IOM 2019, Теория — Задача 3. «Monopoles and Dyons»

• 1.1. Graph (Fig. 1 of the official solution): $I(t) = \frac{\mu_0 M}{2L}\left(1+\frac{vt}{\sqrt{a^2+v^2t^2}}\right)$, a monotonic increase of current reaching a constant positive value as $t \to +\infty$.

• 1.2. 2 mA

  Table of answers: $I^S_{max} = \frac{\mu_0 M}{2L} \approx 2$ mA and $\Delta v_{max} \approx \frac{(\mu_0 M)^2}{8mLv_0} \approx 2 \cdot 10^{-5}$ m/s. The text of the official solution instead prints $I^S_{max} = \frac{\mu_0 M}{L} \approx 4$ mA and $\Delta v_{max} \approx \frac{(\mu_0 M)^2}{2mLv_0} \approx 8 \cdot 10^{-5}$ m/s; both are kept as printed.

• 1.3. Qualitative graphs, Fig. 2, Fig. 3 (or Fig. 3') and Fig. 4 of the official solution.

• 1.4. 0.000006 m/s

  $I(x) = \frac{\mu_0 M}{2R}\frac{a^2}{(a^2+x^2)^{3/2}}v(x)$; $\Delta v \approx (6 \pm 2) \cdot 10^{-6}$ m/s.

• 2.1. $r = \frac{1}{4\pi\varepsilon_0}\frac{|q|Q}{mv^2} = \mathrm{const}$

• 2.2. $R = \frac{1}{4\pi\varepsilon_0}\frac{|q|Q^2c^2}{mv^2\sqrt{c^4Q^2+v^2M^2}},\quad h = \frac{1}{4\pi\varepsilon_0}\frac{|q|QM}{mv\sqrt{c^4Q^2+v^2M^2}},\quad \alpha = \mathrm{arctg}\left(\frac{c^2Q}{vM}\right)$

• 2.3. $E \equiv \frac{m\vec{v}^2}{2} + \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r},\quad \vec{J} \equiv m[\vec{r}\times\vec{v}] + \frac{\mu_0 M|q|}{4\pi}\frac{\vec{r}}{r};\quad E = -\frac{mv^2}{2},\quad \vec{J} = \frac{\mu_0|q|}{4\pi v}\sqrt{c^4Q^2+v^2M^2}\cdot\vec{n}$

• 3.1. $v' = v_0$

• 3.2. $\theta_1 = \arcsin\left(\frac{4\pi mv_0d}{\mu_0|Mq|}\right),\ \theta_2 = \pi - \arcsin\left(\frac{4\pi mv_0d}{\mu_0|Mq|}\right)\ \text{or}\ \theta_1 = 0,\ \theta_2 = \pi$

• 3.3. $b' = \sqrt{b^2+d^2} = b$

  The table of answers prints $b' = \sqrt{b^2+d^2} = b'$; the solution text derives $b' = \sqrt{b^2+d^2}$ and $b' = b$ (hence $d = 0$).

• 3.4. $3.3 \times 10^{-9}$ C·m/s

  $qM = k \cdot \frac{h}{\mu_0}$ ($k$ an integer); $M = \frac{h}{\mu_0 e} \approx 3{,}3 \cdot 10^{-9}$ C·m/s.

Proposed Solution

Part I

1.1. There are two possible methods of solution.

Method 1: Magnetic charges must be introduced into electrodynamics so that the "symmetry" of the electrodynamic equations is preserved when replacing a magnetic charge with an electric one and at the same time a magnetic field with an electric one (as can be seen from the task, the magnetic field of a stationary magnetic charge fully coincides with the electric field of a stationary electric charge if necessary replacements are made: $M \leftrightarrow q$, $\mu_0 \leftrightarrow \frac{1}{\varepsilon_0}$ and $\vec{B} \leftrightarrow \vec{E}$). As you know, moving electric charges create magnetic fields. The electric field of a point electric charge is $\vec{E} = \frac{q}{4\pi\varepsilon_0 r^3}\vec{r}$, where vector $\vec{r}$ connects the point of charge location and the observation point P, that is, $\vec{r} \equiv \vec{r}_P - \vec{\rho}(t)$ (here $\vec{\rho}(t)$ is the radius vector that sets the position of the charge relative to the origin of the coordinate system). The magnetic field of a moving charge can be determined from the Bio-Savard-Laplace law: $\vec{B} = \frac{\mu_0 q}{4\pi r^3}[\vec{v}\times\vec{r}]$ (taking into account that $I \cdot d\vec{l} = q\frac{d\vec{l}}{dt} = q\vec{v}$). Thus, for a point charge moving at a speed $\vec{v}$, $\vec{B} = \frac{1}{c^2}\left[\vec{v}\times\vec{E}\right]$. Keeping in mind dimensional considerations and the mentioned symmetry we can conclude that a moving magnetic charge creates an electric field $\vec{E} = \left[\vec{v}\times\vec{B}\right]$. We emphasize that this is a total field created by a uniformly moving charge (at variable speed, radiation also emerge).

