# Tag Info

## New answers tagged electromagnetism

0

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

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 ...

-3

The electromagnetic induction of a moving charge in a magnetic field is based on the electron’s magnetic moment. The magnetic field turns the magnetic moment of the electron in the direction of the magnetic field. The motion of the electron undergoes a - predictable and perpendicular on the two vectors of the velocity and the magnetic field - acceleration ...

0

Well there is energy in the magnetic field. (Search for energy density in magnetic field.) It's B^2/(2*mu_sub_zero) in MKS units. (And as long as no magnetic materials are around.) The act of turning on the magnetic field creates a changing B field for a short time. A changing B field creates an electric field, and this can cause current to flow in ...

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.

0

I agree that this seems mysterious on first meeting. Multiplication by $n$ is shown to be correct by something called Stokes's Theorem, which you won't have met yet, and which lets us translate the basic equations of electromagnetism (called their Maxwell Equations) between their local and "spread out" forms. But at an easier level, think of a very long, ...

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 $$\lambda = n \Phi$$

0

As CuriousOne says, look carefully for a test suite within your software installation. MEEP is widespread, notwithstanding the LISP interface (gotta love MIT's confidence in its own creations), so if you seek carefully, you are bound to find MEEP analysed examples on the web. As for your proposed simple test: it is a good idea, and there are many, well ...

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 $$\vec{E} + \vec{v} \times \vec{B} \equiv \vec{E} + (\vec{v} \times \vec{B}).$$

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 ...

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 ...

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

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 ...

0

Almost all electrical machines can be run in both ways (generator or motor). If you're talking about the direction of rotation, it will work too. However, your message is so vague, we can't help you : we don't have any clue about the kind of machine, about the frequency, voltage, use...

0

Electrons don't spin. It is just what they decided to call a certain intrinsic property. They could of called it the X factor or magnetic factor but they called it spin for some reason

0

Let us start with the question, how can one determine the poles of a magnet. It is purely an agreement which pole of a permanent magnet is the north pole and which is the south. But fortunately it seems that all other the world the poles are marked in the same way. So you carry your permanent magnet to the pole of an electromagnet and you feel if this is the ...

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

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

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

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]"

0

My answer will probably be a little off topic, but why do you need this kind of analytical function ? Since you have the magnetic flux density (FEM simulation I guess), you can use any interpolation to get the B value anywhere... Unless you have a very specific need, if you only want to get the value of B anywhere, that is probably the easiest solution. I ...

0

I don't know that you could produce such a ring by putting individual neodymium magnets next to each other — the inside/south of one would repel the inside/south of its neighbor, and so they'd want to jump around until they were at different radius and the south and north parts could overlap. However you can certainly produce a ring of radially ...

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 ...

0

The polarisation field can be imagined as due to lots of tiny dipoles within the medium, that are created in response to an applied electric field. The dipoles reduce the electric field within the medium (at least for materials with a dielectric susceptibility >0). Imagine the situation where you have a charged, plane-parallel capacitor, with a dielectric ...

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 ...

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

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 ...

4

Divergence is an operation that maps a vector field $\vec D(x,y,z)$ to a scalar field ${\rm div}\,\vec D(x,y,z)$. How do you calculate ${\rm div}\,\vec D(x,y,z)$? Either you follow the definitions using derivatives, which you can't if you don't know what a derivative is. Or you imagine the following: the vector field $\vec D(x,y,z)$ tells you about the ...

1

In addition to @Floris response: You have missed a lot of wavelengths in your list of wavelengths that would experience interference. Take your example of a $6,000,000 \text{ nanometer}$ pane of glass, and consider that 15,000 waves of $400 \text{ nanometer}$ wavelength light exactly fills this space. So, indeed, this light will experience some sort of ...

4

The wikipedia article is correct. The relation between the actual response function and the conductivity is not immediately obvious, however. Let us consider the case of the longitudinal conductivity for example. The susceptibility, $\chi^L(\textbf{q},\omega)$, which is the true response function, is related to the conductivity using the following equations: ...

3

Very simply, when a plate is quite thick, the fringe patterns will be very close together - because a tiny change in angle will result in an additional wavelength's worth of path difference. Different colors will have a different repeat distance (because of different wavelengths); and light will typically arrive at the eye from more than one direction ...

0

There is a deeper meaning of the magnetic length $l_B$. It is the lattice constant of a two-dimensional (2D) artificial structure. Let's consider a particle moving in 2D plane under a perpendicular uniform magnetic field of strength $B$. In classical mechanics, the system is obviously translationally invariant under any translation. However, in quantum ...

2

A motor-generator can be used to convert electric power from one voltage x current combination to another voltage x current combination. Such systems have been used, and are sometimes still used, for exactly this purpose. One advantage is that a large flywheel can be added to the shaft, effectively low pass filtering the average power. This can be useful ...

0

From the comments of @Alfred Centauri and @Mateus, I think I've figured it out: In the stationary frame the B-field is static so there is no E-field, but we still measure an emf because the B-field is acting on the electrons in the wire by the Lorentz force law. In the primed frame the loop appears to be stationary, so the emf comes instead from the ...

