s B_0\frac{e\hbar}{2m_e}$	0.2	1.0
	Formula (16): $E = \hbar\omega$	0.2
	Formula (17): $E = A$	0.2
	Formula (18): $\omega = \frac{eB_0}{m_e}$	0.2
	Numerical value in formula (18): $\omega = 6.15 \cdot 10^{10}\ \mathrm{s^{-1}}$	0.2
3.6	Formula (19): $B_0 \propto I_0$	0.2	0.8
	Formula (20): $B \propto \mu I$	0.2
	Formula (21): $I = \frac{I_0}{\mu}$	0.2
	Numerical value in formula (21): $I = 1.2\ \mathrm{A}$	0.2
3.7	Formula (22): $\frac{N_2}{N_1} = \exp\left(-\frac{\hbar\omega}{k_B T}\right)$	0.2	1.0
	Formula (23): $N = N_1 + N_2$	0.2
	Formula (24): $n_0 = \frac{\hbar\omega}{2k_B T}N$	0.3
	Numerical value in formula (24): $n_0 = 4.68 \cdot 10^{17}$	0.3
3.8	Formula (25): $-B_{12}\rho N_1 + A_{21}N_2 + B_{21}\rho N_2 = 0$	0.4	1.0
	Formula (26): $\rho = \frac{A_{21}}{B_{21}\left(\frac{N_1B_{12}}{N_2B_{21}} - 1\right)}$	0.4
	Formula (22) is used	0.2
3.9	Formula (28): $\frac{dN_1}{dt} = -kN_1 + kN_2$	0.2	0.6
	Formula (29): $\frac{dn}{dt} = -2kn$	0.2
	Formula (30): $n(t) = n_0\exp(-2kt)$	0.2
3.10	Formula (31): $k = \frac{\ln 2}{2\tau}$	0.2	0.8
	Formula (32): $\frac{dE}{dt} = kn\hbar\omega$	0.2
	Formula (33): $P = \frac{n_0\hbar\omega\ln 2}{2\tau}$	0.2
	Numerical value in formula (33): $P = 1.05\ \mathrm{mW}$	0.2
3.11	Formula (34): $\frac{dn}{dt} = N(\alpha_2 - \alpha_1) - n(\alpha_1 + \alpha_2)$	0.3	2.0
	Formula (35): $N = n_0\frac{(\alpha_1 + \alpha_2)}{(\alpha_2 - \alpha_1)}$	0.3
	Formula (36): $\frac{dn}{dt} = -2kn - (n - n_0)(\alpha_1 + \alpha_2)$	0.3
	Formula (37): $n = \frac{n_0}{1 + \frac{2k}{(\alpha_1 + \alpha_2)}}$	0.3
	Formula (38): $P = \frac{1}{2}n_0\hbar\omega(\alpha_1 + \alpha_2)$	0.3
	Numerical value in formula (38): $P = 1.01\ \mathrm{\mu W}$	0.5
Total			10.0

Mathematical hint for the theoretical competition

The following formulas may be useful:

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$ , where $n \neq -1$ is a fixed number, $C$ refers to an arbitrary constant;

$\int \frac{dx}{x} = \ln|x| + C$ , where $C$ stands for an arbitrary constant;

$\int \frac{dx}{(a^2+x^2)^{3/2}} = \frac{x}{a^2\sqrt{a^2+x^2}} + C$ , where $a$ – constant, $C$ – arbitrary constant;

$(1+x)^{\gamma} \approx 1 + \gamma x + \frac{\gamma(\gamma-1)}{2}x^2$ , for $|x| \ll 1$ and any value of $\gamma$ ;

$\tan x \approx \sin x \approx x$ , for $|x| \ll 1$ ;

$\ln(1+x) \approx x$ , for $|x| \ll 1$ .

List of physical constants

Acceleration of gravity $g = 9.80\ \mathrm{m/s^2}$ ;

Universal gas constant $R = 8.31\ \mathrm{J/(mole\cdot K)}$ ;

Speed of light in vacuum $c = 3.00 \cdot 10^{8}\ \mathrm{m/s}$ ;

Reduced Planck constant $\hbar = 1.05 \cdot 10^{-34}\ \mathrm{J\cdot s}$ ;

Magnetic constant $\mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{H/m}$ ;

Electric constant $\varepsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{F/m}$ ;

Stefan-Boltzmann constant $\sigma = 5.67 \cdot 10^{-8}\ W/(m^2 \cdot K^4)$ ;

Elementary charge $e = 1.60 \cdot 10^{-19}\ \mathrm{C}$ ;

Electron mass $m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}$ .

Оригинал в Архива: IZhO-2026-Theory_eng.pdf · официални решения: IZhO-2026-Theory_eng_sol.pdf