Now consider the passage of a monopole through a superconducting ring. We introduce the coordinate system shown in the figure: origin is in the center of the ring, the X axis is along the symmetry axis of the ring, the Y axis is directed to the selected element P of the ring. The monopole velocity $\vec{v} = v \cdot \vec{e}_x$, the law of its motion $\vec{\rho}(t) = vt \cdot \vec{e}_x$, and $\vec{r}_P = a\vec{e}_y$. The electric field created by the monopole at point P is $\vec{E} = \left[\vec{v}\times\vec{B}\right] = \frac{\mu_0 M}{4\pi r^3}\left[(v\vec{e}_x)\times\left(a\vec{e}_y - vt\vec{e}_x\right)\right] = \frac{\mu_0 M}{4\pi} \cdot \frac{av}{(a^2+v^2t^2)^{3/2}}\vec{e}_z$. Since this field is symmetric about the $x$ axis, the monopole field in the ring creates an emf equal to $U_E = E \cdot 2\pi a = \frac{\mu_0 M}{2} \cdot \frac{a^2}{(a^2+v^2t^2)^{3/2}}v$. The emf created in the superconducting ring should remain zero for any finite current. The emf in the ring is composed of the “external” EMF created by external fields, and the self-induction emf $U_L = -L\frac{dI}{dt}$. The expression for the EMF of the electric field of the monopole obtained above can be written in the form of the time derivative: $U_E = \frac{d}{dt}\left(\frac{\mu_0 M}{2} \cdot \frac{vt}{\sqrt{a^2+v^2t^2}}\right)$. Therefore, the condition that the total emf is equal to zero leads to the relation

The constant is calculated as $t \to -\infty$. Thus, $I(t) = \frac{\mu_0 M}{2L} \cdot \left(1 + \frac{vt}{\sqrt{a^2+v^2t^2}}\right)$. The graph of this function is shown in the figure $\left(\tau \approx \frac{a}{v}\right)$.

The correctness of plot is determined by the following features: a monotonic increase of current $I$, reaching a constant positive value as $t \to +\infty$.

Method 2: Another method is based on the Faraday's law of induction and the law of total current. In this case, it is important to take into account that the electric field of the moving monopole, which we calculated above, can be represented as the sum of the vortex induction field (which creates the induction emf $U_i = -\frac{d\Phi}{dt} = -\frac{d}{dt}(\Phi_{ex} + LI)$) and the field magnetic current. We start with calculating $\Phi_{ex}$, the flux from the magnetic field of the monopole through the loop. Since the magnetic field of a monopole is similar to the electric field of a point electric charge, this magnetic flux is $\Phi_{ex} = \frac{\Omega}{4\pi}\mu_0 M$. Here, $\Omega$ is the solid angle enclosed by the cone at which the vertex is at the monopole location and the base is the area of the ring. The half-angle α of this cone is determined from the relation $\tan\alpha = \frac{a}{x}$. The value of the solid angle (according to the mathematical hint)

Therefore:

To take into account the contribution from the magnetic current, we use the total current law, according to which the circulation of the magnetic induction vector along an arbitrary circuit in vacuum is equal to the total electric current penetrating this circuit times $\mu_0$. Note that the circulation of vector $\vec{B}$ is calculated as follows: the contour is divided into infinitesimal elements $d\vec{l}$ and scalar products $(\vec{B} \cdot d\vec{l})$ are calculated. The sum of these scalar products is circulation. In fact, such an operation (summation of an infinitely large number of infinitesimal contributions) is integration along a closed circuit, therefore it is denoted by the symbol $\oint$, thus, the law of the total current can be written as follows: $\oint(\vec{B} \cdot d\vec{l}) = \mu_0 I$. The circulation of the electric field $\vec{E}$ is the EMF. Therefore, the contribution to the emf of the magnetic field can be written as $U_M = \mu_0 I_M = \mu_0\frac{dM}{dt}$. The EMF is proportional to the rate of change of the magnetic charge $M(t)$ passed through the circuit at time $t$. In case of the magnetic monopole passage , the law of change of the magnetic charge is $M(t) = \begin{cases} 0, t < 0 \\ M, t \ge 0 \end{cases}$, and this expression can also be written in the following form $M(t) = \frac{M}{2}\left(1 + \frac{t}{|t|}\right)$. Thus, the condition that the total EMF is equal to zero leads to the relation $\frac{d}{dt}\left(-\Phi_{ex} - LI + \frac{\mu_0 M}{2} \cdot \left(1 + \frac{t}{|t|}\right)\right) = 0$, and taking into account the fact that as $t \to -\infty$, all three terms in brackets are equal to zero, we obtain: $LI = \frac{\mu_0 M}{2} \cdot \left(1 + \frac{t}{|t|}\right) - \Phi_{ex}$. Substituting here the expression for $\Phi_{ex}$, we finally find: $I(t) = \frac{\mu_0 M}{2L} \cdot \left(1 + \frac{vt}{\sqrt{a^2+v^2t^2}}\right)$. We got the same expression as in the first method.

1.2. The expression for the current in the ring obtained in the previous paragraph can be written as a function depending on $x$ if we assume that the monopole velocity changes very insignificantly ($x \approx vt$). In fact, we have already used this approximation, since we used the formula for the field of a uniformly moving charge, so in the end its applicability will need to be checked. If it is valid, $(x) = \frac{\mu_0 M}{2L} \cdot \left(1 + \frac{x}{\sqrt{a^2+x^2}}\right)$. The maximum current is achieved at $x \to +\infty$, that is, $I^S_{max} = \frac{\mu_0 M}{L} \approx 4$ mA (the inductance is determined by the Grover formula: $L \approx 10^{-5}$ H).

In the process of movement, the monopole expends its kinetic energy to increase the energy of the magnetic field of the ring. The minimum speed corresponds to the position of the monopole at which the current is maximum, i.e.

Even very rough numerical estimates show that $\frac{L\left(I^S_{max}\right)^2}{2} \ll \frac{mv_0^2}{2}$, that is, the decrease in the kinetic energy of the monopole is really very small and its speed changes insignificantly (we can assume that $v_0 \approx v_{min}$). Therefore, $\Delta v_{max} = \frac{v_0^2 - v_{min}^2}{v_0 + v_{min}} \approx \frac{v_0^2 - v_{min}^2}{2v_0} \approx \frac{(\mu_0 M)^2}{2mLv_0} \approx 8 \cdot 10^{-5}$ m / s.

Note: since taking the magnetic current field into account is a rather non-trivial part of the problem, at the Olympiad a positive mark for points 1.1 and 1.2 was put up for participants who, without making other errors, got the result using a model that does not take this contribution into account.

1.3. A) For a large resistance $R \gg \frac{Lv_0}{2r}$ any effect of the EMF due to self-induction is small and the induction current is proportional to the EMF of «external» induction which, in turn, is proportional to the rate of magnetic flux variation. During the magnetic monopole passage the flux increases at first, then drops «abruptly» reversing the sign, then leaps upward and reverses the sign again, and finally returns to the initial zero value. The induction current is shown in Fig.2. Notice that this plot is similar to the derivative of the function in Fig.1.

B) If $\frac{Lv_0}{a} \ll R \ll \frac{Lv_0}{2r}$, the EMF of self-induction is negligible only «far outside» from the time interval $|t| \le \frac{r}{v_0}$ (t = 0 is the moment of passage through the ring) when the monopole is far away from the ring. Within the time interval the EMF of self-induction is, on the contrary, very large thereby opposing a change of induction current in the ring. Therefore, the maximum current attained when the monopole was approaching the ring remains approximately constant when the monopole passes through the ring. After the monopole leaves this interval, the current returns to the «quasi-stationary» dependence in a time which is small compared to $\frac{a}{v_0}$ (see Fig.3).

