Electromagnetism Physics Problems: Get Expert Help with Plainmath

Magnetic force on open circuit?
Let's say we have a straight horizontal wire and we let it drop inside a magnetic field which will be parallel to the ground(coming out of the screen). Charges inside the wire feel a force due to their movement inside the magnetic field .
Let's say the field is coming out of the screen. Positive charges gather on the left and negative on the right. If we had a loop I would have no trouble with this but during the charges' movement do we consider that we have a current? Therefore leading to a magnetic force opposing the bar's drop or not?

Direction of magnetic force
Does magnetic force act along the line joining the centres like gravitational and electric forces do?
Are the directions of magnetic field and magnetic lines of force the same? I have read that the direction of the field is tangential to the direction of the line of force.

What are the specific electronic properties that make an atom ferromagnetic versus simply paramagnetic?

The area of a rectangular loop is $200c{m}^2$, and the plane of the loop makes an angle of ${41}^{\circ <!--\; \circ \; -->}$ with a 0.28-T magnetic field. What is the magnetic flux penetrating the loop?

Is magnetic force pseudo?
Magnetic force exist only if charge is moving, so it must be pseudo. Imagine, a positively charged man who has the same speed as electron (charge). So, he doesn't feel any magnetic force as charge is at rest with respect to him. Therefore, he only experience electric force.
However a man who is at rest or has different speed than electron feels a magnetic force
Therefore magnetic force must be pseudo. Pls answer me

Given the quantum Heisenberg model with Hamiltonian
$\stackrel{^<!--\; ^\; -->}{H}=-<!--\; -\; -->\frac{1}{2}\underset{i,j}{\sum <!--\; \sum \; -->}{J}_{ij}{\stackrel{^<!--\; ^\; -->}{\mathbf{\text{S}}}}_i\cdot <!--\; \cdot \; -->{\stackrel{^<!--\; ^\; -->}{\mathbf{\text{S}}}}_j$,
the uniform mean-field approximation ${\stackrel{^<!--\; ^\; -->}{\mathbf{\text{S}}}}_i=⟨<!--\; ⟨\; -->\stackrel{^<!--\; ^\; -->}{\mathbf{\text{S}}}⟩<!--\; ⟩\; -->+({\stackrel{^<!--\; ^\; -->}{\mathbf{\text{S}}}}_i-<!--\; -\; -->⟨<!--\; ⟨\; -->\stackrel{^<!--\; ^\; -->}{\mathbf{\text{S}}}⟩<!--\; ⟩\; -->)$ allows to rewrite it as
${\stackrel{^<!--\; ^\; -->}{H}}_{MF}=\underset{i}{\sum <!--\; \sum \; -->}{\mathbf{\text{B}}}_{eff}\cdot <!--\; \cdot \; -->{\stackrel{^<!--\; ^\; -->}{\mathbf{\text{S}}}}_i+\text{const.}$,
in order to perform a diagonalization by means of a Fourier transform. To do so, I am told to choose the z-axis so to align with the effective field ${\mathbf{\text{B}}}_{eff}$, which I find to be ${\mathbf{\text{B}}}_{eff}=-<!--\; -\; -->⟨<!--\; ⟨\; -->\stackrel{^<!--\; ^\; -->}{\mathbf{\text{S}}}⟩<!--\; ⟩\; -->\underset{j}{\sum <!--\; \sum \; -->}(J_{ij}+J_{ji})$. That's where I remain stuck. First of all, how should the Fourier transform which allows me to diagonalize ${\stackrel{^<!--\; ^\; -->}{H}}_{MF}$ look like? And how is this related to the alignment of the effective field?

Gauss's law for magnetism is stated as followed with the beautiful closed surface double integral :
$\underset{S}{\text{∯<!-- ∯ -->}}<!--\; \; -->\mathbf{B}\cdot <!--\; \cdot \; -->\text{d}\mathbf{A}=0$
As I understand, the idea is to say that if we sum (continuous sum since integral) all the scalar products between the vector field B (i.e., magnetic field) and surface elements dA defined by their surface normals, we get 0?

Can magnetic force do work?
I have been told numerous times that magnetic force do no work at all but I have some trouble digesting this fact. Now suppose we have two straight wire with some current, they certainly can feel force which may be repulsive or attractive depending upon current direction, can magnetic force do work? We also have magnetic potential energy defined to $U=-<!--\; -\; -->\stackrel{\to <!--\; \to \; -->}{\mathrm{\mu <!--\; \mu \; -->}}\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{B}$ which suggests magnetic field can store energy and hence do some kind of work.

Gauss's Law of Magnetism shows us that the divergence of Magnetic field is 0, $▽<!--\; ▽\; -->\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{B}=0$
Then how do you derive that statement by showing the divergence of a magnetic field upon an axis of a current carrying coil where radius is much smaller that distance so that we can use,
$B_z=\frac{{\mathrm{\mu <!--\; \mu \; -->}}_{o}I}{2z^3}\stackrel{^<!--\; ^\; -->}{z}$
$\therefore <!--\; \therefore \; -->$
$▽<!--\; ▽\; -->\cdot <!--\; \cdot \; -->B\equiv <!--\; \equiv \; -->\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}z}\cdot <!--\; \cdot \; -->\frac{u_{o}I}{2z^3}\stackrel{^<!--\; ^\; -->}{z}\ne <!--\; \ne \; -->0$
This doesn't equal zero? What am I missing?

In explaining/introducing second-order phase transition using Ising system as an example, it is shown via mean-field theory that there are two magnetized phases below the critical temperature. This derivation is done for zero external magnetic field B=0 and termed spontaneous symmetry breaking The magnetic field is then called the symmetry breaking field. But, if the symmetry breaking occurs "spontaneously" at zero external field why do we need to call the external magnetic field the symmetry breaking field? I am confused by the terminology.

