What do we mean with magnetic monopole and dipole? I can not find a
way to relate magnetic monopoles and dipoles with electric ones. I do
not understand their outcomes.
Luckily, there exists a truly amazing one-to-one correspondence between magnetism and electricity.
Monopole in magnetism is analogous to charge in electrostatics/electricity. Just like we have two types of charges (+ve & -ve), we have two types of monopoles (north & south). The north magnetic pole is also known as the positive pole and the south magnetic pole is known as the negative pole.
We refer to the size/magnitude of the charge as charge itself, but for a monopole, we refer to its size/magnitude as pole strength.
Electric charge is denoted by the letter 'q', the pole strength of a magnetic pole is denoted by the letter 'm'.
Electric charges produce electric fields. The magnetostatics analog is the magnetic field.
We make use of electric field lines to represent the electric field visually. By convention, electric field lines start from a positive charge and terminate at a negative charge. So is the case with magnetic poles as well. Magnetic field lines originate from the north pole (+ve pole) and terminate at the south pole (-ve pole).
The strength of the magnetic field produced by a monopole is given by
$$\vec{B} = (\frac{\mu_o}{4\pi})\frac{m}{r^3}\vec{r}$$
You probably noticed that the equation is identical to Coulomb's law except for the fact that we have magnetic pole strengths instead of the magnitude of charge.
As of 2016, we aren't sure if magnetic monopoles exist (we haven't found one yet). When we try to make a monopole by slicing a bar magnet exactly in the middle, we end up creating two new bar magnets where each magnet has both north and south pole.
Well, you might ask how do I know about the behavior of monopoles if monopoles don't exist. Whatever I said above is just a hypothesis and the hypothesis is consistent with the reality.
When two magnetic poles of equal pole strength are kept close to each other, we call it a dipole. A bar magnet is an example of a dipole.
We have an electric dipole moment equivalent in magnetostatics. We call it magnetic dipole moment (usually represented by '$\vec{M}$').
$$M = md$$
where d is the distance between the two poles and m is the pole strength of the poles. The direction of the magnetic dipole momentum is from negative pole to the positive pole (In the case of an electric dipole, it is from the negative charge to the positive charge).
The field along the axis of the magnetic dipole is given by,
$$B = 2(\frac{\mu_o}{4\pi})\frac{M}{r^3}$$
and the field along the equatorial line of the dipole is given by
$$B = (\frac{\mu_o}{4\pi})\frac{M}{r^3}$$
If you open your textbook and look for the formulae of the electrostatics analogue of the dipole, you will find that the formulae are a perfect match.
The formulae given above can also be derived from the magnetic monopole hypothesis.
The similarities don't end here.
Torque on an electric dipole in an external uniform electric field is given by
$$\tau = \vec{p}\times\vec{E}$$
Torque on a magnetic dipole in an external uniform magnetic field is given by
$$\tau = \vec{M}\times\vec{B}$$
You shouldn't be surprised if you find more exact matches. The magnetic monopole behaves just like electric charge and the formulae are identical. The derivations for magnetic dipoles are identical to derivations for the electric dipoles so you should end up with the same formulae.
A circulating electric current behaves like a magnetic dipole.
If you derive the formula for magnitude of magnetic field along the axis of the circular coil, you should get something similar to the following,
$$B = \frac{\mu_o NiR^2}{2(R^2 + x^2)^{\frac{3}{2}}}$$
where $N$ is the number of turns in the coil, $i$ is the current passing through the wire, $R$ is the radius of the coil and $x$ is the distance along the axis of the coil.
If we use the approximation $x>>R$, we get,
$$B = \frac{\mu_oNiR^2}{2x^3}$$
Rearranging further, we get
$$B = 2(\frac{\mu_o}{4\pi})\frac{(iN\pi R^2)}{x^3}$$
If you look carefully, you will notice that the formula looks very similar to the formula which gives the strength of the field along the axis of a magnetic dipole.
We can define magnetic dipole moment as
$$M = iN(\pi R^2)$$
The direction of the magnetic moment is the direction of the magnetic field at the center of the coil.
Now the formula looks pretty much identical to the formula given earlier for the strength of the field along the axis of a magnetic dipole.
In general, magnetic moment for any closed loop circuit can be defined as
$$\vec{M} = iN\vec{A}$$
where $\vec{A}$ is the unit normal vector for the loop.
- Also,what is their role in Gauss' law for magnetism (the net magnetic
flux through a closed surface is zero)?
I read that the magnetic dipoles are essential for the meaning of this law. Why?
If magnetic monopoles existed,then why would the law not be valid?
- Lastly,why do we say that the magnetic field is divergeless?
One of the consequences of monopoles not existing is that magnetic field lines are always closed. Consider any dipole, a field line which emanates from the north pole must end up back at the south pole.
