Electromagnetism Physics Problems: Get Expert Help with Plainmath

I am currently studying magnetism as a part of AP Physics 2, and what confused me was that a Tesla was the unit for magnetic field strength(which decreases with distance) but it was also used to measure the strength of magnets. Why is this the case? Is there not another metric that could be used to measure the total strength of the magnet? Also, when the strength of a magnet is given in Teslas, at what point from the magnet is that number derived?

How mass (a magnet) affects magnetic force?
I'm here asking this as a student. Magnetic force can be effected by distance, the longer the distance, the weaker, the shorter the stronger. But does mass also effect magnetic force? Does a heavier magnet produce stronger magnetic force, compare to lighter one. when they're measured in the same distance? If so, is there an equation or formula to calculate it, the strength of the magnetic force? I've been looking for calculation formulas about magnetic forces for magnets on google, but I can't find anything about how mass affects magnetic force for a magnet. (Maybe I'm just using the wrong key word.)
Ps: This has nothing to do about electromagnetism, just normal magnets.

Which magnetic force is stronger? Electron or proton?
Assume the particles generates the magnetic field with their own spin. No motions except spin.(they have their intrinsic angular momentum.) Then which one generates stronger magnetic force?
I've seen the formula which represents the magnetic moment of the 1/2 spin particles with charge $q$, mass $m$, $\stackrel{\to <!--\; \to \; -->}{\mathrm{\mu <!--\; \mu \; -->}}=\frac{g_{s}q}{2m}\stackrel{\to <!--\; \to \; -->}{S}$ where $g_s$ called g-factor.
But I can't understand the formula.
Let me ask a question, which magnetic force is stronger between electron and proton?

What I understand is that Synthetic magnetism is just a fancy name of a method to make a charge neutral particle act like it is in a magnetic field.
A charged particle in a magnetic field acquires a geometric phase, so a neutral particle if by any method is able to acquire this geometric phase, then that method is said to create a synthetic magnetic field.

In Three Lectures On Topological Phases Of Matter section 2.1 mentioned, that:
$I^{\mathrm{\prime <!--\; \prime \; -->}}=\int <!--\; \int \; -->dt{d}^{3}x(\stackrel{\to <!--\; \to \; -->}{a}\stackrel{\to <!--\; \to \; -->}{E}+\stackrel{\to <!--\; \to \; -->}{b}\stackrel{\to <!--\; \to \; -->}{B})$
correspond to ferromagnetism and ferroelectricity. And that
$I^{\mathrm{\prime <!--\; \prime \; -->}\mathrm{\prime <!--\; \prime \; -->}}=\int <!--\; \int \; -->dt{d}^{3}x(a_{ij}{E}^i{E}^j+b_{ij}{B}^i{B}^j)$
correspondence to electric and magnetic susceptibility.
Could somebody clarify, why?

The given equation $\int <!--\; \int \; -->\stackrel{\to <!--\; \to \; -->}{E}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{ds}=-<!--\; -\; -->\frac{d{\mathrm{\Phi <!--\; \Phi \; -->}}_m}{dt}$ is called dt
(a) Faraday's law
(b) Ampere - Maxwell law
(c) Gauss's law in electricity
(d) Gauss's law in magnetism

A microwave oven operates at 2.4 GHz with an intensity inside the oven of $2700W/m^2.$
What is the amplitude of the oscillating electric field?
What is the amplitude of the oscillating magnetic field?

Magnetic force in relation to velocity
I'm misunderstanding something very important regarding magnetic force and its relation to velocity.
According to the Lorentz force, the magnetic force ${\mathbf{F}}_{\mathbf{B}}=q\mathbf{v}\times <!--\; \times \; -->\mathbf{B}$. Assuming the charge is not moving, then $\mathbf{v}=\left(\begin{array}{ccc}0& 0& 0\end{array}\right)$. Therefore, ${\mathbf{F}}_{\mathbf{B}}=0$
So why when I hold a magnet close to a piece of metal, I can feel the magnet applying a force on the piece of metal? Since the piece of metal and the magnet are not moving, shouldn't the net magnetic force be 0? I assume the force I am feeling is from the magnetic field from the magnet and the charges in the piece of metal.