C) A thin linear cylindric magnet is a magnetic dipole. The lines of magnetic induction of such a magnet come out of one pole and enter the other one. When the dipole passes through the ring the magnetic flux grows at first, then rapidly drops, and then changes smoothly near the minimum (goes through 0 when the magnet middle point coincides with the ring plane). Then the flux reaches the maximum again and after that slowly decays. A time dependence at a large resistance corresponds to the rate of change of magnetic flux (see Fig.4).

Correctness of a plot should be judged according to the following criteria: the number of «outbursts», their «polarity», symmetry or antisymmetry. The plot in Fig.3 can be «distorted» near the maximum: it can be «cut off» or exhibit a small variation of the current as in Fig.3'.

1.4. First, notice that the kinetic energy of monopole partially converts into the energy of magnetic field of the induction current ($W_L = \frac{LI^2}{2}$) and into the heat released in the conducting ring ($dQ = I^2R \cdot dt$). The ring inductance in our case is $L \approx 10^{-5}$H and the electrical resistance equals $R = \frac{2\rho a}{r^2} = 0{,}08$ Ω. For $x \gg r$ the induction losses are negligible compared to the thermal ones. The former are significant only for $x \lesssim r$ where the flux changes rapidly, so a calculation which does not take into account the induction losses results in a current exceeding $I^S_{max}$, which is obviously impossible (electrical resistance of the ring cannot increase the current). The “external” EMF (see 1.1) is determined by the expression $U_E \approx \frac{\mu_0 M}{2}\frac{a^2}{(a^2+x^2)^{3/2}}\frac{dx}{dt}$. Thus, the magnitude of the induction current at the time when the monopole is at a distance $x$ from the circuit (we assume that $x < 0$ when approaching the ring and $x > 0$ after flying through it) and moves with speed $v$ is equal to $(x) = \frac{|U_E|}{R} = \frac{\mu_0 M}{2R}\frac{a^2}{(a^2+x^2)^{3/2}}v(x)$. For the given ring dimensions $R \approx \frac{1}{125}\frac{Lv_0}{2r} \approx 8\frac{Lv_0}{a}$, therefore it is reasonable to consider that $\frac{Lv_0}{a} \ll R \ll \frac{Lv_0}{2r}$. Hence, the plot in Fig.3 approximately corresponds to the time dependence of induction current and we can use the obtained expression for the current as a reasonable approximation for any moment of time. In this approximation the instantaneous power of heat losses becomes $P = I^2R = \frac{(\mu_0 M)^2}{4R}\frac{a^4}{(a^2+x^2)^3}v^2(x)$. It is also clear that in the case at hand monopole acceleration is very small like in the previous cases (see 1.2). Therefore, a net heat released in the ring is approximately equal to $Q = \int_{-\infty}^{+\infty} P\,dt \approx \frac{(\mu_0 M)^2 v_0}{4R}\int_{-\infty}^{+\infty}\frac{a^4}{(a^2+x^2)^3}dx$. Since $x = a \cdot \mathrm{ctg}\,\alpha$, we have $dx = a\sin^{-2}\alpha d\alpha$, and

Now, using the law of conservation of energy one can write:

Since the change of velocity is small, we obtain $\Delta v \approx \frac{v_0^2 - v_{min}^2}{2v_0} \approx \frac{3\pi(\mu_0 M)^2}{32mRa} \approx 6 \cdot 10^{-6}$ m/s. Note that in questions 1.3 and 1.4, the absence of taking into account the contribution of the magnetic current field does not lead to a significant distortion of the results. Therefore, in these questions for solutions without taking into account the specified contribution, a full score was marked.

Part II

2.1. Notice that motion of a charged particle with a mass m and a charge q in the field of a heavy dyon conserves the sum of the particle kinetic energy and the potential energy of its interaction with the electric field of the dyon. The work done by the magnetic component of the Lorentz force $\vec{F}_M = q\left[\vec{v}\times\vec{B}\right]$ is identically zero and the work of the electrical component $\vec{F}_E = q\vec{E}$ can be computed as a difference of potential energy $U = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r}$ (r is a distance between the particle and the dyon). Thus, $\frac{mv^2}{2} + \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r} = \mathrm{const}$. Therefore, if the particle velocity is constant, the distance to the dyon must be constant as well. This implies that the dyon must reside at the axis of symmetry of a particle circular orbit (the axis passes through the orbit center perpendicular to the orbit plane, see the Figure). The force exerted on the particle, $\vec{F} = q\vec{E} + q\left[\vec{v}\times\vec{B}\right]$, must be directed at the orbit center. Recall that magnetic induction $\vec{B}$ of the dyon points in the radial direction from it. By now it should be clear that such a motion is possible only for $q < 0$ providing the magnetic and electric components of the Lorentz force balance each other in the direction perpendicular to the orbit plane:

The radius r can be determined from the equation of centripetal acceleration,

yielding $r = \frac{1}{4\pi\varepsilon_0}\frac{|q|Q}{mv^2}$. It is interesting that this formula does not include M: the distance to the points of the circular orbit of a charged particle in the electromagnetic field of the dion coincides with the radius of its circular orbit in the field of only the electric charge of the dion. However, the magnetic field of the dyon displaces the particle orbital plane so that the dyon is not at this plane anymore.

2.2. According to the above analysis, the orbit of a radius $R = r \cdot \sin\alpha = \frac{1}{4\pi\varepsilon_0}\frac{|q|Q^2c^2}{mv^2\sqrt{c^4Q^2+v^2M^2}}$ is at the plane separated from the dyon by a distance $h = r \cdot \cos\alpha = \frac{1}{4\pi\varepsilon_0}\frac{|q|QM}{mv\sqrt{c^4Q^2+v^2M^2}}$ where the dyon is at the orbit symmetry axis. The orbit subtends an angle $\alpha = \mathrm{arctg}\left(\frac{c^2Q}{vM}\right)$ from the dyon location. Obviously, any two of these three quantities completely determine the orbit.

2.3. Now it should be clear that the scalar integral of motion is given by the total mechanical energy of the particle: $E \equiv \frac{m\vec{v}^2}{2} + \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r}$. The vector integral of motion must be related to the angular momentum $\vec{L} = m[\vec{r}\times\vec{v}]$ which is obviously conserved in the field of an immobile electric charge if M = 0. The rate of change of the particle angular momentum during motion in a circular orbit is only due to the magnetic component of the Lorentz force (the torque of the electric force vanishes):

Since $\vec{v} = \frac{d\vec{r}}{dt}$ and $\frac{(\vec{v}\vec{r})}{r} = \frac{vr\cos(\vec{v},\vec{r})}{r} = v\cos(\vec{v},\vec{r}) = v_r = \frac{dr}{dt}$, we obtain

Thus, $\frac{d}{dt}\left(\vec{L} + \frac{\mu_0 M|q|}{4\pi}\frac{\vec{r}}{r}\right) = 0$! A vector quantity $\vec{J} \equiv \vec{L} + \frac{\mu_0 M|q|}{4\pi}\frac{\vec{r}}{r}$ is conserved for a charged particle traveling in the dyon field. Evaluating $E$ and $\vec{J}$ (from symmetry considerations $\vec{J}$ is directed along the orbit axis) by means of the derived expressions for $r$ and $\alpha$ yields: $E = -\frac{mv^2}{2}$ and $\vec{J} = \frac{\mu_0|q|}{4\pi v}\sqrt{c^4Q^2+v^2M^2} \cdot \vec{n}$, where the unit vector $\vec{n}$ points from the dyon to the center of particle orbit.

Comments. Electromagnetic field is an independent type of matter which carries energy, momentum (its energy flux and momentum flux are related by the Pointing vector $\vec{S}_E = c \cdot \vec{S}_p = \frac{1}{\mu_0}\left[\vec{E}\times\vec{B}\right]$), and, accordingly, angular momentum. Angular momentum of an isolated system must be conserved, in our case this conserved angular momentum equals to the sum of angular momenta of the particle and electromagnetic field in the reference frame of the dyon (one can see that for our system $\vec{S}_p \ne 0$ !). Therefore, the second term in $\vec{J}$ is the angular momentum of electromagnetic field and $\vec{J}$ is the net angular momentum of the system consisting of a dyon, a charged particle, and electromagnetic field.

Part III:

3.1. According to 2.3, motion of a charged particle in the field of magnetic monopole conserves the particle kinetic energy $E = \frac{mv^2}{2}$ and the vector $\vec{J} \equiv \vec{L} - \frac{\mu_0 Mq}{4\pi}\frac{\vec{r}}{r}$.