Direction of magnetic force on a permanent magnet
How would one calculate the direction of the magnetic force on a permanent magnet? I have read about the poles of ferromagnets aligning in a magnetic field, so would the magnetic force just create a torque on the ferromagnet until it aligns with the field?

Why are the electric force and magnetic force classified as electromagnetism?
I confuse the four kinds of fundamental interactions, so I think the electric force and magnetic force should not be classified as a big class called electromagnetism.
Here is my evidence:
1.The Gauss law of electric force is related to the surface integration but the Ampere's law corresponds with path integration.
2.The electric field can be caused by a single static charge while the magnetic force is caused by a moving charge or two moving infinitesimal current.
3.The electric field line is never closed, but the magnetic field line (except those to infinity) is a closed curve.

If Div B = 0, where B = magnetic field intensity, then B must be a Curl of a some vector function. What is that vector function?

Solve the equations for $v_x$ and $v_y$ :
$m\frac{d(v_x)}{dt}=q{v}_{y}Bm\frac{d(v_y)}{dt}=-<!--\; -\; -->q{v}_{x}B$
by differentiating them with respect to time to obtain two equations of the form:
$\frac{d^{2}u}{d{t}^2}+{\mathrm{\alpha <!--\; \alpha \; -->}}^{2}u=0$
where $u=v_x$ or $v_y$ and ${\mathrm{\alpha <!--\; \alpha \; -->}}^2=qB/m$. Then show that $u=C\cos <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}t$ and $u=D\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->}t$, where C and D are constants, satisfy this equation
Whenever I differentiate the first equation with respect to time, I get a resulting equation with the form:
$\frac{d^{2}u}{d{t}^2}+{\mathrm{\alpha <!--\; \alpha \; -->}}^2\frac{du}{dt}=0$

The current in a conductor is found to be moving from (-1, 2, -3) m to (-4, 5, -6) m. Given that the force the conductor experiences is $\stackrel{\to <!--\; \to \; -->}{F}=12(a_y+a_z{)}^N$ and the magnetic field present in the region is $\stackrel{\to <!--\; \to \; -->}{B}=2a_{x}T$, determine the current through the conductor. a.) -1 A;
b.) -2 A;
c.) This problem has no solution.
d.) 2 A;

Magnetic force between moving charges
Given two infinite parallel charged rods with equal charge density $\mathrm{\lambda <!--\; \lambda \; -->}$. They are moving with same constant velocity $\stackrel{\to <!--\; \to \; -->}{v}$ parallel to the rods. Find the speed $v$ for which the magnetic attraction is equal to the electrostatic repulsion.
Well, I know how to solve this problem: we first find out the magnetic field created by one rod on the other using Biot's and Savart's law, then we use the definition of $\stackrel{\to <!--\; \to \; -->}{B}$ ( $d\stackrel{\to <!--\; \to \; -->}{F}=\stackrel{\to <!--\; \to \; -->}{v}dq\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{B}$ ) to find the magnetic force, then equate magnetic and electrostatic forces to find $v$, which will be greater than or equal to $c$, thus conclude it is impossible for the forces to be equal.
However, one can argue as the following:
We all know that "same laws of physics apply in all inertial frames". With a constant velocity $\stackrel{\to <!--\; \to \; -->}{v}$,the rest frame of the rods is an inertial frame. Therefore, if Biot-Savart law applies in our frame, it has to apply in the rest frame. If so, none of the rods will feel a magnetic field from the other one because their relative speed is zero, and there will be no magnetic force between the rods.
I've seen this question several times before in references, exams, exercise sheets,and in many different forms (parallel planes, beam of electrons ...),but no one ever used this argument.What is the problem in it ? Is it something related to Maxwell's equations or special relativity ? Or what else ?
I know a similar question was asked before, but the answers weren't satisfying. Please provide your answers with necessary mathematics.

In Maxwell's equations, I understand intuitively how: $\oint <!--\; \oint \; -->B\cdot <!--\; \cdot \; -->da=0$ (because there are no monopoles and so equal number of field lines going in and coming out of the surface).
And then using the divergence theorem:
${\int <!--\; \int \; -->}_V(\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->B)d\mathrm{\tau <!--\; \tau \; -->}={\oint <!--\; \oint \; -->}_{S}B\cdot <!--\; \cdot \; -->da$
Then ${\int <!--\; \int \; -->}_V(\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->B)d\mathrm{\tau <!--\; \tau \; -->}$ must be = 0.
But then I'm not sure why I can say: $\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->B=0$ and forget about the integral.
Does it just mean that $\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->B$ must be zero everywhere?

Electric Force is to Magnetic Force as Gravitational Force is to ...?
One can no nothing about the magnetic force and yet arrive at it by taking the relativistic effects of a current and a moving charge system into account. I ask whether there exists such an inherent force in case of gravity.\

Magnetic force and work
If the magnetic force does no work on a particle with electric charge, then: How can you influence the motion of the particle? Is there perhaps another example of the work force but do not have a significant effect on the motion of the particle?

$M=\sqrt{{\displaystyle \frac{4U}{3}}}\mathrm{\varphi <!--\; \varphi \; -->}$
where M is the ferromagnetic order parameter and $\mathrm{\varphi <!--\; \varphi \; -->}$ is the auxiliary field from the Hubbard-Stratonovich transformation. The book argues that because the above equation is correct, the mean field theory which is derived from the Hartree-Fock approach is equivalent to the the saddle point approximation formalism for H-S transformation auxiliary field Lagrangian. But I can not understand the equation.