$|u|=i\stackrel{\to <!--\; \to \; -->}{A}$
$\stackrel{\to <!--\; \to \; -->}{B}=\left(2|u|\frac{\mathrm{\mu <!--\; \mu \; -->}\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{4\mathrm{\pi <!--\; \pi \; -->}|\stackrel{\to <!--\; \to \; -->}{R}{|}^3}\right)\stackrel{^<!--\; ^\; -->}{r}+\left(|u|\frac{\mathrm{\mu <!--\; \mu \; -->}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{4\mathrm{\pi <!--\; \pi \; -->}|\stackrel{\to <!--\; \to \; -->}{R}{|}^3}\right)\stackrel{^<!--\; ^\; -->}{\mathrm{\theta <!--\; \theta \; -->}}+0\stackrel{^<!--\; ^\; -->}{\mathrm{\varphi <!--\; \varphi \; -->}}.$
Gauss's Law for Magnetism stats that $▽<!--\; ▽\; -->\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{B}=0$. However, this:
$(\stackrel{^<!--\; ^\; -->}{r}\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}r}+\frac{\stackrel{^<!--\; ^\; -->}{\mathrm{\theta <!--\; \theta \; -->}}}{r}\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\theta <!--\; \theta \; -->}}+\frac{\stackrel{^<!--\; ^\; -->}{\mathrm{\varphi <!--\; \varphi \; -->}}}{r\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}\mathrm{\varphi <!--\; \varphi \; -->}})\cdot <!--\; \cdot \; -->[\left(2|u|\frac{\mathrm{\mu <!--\; \mu \; -->}\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{4\mathrm{\pi <!--\; \pi \; -->}|\stackrel{\to <!--\; \to \; -->}{R}{|}^3}\right)\stackrel{^<!--\; ^\; -->}{r}+\left(|u|\frac{\mathrm{\mu <!--\; \mu \; -->}\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{4\mathrm{\pi <!--\; \pi \; -->}|\stackrel{\to <!--\; \to \; -->}{R}{|}^3}\right)\stackrel{^<!--\; ^\; -->}{\mathrm{\theta <!--\; \theta \; -->}}+0\stackrel{^<!--\; ^\; -->}{\mathrm{\varphi <!--\; \varphi \; -->}}]\ne <!--\; \ne \; -->0.$
All terms should cancel, but they do not. Where am I going wrong?
What I got is:
$▽<!--\; ▽\; -->\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{B}=-<!--\; -\; -->\frac{6|u|\mathrm{\mu <!--\; \mu \; -->}\cdot <!--\; \cdot \; -->\stackrel{^<!--\; ^\; -->}{r}}{4\mathrm{\pi <!--\; \pi \; -->}|R{|}^4}+\frac{|u|\mathrm{\mu <!--\; \mu \; -->}\cdot <!--\; \cdot \; -->\stackrel{^<!--\; ^\; -->}{r}}{4\mathrm{\pi <!--\; \pi \; -->}|R{|}^4}+0\therefore <!--\; \therefore \; -->\ne <!--\; \ne \; -->0.$

Magnetic force on a stationary body
A charged particle near a current-carrying wire does not experience a magnetic force when its velocity is equal to $0$. So why does a compass needle kept near a current carrying wire experience a force when the compass needle is at rest?

Earth has a magnetic dipole moment of $8.0{10}^{22}J/T$. Find the magnitude of current in a single turn of wire extending around Earth at its geomagnetic equator to produce a dipole having the same dipole moment. Could such an arrangement be used to cancel out Earth's magnetism?

Question about magnetic force
I have some confusion regarding the magnetic force. I know that the magnetic field created by a moving charge or current EXERTS a force on any moving charge or current that is present in the field. But when trying to understand the motion of a charged particle in an uniform magnetic field, the youtube video I saw explained it like this: "The magnetic force on a charged particle ALWAYS points perpendicularly with respect to the velocity and magnetic field. Whenever a force acts on an object perpendicular to its motion, the object will undergo circular motion--this creates a centripetal acceleration)"
I am confused. Is the charged particle exerting a force on itself? Or what is the force that acts on the charged particle that is moving? If the charge particle creates a force due to the magnetic field, is the force it creates itself the force that makes it undergo a circular motion?

