APhO 2015, Теория — Задача 2. How are aurora ignited by the solar wind?

16th ASIAN PHYSICS OLYMPIAD 2015 — Question 2 · 3 май 2015 г.

Условие

Photograph of a bright green aurora filling the night sky above a forested lake shore, reflected in the still water.
Figure 1

The following questions are designed to guide you to find the answer one step by one step.

Background information of the interaction between the solar wind and the Earth's magnetic field

It is well known that the Earth has a substantial magnetic field. The field lines defining the structure of the Earth's magnetic field is similar to that of a simple bar magnet, as shown in Figure 2. The Earth's magnetic field is disturbed by the solar wind, which is a high-speed stream of hot plasma. (The plasma is the quasi-neutral ionized gas.) The plasma blows outward from the Sun and varies in intensity with the amount of surface activity on the Sun. The solar wind compresses the Earth's magnetic field. On the other hand, the Earth's magnetic field shields the Earth from much of the solar wind. When the solar wind encounters the Earth's magnetic field, it is deflected like water around the bow of a ship, as illustrated in Figure 3.

Dipole field of the Earth: a blue circle at the origin of axes z (up) and x (right), surrounded by closed field-line loops on both sides with shaded blue and pink belts; label 'Magnetic Field Lines'.
Figure 2
Magnetosphere diagram: the Sun (red dot, 'sun') on the left sends yellow arrows labelled 'Solar Wind' through the 'Interplanetary Medium' to a curved 'Bow Shock'; inside are the 'Magnetosheath', 'Magnetopause', 'Polar Cusp', 'Magnetosphere', 'Plasmasphere', 'Plasma mantle', 'North Lobe', 'South Lobe', 'Magnetotail' and 'Neutral point'; axes z (up) and x (right) through the Earth.
Figure 3

The curved surface at which the solar wind is first deflected is called the bow shock. The corresponding region behind the bow shock and front of the Earth's magnetic field is called the magnetosheath. The region surrounded by the solar wind is called the magnetosphere. The Earth's magnetic field largely prevents the solar wind from entering the magnetosphere. The contact region between the solar wind and the Earth's magnetic field is named the magnetopause. The location of the magnetopause is mainly determined by the intensity and the magnetic field direction of the solar wind. When the magnetic field in the solar wind is antiparallel to the Earth's magnetic field, magnetic reconnection as shown in Figure 4 takes place at the dayside magnetopause, which allows some charged particles of the solar wind in the region "A" to move into the magnetotail "P" on the night side as illustrated in Figure 5. A powerful solar wind can push the dayside magnetopause to very close to the Earth, which could cause a high-orbit satellite (such as a geosynchronous satellite) to be fully exposed to the solar wind. The energetic particles in the solar wind could damage high-tech electronic components in a satellite. Therefore, it is important to study the motion of charged particles in magnetic fields, which will give an answer of the aurora generation and could help us to understand the mechanism of the interaction between the solar wind and the Earth's magnetic field.

Three panels of magnetic reconnection at t=t0, t=t1 and t=t2: antiparallel green (downward) and blue (upward) field lines carrying points A, B, C, D bend towards each other and reconnect into an X shape.
Figure 4
Illustration of the Earth's magnetosphere with green reconnecting field lines; region 'A' (red dot) at the dayside magnetopause and region 'P' (red dot) in the pink magnetotail on the night side; insets show field-line details.
Figure 5

Numerical values of physical constants and the Earth's dipole magnetic field:

Speed of light in vacuum: c=2.998×108 m/sc = 2.998\times10^{8}\ \mathrm{m/s};

Permittivity in vacuum: ε0=8.9×10−12 C2/(N⋅m2)\varepsilon_0 = 8.9\times10^{-12}\ \mathrm{C^2/(N\cdot m^2)};

Permeability in vacuum: μ0=4π×10−7 N/A2\mu_0 = 4\pi\times10^{-7}\ \mathrm{N/A^2};