0

Wonderful question! And it relies on one central concept: virtual particles. Particles that do not necessarily exist, but explain (or would explain) phenomena wonderfully. By assuming that the particles (electrons, protons, etc) are exchanging photons, we can predict their behavior with great accuracy. Imagine that two protons are people, and they are ...

3

I think the better way to derive this is to first observe the Biot-Savart law, $$\mathbf B(\mathbf r)=\frac{\mu_0}{4\pi}\int\mathbf J(\mathbf r')\times\frac{\hat{r}}{r^2}dV'\tag{1}$$ Since $$\frac{\hat r}{r^2}=-\nabla_r\left(\frac1r\right)$$ (your text may derive this, if not you can prove it by starting with the RHS), we can write (1) as $$\mathbf ... 1 I also dislike when authors claim things to be obvious. If it's so simple, then why not just write it out. Anyhow, regarding this specific case. If you go to the definition of the curl you will see that this is a collection of partial derivatives with respect to position. So to claim that the curl is zero is to claim that the velocity is independent of the ... 1 I believe it is a good approach to use your intuition and real physics to guess how the effects should work. However, electromagnetism is the place where you start to need to understand the fields and other less tangible concepts. All the pictures you show describe the situation from a point of view where your distance d from the oscillating charge is ... 0 Unfortunately I have not met the reputation threshold to post comments, so I'm resorting to providing an answer. I would assume this would have to do with a common misconception you may have of an electron being a billiard-ball-like spherical ball of matter traveling along a wave. This is not quite accurate, though you may very well already be aware of ... 2 EMP is typically a high frequency signal. It is possible to allow low frequency signals to enter your Faraday cage by decoupling the signal - a choke in series and a capacitor in parallel. You need to make sure that the choke does not saturate at the current spikes expected - and that the voltage rating of the capacitor is sufficient to absorb the energy. A ... 0 If you're concerned about Ohm's law, you might be getting a little muddled in your thoughts (don't worry, I only know this because I've had my turn at being muddled before on this point!) Ohm's law doesn't describe a transformer: Ohm's law may apply to whatever load you connect to the transformer; if so, then Ohm's law combines with the transformer law ... 0 My thought would be: don't use cosine rule to solve this problem. Instead resolve the forces due to each charge along the x and y axes and write them as vectors. Hence:$${\bf F}_{21} = -k\frac{q_{1} q_{2}}{r_{21}^{2}} [\cos(23^{\circ}) \hat{\bf x} + \sin(23^{\circ}) \hat{\bf y}]{\bf F}_{31} = -k\frac{q_1 q_3}{r_{31}^{2}} [\cos(23^{\circ}) \hat{\bf ...

1

As Rod notes in his answer, any purely linear circuit will certainly behave the same way if all the currents (and voltages) are scaled by any (non-zero) constant value $\alpha$, including the case $\alpha = -1$. In fact, a typical feature of linear circuit elements, like resistors, inductors and (non-electrolytic) capacitors, is that they'll work equally ...

0

To add to Rijul Gupta's answer: depending on the exact symmetry of the arrangement (see below), theoretically, the needle can move freely to point in any direction. Practically, there will be some direction which is one of minimum energy, so the needle will have a weak tendency to point in this minimum energy direction. Try analysing your situation with some ...

0

Take a infintesimally small element 'dl' ,now the potential due to this element is dv=1/4piεo λdl/r^2 and r =sqrt(z^2 +R^2) ,(which is const) And thus integrate 'dv' And since 'r' and all others in the equation are const except 'l' thus integrating 'dl' will give 2piR (problem solved)

4

I suggest you should look at all your arguments to date and see what they have in common. I'd almost be willing to bet that they can be reduced to something like the following: The equations governing a circuit's behaviour are linear in the vector of state variables, amongst whose members is the current. That is, if $\vec{U}$ is a column vector of state ...

1

From Magnet University You may follow an imagined direction of current (conventional flow) or the actual (electron flow) with equal success insofar as circuit analysis is concerned. Concepts of voltage, current, resistance, continuity, and even mathematical treatments such as Ohm's Law and Kirchhoff's Laws remain just as valid with either style of ...

0

This question is using spherical coordinates, and in spherical coordinates the divergence of a vector (like $\nabla\cdot \textbf{E}$) is: $\nabla\cdot \textbf{A}=\frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2 A_r \right) + \frac{1}{r\sin\theta}\frac{\partial}{\partial \theta} \left( A_\theta\sin\theta \right) + \frac{1}{r\sin\theta}\frac{\partial ... 0 The electric field given in the problem only depends on the radius$r$. Therefore, the divergence is most conveniently computed in spherical coordinates. The factors of$r^2$and$1/r^2$come from the transformation of$\nabla$to spherical coordinates, see here. For a function$f(r)\hat{r}$the divergence is$\$\nabla\cdot ...

0

Look at the expression for the divergence in the spherical coordinates.

Top 50 recent answers are included