New answers tagged

0

The "entanglement" is that "red" or "blue" or such things are dependent on the wavelength of the light, and the "length" part of a wavelength is a spatial measure. In our universe it appears that space is "stretching" (expanding, in the usual terminology) and it stretches the wavelengths of various waves along with it. The stretching is very slight, though, ...


0

There are many many misconceptions tied into knots in your question. Firstly, the electric force between two charges doesn't depend on just the distance between them and their charges, it also depends on their velocity. When charge A moves then its electric field is different, so the electric force charge B feels is different. This means you can't reason ...


1

The potential energy is the integral of the force from infinity to the position of the charge: $$ V(\mathbf{x}) = \int_\infty^\mathbf{x} \mathbf{F}(\mathbf{x}') \cdot d\mathbf{x}' $$ where you need to note that both $\mathbf{F}$ and $d\mathbf{x}'$ are vectors so the direction of the force matters. The easy way to do this is to note that the net force on ...


1

First of all, try to imagine a simple MRI experiment with an perfectly homogeneous main magnetic field in the order of 1-3 T and an object to image that only contains protons of the same kind (i.e. the signal-baring protons are chemically all of the same kind in the object, i.e. there is only water in your object). When the object is brought into the main ...


0

The point that you are missing is that light which is a type of electromagnetic wave consists of oscillating electric and magnetic fields. So your assertion is not correct. In the case of a metal the light causes the free electrons to oscillate and reradiate what you call the reflected light. I am not sure about the use of the term reflection in the ...


0

It might help you to redraw the arrangement? Draw vertically a stack of 4 metal plates with dielectric in between but without any connections between the plates. Now redraw the diagram but change the middle metal plates into two plates connected by a wire like a capital $I$ so it now looks like that you have 3 capacitors in series. Now make the ...


0

What you're saying is mostly correct, but your language is a bit imprecise, which makes me think your understanding is a bit incorrect. To be clear: there are both static electric fields, as well as time varying electric fields, which occur together with magnetic fields electromagnetic (EM) waves. The equations you're using describe reflection and ...


1

Trackpy is a Python package for particle tracking in 2D, 3D, and higher dimensions. http://soft-matter.github.io/trackpy/stable/ https://github.com/soft-matter/trackpy The Matlab Particle Tracking Code Repository Daniel Blair and Eric Dufresne http://site.physics.georgetown.edu/matlab/ Particle tracking using IDL John C. Crocker and Eric R. Weeks ...


0

http://physics.stackexchange.com/a/65392/101895 Brilliantly explained there, relate the contracting effect with classical sound Doppler effect if you could. Again, only force acting on a particle due to another particle is solely coulomb's force.The most important thing is that concept of exchange of particles is supposed to create the force {basically, a ...


0

$\displaystyle \nabla \times \frac{dE}{dt}$ is equivalent to $\displaystyle \frac{d}{dt} \left( \nabla \times E \right)$ This can be shown by breaking down the curl. For example, $\displaystyle \frac{d}{dx} \frac{d}{dt} E = \frac{d}{dt} \frac{d}{dx} E$ since $x$ is independent of $t$ (Newtonian speeds) Take the curl of one of the existing equations (same ...


1

I think I have a proof for you (though you may find it unsatisfying). For the most part, I'm following the proof on the wikipedia page you link to. I do avoid the dirac delta function however. Starting from the Biot-Savart law: $$\mathbf{B}(\mathbf{r})=\frac{\mu_0}{4\pi}\int_V ...


1

$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\div}[1]{\nabla \cdot #1} \newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}} \newcommand{\curl}[1]{\nabla \times \vec{ #1}}$I assume you know how to solve the Maxwell's equations in vacum. In the end you get an expression of the ...


0

I am not a physicist but i am a curious person like you. So i seached for the same answer for so long, finally i got my answer from this Richard Feynman video in which he describes it perfectly for an uneducated mind like mine. http://youtu.be/3D2RaDVkylY


2

Comments from @aquirdturtle have led me to rewrite my answer and to realise that it was a question worth asking. @ACuriousMind has likened the situation to a mass falling on the Earth. In that case the mass and Earth system loses gravitational potential energy and they both gain kinetic energy although almost all of it resides with the mass. Carrying on ...


0

Electromagnetic induction indeed has three possibilities of interaction between two of the three constituents (when non-parallel to each over) to get the third constituent: - a moving charge in a magnetic field induce a sideway movement (deflection) of the charge (Lorentz force) - an electric charge, accelerated in a coil (deflected sideways) induce a ...


1

The basic principle: 1 volt is induced in a loop when the magnetic flux through the loop changes at a rate of 1 weber per second, i.e. 1 tesla times square meter per second. Provided you have a 1x1 cm loop, in a 0.5 T field of a pair of very strong neodymium magnets, and the loop has 100 turns, the overall magnetic flux through all the coil turns would be ...


2

Below I'll use Planck units, for which, in particular, $c = \epsilon_{0} = =1$. In fact, the full system of Maxwell's equations provides the statement that the only two vector components of the EM field $\mathbf E, \mathbf B$ are independent (in general, due to a deep symmetry reason, namely that a massless particle has only two polarizations). Next, if we ...


1

The reason to use the alloy is because it has a much higher resistivity than copper and so the alloy wire will have a higher resistance which with standard laboratory apparatus can be measured more accurately. I also seem to remember that the temperature of coefficient is lower for some of these alloys ie for a given increase in temperature the resistance ...


0

The choice of a non-metallic sample is to get the resistance high enough to measure. The resistance of the rheostat, the voltage supply, and the hookup wires can make the measurement inaccurate if you do not account for them. If they are much smaller than the resistance of the sample they will not matter much. You also will not try to draw too much current ...