The number of field lines that leave the surface is equal to the number of field lines entering the surface. Hence, there is no net flux entering or leaving the surface.
$$\oint{\vec{B}.d\vec{A}} = 0$$
In the Gauss Law for electrostatics, we consider the volume charge density or the net charge enclosed within the gaussian surface. The same goes for Gauss Law for magnetism.
$$\oint{\vec{B}.d\vec{A}} = \mu_o m_{enclosed}$$
since there can never be any isolated monopole, $m_{enclosed}$ is always zero.
If monopoles existed, then $m_{enclosed}$ needn't be zero. Hence, the Gauss law would be invalid.
Some physicists try to find a magnetic monopole (something Dirac tried
to explain i think). So if they actually find one, what does that
mean?
If we ever find a monopole, it would imply that Gauss law for magnetism would be invalid. We will need to amend the law to make allowance for monopoles. (the amended version of the law was presented in the answer to your previous question)
Discovery of magnetic monopoles would imply the existence of magnetic currents. We would have a new chapter in our physics textbooks. Possibly a new engineering branch might appear (magnetronics?).
For a classic magnet(N-S) we know that we have charges inside moving
in a circular motion and they althogether form that magnetic field of
the magnet. So how does finding a magnetic monopole change this? What
does it mean for those small currents?
Every electron has two types of magnetic dipole moment, namely spin magnetic dipole moment and orbital magnetic dipole moment.
There are several electrons in an atom and most electrons cancel out each other's magnetic moment. When all the electrons cancel out neatly, the total magnetic moment of the atom is zero. The substances where the total magnetic moment of the constituent atoms is zero are known as diamagnetic substances.
There are many substances with unpaired electrons. This gives rise to a net magnetic moment for each atom. When these atoms collectively align in the same direction. This kind of collection of atoms is known as a domain. When domains align to give a net magnetic moment to a substance, the substance is said to be ferromagnetic.
Discovery of a magnetic monopole shouldn't affect the existing theories. However, it might force us to investigate further at the fundamental level.
For the physical interpretation of the limit: imagine you had a current loop whose size you could decrease easily, like pulling on a drawstring. If you just make it smaller, you decrease the magnetic moment $\mu = IA$; to keep the field the same, you'd have to increase the current. If you cut the area $A$ of your loop in half, but doubled the current $I$, you'd have the same magnetic field far from the loop.
The ideal dipole source has zero size and infinite current, and the ideal dipole field is therefore infinitely strong at the origin. That's annoying. But we have the same problem with the monopole field, like the electric field from a point charge, which is proportional to $1/r^2$ and is infinite at the origin. For the electric field we get around this by inventing quantum mechanics and discovering that a "point charge" is not actually a thing that exists. The proton has a finite size; while the electron is a structureless "point particle," for computing its electric field you actually care about the finite charge density described by its wavefunction.
I might rephrase your second question as "what do we mean by 'large distance'?" Suppose you have a real cylindrical solenoid, made out of wires, with length $L$ and radius $R$. The dipole approximation is only good if your distance $r$ from the solenoid is much larger than $L$ or $R$. If you're inside the core of the solenoid you see a uniform field; if you are a tiny gnat tunneling through the wall you might prefer to treat the local field as due to a locally-flat sheet of current. The local field is complicated.
For a complicated field, it's helpful to describe it using a multipole expansion. I've already hinted at this by reminding you about the monopole field, which is produced by a point charge, and gets weaker like $1/r^2$ as you move away. If two opposite-sign charges are near each other, their monopole fields approximately cancel, and most of what's left over is described by the dipole field. The dipole field gets weaker with distance like $1/r^3$ — that's sort of what we mean when we say that the monopole fields approximately cancel out. The dipole field also has the more complicated shape, which leaks information about the orientation of the charges at the source.
Two back-to-back dipoles also approximately cancel out. What's left there is called a "quadrupole field," which gets weaker like $1/r^4$ and has an even more complicated shape than the dipole field. There's an infinite series of these higher-order corrections, which get weaker more rapidly as you move away.
A current loop with radius $R$ produces a magnetic field with nonzero dipole moment, but also nonzero quadrupole, octupole, hexadecapole, etc. moments, all of which are parameterized by $R$. If you move from $r$ to $2r$, the dipole field gets weaker by a factor of $2^3=8$, but the quadrupole field gets weaker by $2^4=16$. If you move many $R$ away (or equivalently, rebuild your current loop so that $R$ is very small), eventually only the dipole shape of the field will be measurable.