Charge of a proton: e=1.6×10−19 Ce = 1.6\times10^{-19}\ \mathrm{C};

Mass of an electron: m=9.1×10−31 kgm = 9.1\times10^{-31}\ \mathrm{kg};

Mass of a proton: mp=1.67×10−27 kgm_p = 1.67\times10^{-27}\ \mathrm{kg};

Boltzmann's constant: k=1.38×10−23 J/Kk = 1.38\times10^{-23}\ \mathrm{J/K};

Gravitational acceleration: g=9.8 m/s2g = 9.8\ \mathrm{m/s^2};

Planck’s constant: h=6.626×10−34 J⋅sh = 6.626\times10^{-34}\ \mathrm{J\cdot s}

Earth's radius RE=6.4×106 mR_E = 6.4\times10^{6}\ \mathrm{m}.

The Earth's dipole magnetic field can be expressed as

B⃗d=B0RE3r5[3xzx^−3yzy^+(x2+y2−2z2)z^], (r≥RE)(1)\vec{B}_d = \frac{B_0R_E^3}{r^5}[3xz\hat{x} - 3yz\hat{y} + (x^2+y^2-2z^2)\hat{z}] \quad ,\ (r \ge R_E) \qquad (1)

where r=x2+y2+z2r = \sqrt{x^2+y^2+z^2}, B0=3.1×10−5 TB_0 = 3.1\times10^{-5}\ \mathrm{T}, and x^,y^,z^\hat{x}, \hat{y}, \hat{z} are the unit vectors in the xx, yy, zz directions, respectively.

Questions:

(a) (3 т.)

(i) Before we study the motion of a charged particle in the Earth's dipole magnetic field, we first consider the motion of an electron in a uniform magnetic field B⃗\vec{B}. When the initial electron velocity v⃗\vec{v} is perpendicular to the uniform magnetic field as shown in Figure 6, please calculate the electron trajectory. The electron is initially located at (x,y,z)=(0,0,0). [1 т.]

Uniform magnetic field B pointing out of the page (red circled dot and a grid of black dots); an electron e- moving downward with velocity V; axes y (up) and x (right).
Figure 6

(ii) Please determine the electric current of the electron motion and calculate the magnetic moment μ⃗=IA⃗\vec{\mu} = I\vec{A}, where A⃗\vec{A} is the area of the electron circular orbit and the direction of A⃗\vec{A} is determined by the right-hand rule of the electric current. [1 т.]

(iii) If the initial electron velocity v⃗\vec{v} is not perpendicular to the uniform magnetic field, i.e, the angle θ\theta between B⃗\vec{B} and v⃗\vec{v} is 00<θ<9000^{0}<\theta<90^{0}, please give the screw pitch (the distance along the z-axis between successive orbits) of the electron trajectory. [1 т.]

(b) (4 т.)

In the uniform background magnetic field as shown in Figure 6, the plasma density is nonuniform in x. For simplicity, we assume that the temperature and the distribution of the ions and electrons are the same. Thus, the plasma pressure can be expressed as

p(x)=kT[ni(x)+ne(x)]=2kTn(x)=2kT(n0+αx),p(x) = kT[n_i(x)+n_e(x)] = 2kTn(x) = 2kT(n_0+\alpha x),

Where B, T, k, n0n_0, and α\alpha are positive constants, ni(x)n_i(x) and ne(x)n_e(x) are the number densities of the ions and electrons.

(i) Please explain the generation mechanism of the electric current by a schematic drawing. [2 т.]