Let us introduce the reference frame shown in the Figure. When the particle is far away from the monopole, $|\vec{v}| \equiv v_0$, $\vec{v} = v_0\vec{e}_z$, $\vec{r} = b\vec{e}_y - r_z\vec{e}_z$, and $r = \sqrt{b^2 + r_z^2}$, where $r \approx r_z \gg b$. Then

After scattering on the monopole (and very far from it) the particle will travel along a straight line at a velocity $\vec{v}'$. Clearly, $v' = v_0$ because $E$ is conserved.

3.2. Let $\vec{n}$ be a unit vector in the direction of $\vec{v}'$ and $\vec{b}'$ be a vector pointing from the monopole to the closest point at the line of particle motion after the scattering. Then the position vector of the particle, when it is at a distance r from the monopole, equals $\vec{r} = \vec{b}' + \vec{n}\sqrt{r^2 - b'^2}$ . According to the frame choice, vector $\vec{n}$ is parallel to (yz), i.e. one can write $\vec{n} = \pm\sin\theta\,\vec{e}_y + \cos\theta\,\vec{e}_z$, where $\theta$ is the angle between $\vec{v}'$ and $\vec{v}_0$. It is also evident that $\vec{b}' \perp \vec{n}$. Using the conservation laws, we obtain:

To simplify derivation, we introduce the notation $\frac{\mu_0 Mq}{4\pi mv_0} \equiv \beta$ and evaluate the cross product of the above equation with $\vec{n}$, which gives: $\vec{b}' = b[\vec{n}\times\vec{e}_x] + \beta[\vec{n}\times\vec{e}_z]$. Substituting the expression for $\vec{n}$ in this formula, we obtain: $\vec{b}' = \pm\beta\,\sin\theta\,\vec{e}_x + b\,\cos\theta\,\vec{e}_y \mp b\,\sin\theta\,\vec{e}_z$. Hence, the distance to the plane of particle motion after the scattering is $d = |d_x| = |\beta\,\sin\theta\,|$. Therefore, there are two possible values of the scattering angle: $\theta_1 = \arcsin\left(\frac{4\pi mv_0d}{\mu_0|Mq|}\right)$ and $\theta_2 = \pi - \arcsin\left(\frac{4\pi mv_0d}{\mu_0|Mq|}\right)$. One can see that the problem is well posed only for $d \le \frac{\mu_0|Mq|}{4\pi mv_0}$ (all allowed values of d fall into this range).

3.3. It follows from the equation for $\vec{b}'$ that its absolute value is $b' = \sqrt{b^2 + \beta^2\sin^2\theta} = \sqrt{b^2 + d^2}$. In addition, when raising the equality $\left[\vec{b}'\times\vec{n}\right] - \beta\vec{n} = b\vec{e}_x + \beta\vec{e}_z$ scalar squared, we find, taking into account the perpendicularity of the vectors $\left[\vec{b}'\times\vec{n}\right]$ and $\vec{n}$, as well as the vectors $\vec{e}_x$ and $\vec{e}_z$ such that $\left[\vec{b}'\times\vec{n}\right]^2 + \beta^2 = b^2 + \beta^2$. Since $\vec{b}'$ and $\vec{n}$ are also perpendicular, then $\left[\vec{b}'\times\vec{n}\right]^2 = b'^2$. It follows that $b' = b$! This means that the restriction on $d$ is even more strong: from the two relations obtained it follows that $d = 0$. At the same time, obviously, $\sin\theta = 0$, i.e., $\theta_1 = 0$ or $\theta_2 = \pi$. In fact, the case $\theta = 0$ is realized only as the limit for very large $b$ and $v_0$. As can be seen, when a charge is scattered by a monopole, in which the charge after scattering moves parallel to the original plane of motion, the charge actually moves in this plane!

3.4. One can argue that the component of the angular momentum of electromagnetic field in the system «magnetic monopole + particle + field» in the direction of $\vec{n} = \frac{\vec{r}}{r}$ must be a multiple of $\frac{h}{4\pi}$. This implies the relation $\frac{\mu_0 Mq}{4\pi} = k \cdot \frac{h}{4\pi} \Rightarrow qM = k \cdot \frac{h}{\mu_0}$, where $k$ is an integer. Therefore, in a self-consistent theory a charge of any particle must be a multiple of the elementary charge provided $\frac{h}{\mu_0 M} = e \Rightarrow M = \frac{h}{\mu_0 e} \approx 3{,}3 \cdot 10^{-9}$ C·m/s.

TABLE OF ANSWERS AND EVALUATION CRITERIA