Magnetic Force Confusing Paradox
Actually I had posted a very similar question, but I wasn't quite satisfied with the answer so I am posting a new variety again.
Imagine an infinitely long wire carrying current $I_1$ from West to East. At a small distance $d$ above the wire there is another small current carrying wire of length $l$ carrying current $I_2$ from East to West (opposite to the direction of current in the below placed wire). Obviously they are magnetically repelling. And if the second wire (carrying current from East to West) is at rest the magnetic force must be equal to $mg$
The magnetic force is upward and $mg$ is down. The magnetic force can be written as
$\frac{Uo{I}_1{I}_2}{2\mathrm{\pi <!--\; \pi \; -->}d}l=mg$
Now, if the magnetic force is greater than $mg$, the wire moves up. Now magnetic force is up and displacement is up too which means that work done by magnetic force should be positive. How is that possible when we know that work done by a magnetic force is always zero? Is it that an e.m.f. is induced which opposes the change?

Work done by the Magnetic Force
The magnetic part of the Lorentz force acts perpendicular to the charge's velocity, and consequently does zero work on it. Can we extrapolate this statement to say that such a nature of the force essentially makes its corresponding work independent of the choice of path, and hence that the magnetic force is conservative?

Lets say you have an uniform prism magnet of Iron for example. How would you calculate the demagnetization field H which the bar magnet produces? As I understand, first you need the magnetization M which is the magnetic moments per volumen. But then what ? Or would you apply and external field to the bar magnet and see how it reacts, if this is the case, how does this work ?

Magnetic force on capacitor
I have this problem: I have a capacitor formed by two parallel square-shaped plates, of side $a$, and with distance $d$ between them. They're asking me out of other things, the magnetic force as a function of some things. I calculate de magnetic force by calculating the gradient of the magnetic energy, given by:
$E=\frac{1}{2}{\iiint <!--\; \iiint \; -->}_{V}B\cdot <!--\; \cdot \; -->HdV$
I have calculated $B$ and $H$, that form, during the charge of the capacitor, loops around the center proportional to $r$, being $r$ the distance from the central axis. The problem is that I have to integrate that to a cubic volume: $a^{2}d$, one of those variables is independent, but the other two are not, and this problems never involve complicated integrals, such that one:
${\int <!--\; \int \; -->}_0^a{\int <!--\; \int \; -->}_0^{a}rdxdy$
If the plates were circles it would be trivial, but being squares, I'm sure I'm missing something.

Gauss's law states that ${\int <!--\; \int \; -->}_S\stackrel{\to <!--\; \to \; -->}{B}\cdot <!--\; \cdot \; -->d\stackrel{\to <!--\; \to \; -->}{S}=0$. But law of induction states that $\mathrm{\xi <!--\; \xi \; -->}=-<!--\; -\; -->\frac{d\mathrm{\varphi <!--\; \varphi \; -->}}{dt}$, where $\mathrm{\varphi <!--\; \varphi \; -->}={\int <!--\; \int \; -->}_S\stackrel{\to <!--\; \to \; -->}{B}\cdot <!--\; \cdot \; -->d\stackrel{\to <!--\; \to \; -->}{S}$
So if Gauss's law was to be correct there should be no induction at all, because then ϕ would be zero through every loop.

Calculate magnetic field from magnetic force on a charged particle
The magnetic force, charge, and velocity of a particle are given. I am asked to find the magnetic field.
$q=-<!--\; -\; -->2\mathrm{\mu <!--\; \mu \; -->}\mathrm{C}$
$\stackrel{\to <!--\; \to \; -->}{v}=(-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{i}+3\stackrel{^<!--\; ^\; -->}{j})\times <!--\; \times \; -->{10}^6\mathrm{m}/\mathrm{s}$
$\stackrel{\to <!--\; \to \; -->}{F}=(3\stackrel{^<!--\; ^\; -->}{i}+\stackrel{^<!--\; ^\; -->}{j}+2\stackrel{^<!--\; ^\; -->}{k})\mathrm{N}$
$B_x=0$

How can the integral form of Gauss's law for magnetism be described as a version of general Stokes' theorem? How does it follow?

My question is: is earth really a magnet? Doesn anyone have any proof that earth is a magnet? Is there a magnetic core at the center of the earth? Has anyone reached the core of the earth?