(ii) If both the ions and electrons have a Maxwellian distribution, the ion distribution is

fi(x,v⊥,v∥)=ni(x)(mi2πkT)3/2e−mi(v⊥2+v∥2)/2kT,f_i(x,v_\perp,v_\parallel) = n_i(x)\left(\frac{m_i}{2\pi kT}\right)^{3/2} e^{-m_i(v_\perp^2+v_\parallel^2)/2kT},

please calculate the constant β in the magnetization M=βn(x)kTBM = \beta n(x)\dfrac{kT}{B}, where the magnetization M is the magnetic moment per unit volume. (Hint: We have ∫0∞xexp⁡(−x)dx=1\int_0^\infty x\exp(-x)dx = 1 and ∫−∞∞exp⁡(−x2)dx=π\int_{-\infty}^{\infty}\exp(-x^2)dx = \sqrt{\pi}.) [2 т.]

(c) (1 т.)

Now let's go back to the Earth's dipole magnetic field. Please apply the result from Question(b) to calculate the ratio of the diamagnetic field and the Earth's dipole magnetic field in Equation (1) at the position (xx=10RE_\mathrm{E},yy=0, zz=1 RE_\mathrm{E}). The plasma pressure is assumed to be p(z)=p0e−(z/a)2p(z) = p_0e^{-(z/a)^2}, where p0=3×10−10 pap_0 = 3\times10^{-10}\ \mathrm{pa} and a=2REa = 2R_E. The magnetic field around this position is also assumed to be uniform. Be aware of the difference in the coordinate systems in Questions (b) and (c). (Hint: The diamagnetic field is given by Bmx=μoMB_{mx} = \mu_o M.)

(d) (4 т.)

From Figures 2, 3, and 5, it can be clearly seen that the Earth's magnetic field strength along a magnetic field line is the largest at the poles and the smallest in the equatorial plane. Since the Earth's dipole magnetic field is axially symmetric and slowly varying along a magnetic field line, it can for simplicity be treated as a magnetic-mirror field as shown in Figure 7. The magnetic field strength along a magnetic field line is the smallest (B0B_0) at the point "P2_2" and the largest (BmB_m) at the points "P1_1" and "P3_3". An electron with an initial velocity v⃗\vec{v} is located at the point "P2_2" and drifts towards the point "P3_3". The angle between the initial velocity v⃗\vec{v} and the magnetic field at the point "P2_2" is 00<θ<9000^{0}<\theta<90^{0}. For the magnetic-mirror field B⃗=Brr^+Bzz^\vec{B} = B_r\hat{r} + B_z\hat{z} (with Br≪BzB_r \ll B_z), we can assume dBdz=dBds\dfrac{dB}{dz} = \dfrac{dB}{ds}, where dBds\dfrac{dB}{ds} is the spatial derivative of BB along a magnetic field line. Since there is no evidence of the existence of magnetic monopoles, we have ⟨Br⟩=−12dBdzrc≪Bz\langle B_r\rangle = -\dfrac{1}{2}\dfrac{dB}{dz}r_c \ll B_z, where ⟨Br⟩\langle B_r\rangle is the gyro-average of BrB_r and rcr_c is the electron gyroradius.

'Magnetic Mirror Field': green B-field lines converging at both ends at points P1 and P3 and spreading in the middle; point P2 on an upper line with velocity v (red arrow); a wavy red 'Electron trajectory' near P1; axes r (up) and z (right).
Figure 7

(i) Please give the gyro-averaged magnetic-field force along the magnetic field lines on an electron and show that the magnetic moment is a motion constant, i.e., dμdt=0\dfrac{d\mu}{dt} = 0, based on the law of the total kinetic energy conservation. [3 т.]

(ii) Based on the motion constant of magnetic moment, please determine what the condition should be satisfied for the angle θ\theta between the initial electron velocity v⃗\vec{v} and the magnetic field at the point "P2_2" if an electron will not escape from the magnetic mirror field. [1 т.]

(e) (1 т.)