Best Answer
The papers you should read for this question are Vaidman, Lev. "Torque and force on a magnetic dipole." Am. J. Phys 58.10 (1990): 978-983 (Paywall-free version) and Haus, H. A., and P. Penfield. "Force on a current loop." Physics Letters A 26.9 (1968): 412-413. (References 8 and 9 in Vaidman's paper are also worth reading for more context.) The gist of the answer is this: if and only if there are no magnetic monopoles, magnetic dipoles and current loops are equivalent.
As you have identified in your point (1), there is indeed a problem with the term magnetic dipole. Let's unpack the term starting with dipole.
The simplest system in electromagnetism is the electric point charge at rest, which produces a spherically symmetric $1/r$ potential (it's easier to work with the potential here). This is called a monopole field, and a point charge is a monopole.
Now consider two equal but opposite charges a distance $a$ from each other. This breaks rotational symmetry so the field will not be spherically symmetric, i.e., it is angle-dependent. Now expand the potential at a distance $r \gg a$ from the charges in powers of $1/r$. Since electromagnetism is linear, the $1/r$ terms are equal but opposite and cancel, but there remains a term of order $1/r^2$. Because the potential is a scalar, it must have the form $\mathbf p \cdot \mathbf r / r^3$ where $\mathbf p$ is a vector called the dipole moment. You can continue the expansion; the next term is $Q_{ij} x_i x_j / r^5$ where the tensor $Q_{ij}$ is the quadrupole moment, and so on. (The names reflect the minimum number of point charges you need to have a nonzero moment of that order: one for a monopole, two for a dipole, four for a quadrupole... All the details of this multipole expansion and what it's good for are in Jackson, of course.)
So by a dipole field we mean a field whose potential looks like $\mathbf p \cdot \mathbf r /r^3$, and its source is an electric dipole. Of course the magnetic field has a vector potential rather than a scalar potential, so by a magnetic dipole field we should mean one where the vector potential is like $\mathbf A \sim \mathbf m \times \mathbf r / r^3$, and its source is a magnetic dipole.
Now, one way to construct a magnetic dipole is by the obvious analogy: take two magnetic charges, i.e., two magnetic monopoles at a small distance. Well, that's easier said than done because no one has found any magnetic monpoles. We'll have to go with currents, then. Since the $1/r$ expansion is possible only if the size of the system is finite and charge is conserved, we'll have to use current loops, and conversely, any current loop will be a magnetic dipole.
In this sense and this sense only magnetic dipoles and current loops are equivalent. If you were to find some magnetic monopoles and arrange them such that the dipole moment is $\mathbf m$, then the force on the system is $$\mathbf F_\text{MM} = (\mathbf m \cdot \nabla) \mathbf B - \frac{1}{c}\dot{\mathbf m} \times \mathbf E$$ where $\dot{\mathbf m}$ is the time derivative, whereas the force on a current loop is $$\mathbf F_\text{CL} = \nabla (\mathbf m \cdot \mathbf B) - \frac{1}{c}\frac{d}{dt} ( \mathbf m \times \mathbf E ).$$ If you expand $\mathbf F_\text{CL}$ and use Ampere's law with Maxwell's current, you see that these forces differ by $ k\mathbf m \times \mathbf J$ where $\mathbf J$ is the current density and $k$ is a constant that depends on your unit system. Clearly these magnetic dipoles are not equivalent.
(However, several authors erroneously calculate the force using the magnetic charge model and think it must be true for current loops, which, as shown by Vaidman, is not the case.)
There is one objection and that is that we know about spin, the intrinsic magnetic moment of particles such as electrons. I don't think it is obvious whether spin should be treated as a magnetic charge dipole, or as a current loop. For an elementary particle to be a current loop certainly seems strange, but it's not really less strange to think about it as a system of magnetic monopoles. One would think it's an experimental question, then, but Bohr and Pauli argued in the 20s that the spin of an individual electron is rather inaccessible to experiment, see Morrison, Margaret. "Spin: All is not what it seems." Studies in History and Philosophy of Science Part B: Studies in History and Philosophy of Modern Physics 38.3 (2007): 529-557 for an account.) In any experiment with electrons, the Lorentz force would anyway dominate the magnetic dipole force, so one would have to turn to neutrons, which come with other difficulties. Vaidman discusses this briefly.
However, theoretically, if the situation is analyzed correctly, that is, using the the Foldy-Wouthuysen transformation (the orignal paper is Foldy, Leslie L., and Siegfried A. Wouthuysen. "On the Dirac theory of spin 1/2 particles and its non-relativistic limit." Physical Review 78.1 (1950): 29 which is a real gem and should be read by everyone studying quantum mechanics.) it is found that the current-loop model is correct. You can square it with the contradiction between current loop and elementary particle by realizing that this is a quantum mechanics thing, and in quantum mechanics you don't have to have point particles. In fact, you can't, by Heisenberg's principle. The electron is always a bit spread out, and in such a way as to produce a current loop magnetic dipole moment.