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11

D.J. Griffith's Introduction to Electrodynamics must be mentioned. To my knowledge this text is ubiquitous in junior-level E&M courses. The writing is extremely friendly and is excellent for self-study. The author frequently tells you what he is doing and provides motivation, unlike the ubiquitous graduate-level text by Jackson. Equations often use a ...


8

Purcell is a good non-Griffiths option. I would judge the completeness of the material between Griffiths and Jackson, but with an intuitive level of understanding close to Griffiths. I used it to study for graduate qual exams when Jackson was making me feel particularly obtuse. Some positives: Touches more ideas than Griffiths Uses some real-world ...


4

Besides Purcell I really like Feynman Vol. II. I finally could understand magnetic materials and electromagnets. (Warning, Feynman uses his own notation for B,H and M.) The lectures are available online and for free, as the New Millenium Edition, at http://www.feynmanlectures.caltech.edu/, in a nice re-mastered edition with re-drawn ...


4

It does matter, and the product comes first. In general any sort of multiplication is understood to have higher precedence than any sort of addition. Thus $$ \vec{E} + \vec{v} \times \vec{B} \equiv \vec{E} + (\vec{v} \times \vec{B}). $$


4

No, different EM waves do (classically) not interact, they just pass through each other. There are (tiny) contributions to a photon-photon diagram in quantum electrodynamics which could be seen as photons scattering off each other, but, at the macroscopic level where we usually talk about EM waves, there isn't any interaction at all.


3

W. K. H. Panofsky and M. Phillips, Classical electricity and magnetism, Addison Wesley, 2nd ed., 1962 Especially the first 14 chapters are very enjoyable yet carefully written study text about both basic and more advanced topics in macroscopic EM theory (including discussion of EM energy from more experimental angle than is usual and of density of force ...


3

This is a hypothetical question, since electrons are elementary particles and protons are composite. The solutions of the potential problem would give stable orbitals with smaller average radii. Here is a Bohr model solution for the muonic hydrogen, where the muon is 200 times heavier than the elecron. The energies become KeV instead of eV. To go to the ...


2

Jackson's classical electrodynamics is very complete, and often seen as the reference on CED. But I also like Rohrlich's classical charged particles that, as the title suggests, puts more emphasis on the subject of particles interacting with EM fields.


2

Here's a derivative-free explanation. For readers who are doing E&M at the college level, the other answers posted here are more comprehensive, but since the OP has stated a high-school knowledge with little math and physics knowledge, here's the primer: A vector is a quantity that, in order to be fully measured and described, needs to include both a ...


2

Floris's answer gives you an excellent description of the forces that would be present in both a magnetic and gravitational field, whilst MaxGraves's Answer gives you a clear and careful discussion of how you should use the word weight. In more the spirit of MaxGraves's Answer, something that seems a little pedantic but may be interesting to you is the ...


2

If I understand correctly you are asking how observer dependent is electromagnetic radiation. The first thing is that non uniformly accelerated charges are described in a inertial frame by Larmor's formula and Abrahm-Lorentz force which take into account the radiated field and the recoil on the particle. Now in Newtonian mechanics and special relativity ...


1

How about this: Think about a loop around the central wire, but inside the outer conductor, with its surface defining a plane perpendicular to the wire. Ampere's law tells us the induced magnetic field curls around the central wire. [Even if there were a time-dependent radial E-field it would not produce any displacement current through the loop.] Having ...


1

Taking your (lack of) knowledge about differential geometry into account, this might be too hard to follow, but here it goes anyway: Let $u_1,\dots,u_n$ be some tangent vectors with base point $p$ and $\omega$ the volume form, ie $V = \omega_p(u_1,\dots,u_n)$ is the (possibly negative) volume of the parallelepiped spanned by these vectors. In case of three ...


1

The two maxwell equations using divergence are $$ div \vec{D} = \rho \\ div \vec{B} = 0 $$ at least in differential form. In integral form they are maybe more clearer for you. They are $$ \iint_{\partial V} \vec{D} \ d \vec{A} = \iiint_{V} \rho \ dV = Q(V) \\ \iint_{\partial V} \vec{B} \ d \vec{A} = 0$$ The first equation just means the electrical flux $D$ ...


1

Ions can indeed carry current (ex. electrolysis). "An electric current is a flow of electric charge. In electric circuits this charge is often carried by moving electrons in a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in a plasma.[1]"


1

Well, technically yes (assuming you mean ionized and not just "charged"). Current (in simple terms) is only the time rate of charge flow, which is not exclusively limited to electrons or any specific charge carrier. Electrolytic conductivity is well documented, naturally being higher for strong electrolytes as compared to the ones that dissociate weakly in ...


1

Yes, electric current is movement of any kind of charges. The problem with your particular example is that most liquids containing ions are also conductive. Electrons will hop between molecules and equalize the ionic charges, then end up providing most of the conduction themselves. There are cases where actual ion migration results in much of the current, ...


1

The sort of magnet most of us immediately think of is a magnetic dipole. This has the characteristic field with north and south poles: This field geometry is produced by current travelling in a loop: Note that the field lines don't begin or end anywhere but pass through the centre of the loop. The South pole is where the field lines enter the loop and ...


1

I'm not really sure if you want to apply this in a classical or quantum mehcanical context, but the application is similar in both cases. Given a gauge choice, the electric and magnetic field are given by $$\mathbf{E}(x,t)=-\nabla V(x,t)-\frac{\partial \mathbf{A}(x,t)}{\partial t};\\ \mathbf{B}(x,t)=\nabla\times\mathbf{A}(x,t),$$ and the classical equations ...


1

The current is not directly due to the magnetic field, rather it is due to the electric field that is induced by the changing magnetic field. It is true that the electrons will experience a force from the magnetic field according to the Lorentz Force Law, but this force will always be perpendicular to the direction of motion and therefore will not produce a ...


1

The short answer is no, the diameter of the wire doesn't affect the bandwidth. Bandwidth can be a tricky subject. If you are talking about injecting a very high frequency sine wave at one of the wire, and seeing if it is detectable at the other end, then wires of all diameters have a surprisingly high bandwidth. But if you are talking about the ability of ...


1

The cross product of two vectors is always perpendicular to both vectors. So, the magnetic force $\vec{F}=q\vec{v}\times\vec{B}$ will always be perpendicular to $\vec{v}$. This can be seen in your special case by realizing thatsince the motion is in the $(x,y)$ plane $v_z=0$, and the only component of $\vec{B}$ that is not null is $B_z=B$, we have ...


1

When current flows in a circle, we can use the right hand rule to find that the magnetic field points in one direction inside the loop, and in the other direction outside the loop (to be precise, the magnetic field lines wrap around the wire, as shown in the second diagram below). Thus, in the case of current carrying loops, if we curl the fingers of our ...


1

$\Phi$ is the flux of $\vec B$ through one surface $$\Phi = BA$$ where A is the area bounded by a loop (turn). But there are $n$ turns and thus $n$ surfaces that are pierced by $\vec B$ so the the flux linkage $\lambda$ is $$\lambda = n \Phi$$



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