Earth's dipole magnetic field lines (blue lines) are shown in Figure 8. The spiral trajectory of a charged particle (red curve) is assumed to be confined in the yy=0 plane since the gradient and the curvature of the magnetic field can be ignored. If a charged particle with the mass mm, charge qq, and velocity v⃗\vec{v} is initially located at the equatorial position [xx=6RE_\mathrm{E}, yy=0, zz=0] and the angle between the electron velocity v⃗\vec{v} and the magnetic field is θ\theta initially, please determine what the condition should be satisfied for θ\theta if the charged particle arrives below 200km of its altitude at the latitude 60060^{0}.

The Earth (blue circle, 'Earth') with a blue dipole 'Magnetic Field Line' loop; a red wavy 'Spiral Trajectory of Charged Particle' follows the line; a charge q with velocity v at the far equatorial point; axes z, y, x.
Figure 8

(f) (5 т.)

As shown in Figure 5, when magnetic reconnection takes place at the dayside magnetopause, reconnected magnetic field lines drift towards the nightside region because the solar wind flows tailward. Thus, some solar wind electrons in the region "A" also move towards the magnetotail in the region "P". After the electrons arrive in the region "P", some electrons can be accelerated to around 1keV. If energetic electrons drift down to the thermosphere (The altitude of the thermosphere is about 85km-800km.), energetic electrons can collide with the neutral atoms, which could cause the neutral atoms to jump into excited states. A photon is emitted when the higher excited state of a neutral atom returns to its lower excited state or ground state. Splendid aurora (Figure 1) is generated in the aurora oval due to photons with different wavelengths. It is found that the aurora is mainly resulted from photons emitted by oxygen atoms. The energy levels in the first and second excited states relative to the ground state are 1.96eV and 4.17eV, respectively. The lifetimes of the two excited states of an oxygen atom are 110s and 0.8s as shown in Figure 9.

Energy-level diagram of oxygen, vertical axis 'Energy (eV)': 'Second excited state' 1S at 4.17 with τ = 0.8s, 'First excited state' 1D at 1.96 with τ = 110s, and the 'Ground state'; downward arrows mark transitions 4.17→1.96 and 1.96→ground.
Figure 9

(i) Please give the atmospheric density as a function of the altitude and the ratio of the oxygen density at the altitudes H=160km and H=220km. For simplicity, we assume that the atmospheric temperature is independent of the altitude and the air is an ideal gas.

(ρ0g/P0=0.13/km\rho_0 g/P_0 = 0.13/\mathrm{km}, where ρ0\rho_0 and P0P_0 are the atmospheric density and pressure at sea level.) [2 т.]

(ii) Please give the colors of auroras at the altitudes H=160km and H=220km. (Hint: The dependence of the collision frequency of atmospheric molecules on the atmospheric density is ν=ν0ρ/ρ0\nu = \nu_0\rho/\rho_0, where ν0≈109/s\nu_0 \approx 10^{9}/s is the collision frequency of atmospheric molecules at sea level. The excited oxygen atom will lose a part of its energy when it collides with other neutral molecules. ) [3 т.]

(g) (2 т.)

As mentioned above, a powerful solar wind can push the dayside magnetopause to very close to the Earth, which could cause a high-orbit satellite to be fully exposed to the solar wind. The energetic particles in the solar wind could damage high-tech electronic components in a satellite. For simplicity, the Earth's dipole magnetic field is assumed to remain unchanged when the solar wind compresses it and that the plasma density is ignorable in the magnetosphere. Please give the minimum solar wind speed to cause a damage of a geosynchronous satellite if the magnetic field strength and the plasma density of the solar wind are Bs=5×10−9 TB_s = 5\times10^{-9}\ \mathrm{T} and ρs=50 proton/cm3\rho_s = 50\ \mathrm{proton/cm^3}, respectively. (Hint: The force per unit area associated with the magnetic field is f=B2/2μ0f = B^2/2\mu_0. We only consider the variation in x for all physical quantities, i.e., the physical quantities are independent of y and z.)

Отговори

Решение


Оригинал в Архива: 2.pdf · официални решения: 2_sol.pdf