0

I have seen Anedars answer and it looks beautiful. Anyhow I have an objection to your question, which is too long for a comment. When an electron moves in uniform external magnetic field ... the magnitude of force is ... perpendicular to both components of velocity. But this force, though only due to the perpendicular component, acts on the particle. ...


0

Magnetic fields is created by the low atomic magnets inside the piece of metal, piece of iron aligning together in a line. The actual magnetism in a piece of iron or in a permanent magnet is actually caused essentially by electrons orbiting in one direction more than the other, and the electrons are going to keep on orbiting forever until something ...


0

You can put the vectors in at the end because this an example of motions in different directions being independent of one another. There is only one force acting on the charged particle and that if due to the component of velocity of the particle which is at right angles to the B-field. Since that force is perpendicular to the direction of motion and of ...


2

Let's assume the magnetic field vectors point in z-direction (or: let's call the direction the magnetic field vector points "z"). Then we have for the magnetic field: $$\vec{B} = \begin{pmatrix}0\\0\\B\end{pmatrix}$$ and for the speed of the electron: $$\vec{v} = \begin{pmatrix}v_x\\v_y\\v_z\end{pmatrix}$$ The Lorentz-force $\vec{F}$ due to a magnetic ...


0

The connection between electric potential and E-field strength can be conceptually difficult usually because some important ideas are forgotten. The first thing is the definition of potential. Potential is defined as the work done by an external force in taking unit positive charge from a arbitrarily chosen zero of potential to the point. For convenience I ...


2

In your last question it is important as to what you mean by WRT the question. If you are trying to find the E-field due to a point charge using Gauss then to make the surface integration easier you choose a surface which has the following properties: the E-field direction is everywhere perpendicular to the surface the E-field has a constant ...


0

Does the electric field at a point on the surface change? Yes, it change. Where distance is decreased Electric field increase. Does the total flux through the Gaussian surface change? No, it remains same. Consider a spherical gaussian surface of radius $2m$ and a charge $q$ inside it. Let this charge emit $10$ field lines. Then, clearly, all the field ...


0

-ve Sign which comes during integration is due to opposite directional motion of test charge. You can't consider it twice, once during integration and next during antiparallel motion. Hint: Why do you think - ve sign comes during integration? Since you know Electric field was positive, How integrating it could be -ve.


0

B should remain since Magnetic field is added in upper part and equally subtracted in lower part. Hint: Wire does not experience its own magnetic force. If you let wire to accelerate in $B$ freely , it won't give any magnetic field.. So, no current flow in wire. Hence EMF is induced in opposite direction.


2

That is not the energy density of the electromagnetic field. That is the energy flow density vector of the field, also known as the Poynting vector. Energy flows in some direction, so its density must be a vector. You're totally right, energy density is not a vector, and it is given in gaussian units as $$ \mathcal{E}=\frac{1}{8\pi}\ (E^{2}+B^{2}) $$ As ...


2

Electromagnetic (EM) radiation between 400nm and 700nm in wavelength is the same thing as light. There is no difference. Neither is there a distinct difference between light, ultraviolet, x-rays, gamma rays, infrared, microwaves, or radio waves. Those names are all just human convention to specify EM radiation in certain frequency ranges. And regardless ...


0

Symmetries as the definition of particle charges Modern realistic particle physics theories are constructed from the requirement that there is such symmetry group which defines the quantities which are conserved in all processes which are described by theory (free propagation, interactions). This symmetry group is given as the direct product of subgroups of ...


2

The helicopter and the power lines are at different potentials, the difference being so great as to cause the air in between to become a conductor. If you applied such a potential difference across a line worker it would probably result in death. You will note that the line worker is holding a metal stake which has a "pointed" end. This increases the ...


0

If you put a charge inside a conducting sphere, you create an electric field inside the sphere. Charges can easily move inside a conductor. The field creates forces, so they will move. If the charge is negative, electrons will flow away. They will leave positive nuclei behind. This will cancel the original charge. Likewise, if the charge is positive, ...


0

The point about the E-field being zero inside is a conductor is that you must be writing about electrostatics, the study of charges when they are not moving. In the ideal world every bit of metal inside your surface (an infinitely thin shell) has no E-field within it. In the real world when the surface is particular in nature it must be very difficult to ...


0

The answer is "it depends what you mean by exactly on the surface". The electric field depends on the amount of charge enclosed. From Poisson's equation: $$\nabla\cdot E = \frac{\rho}{\epsilon_0}$$ If the charge on the surface is an infinitely thin sheet of charge, then the electric field will be zero on one side of the sheet, and a finite value on the ...


1

There is actually a very simple answer to this question: The absolute number of turns which each of the transformer winding should have (with fixed turns ratio) depends entirely upon which voltage and current ratings the transformer is intended to be used for. Let me explain. For instance, measurement transformers for high voltages typically have a ...


0

But if you adopt the different sign convention you have just got the same result! The E-field is a vector; by making the incident and reflected waves have opposite signs on the LHS in your revised formula you have started off by assuming that the reflected E-field is in the opposite direction to the incident E-field. Then when you find that the reflection ...


2

This sticker is indicating caution when you heat a liquid. Microwave can lead to superheater water which can be dangerous. If your water is superheated, it may not explode/boil over until an object is placed inside the water (e.g. a spoon). One way around this is to place a chop stick or object into the water while it is heating. This will prevent that ...


0

Ok, I think I figured it out. We are so used to assuming that a shell can be collapsed to a point charge, that it is easy to miss the subtlety that this is only true for the regular Coulomb law. As soon as we deform the Coulomb law as given above, we get corrections. If you do all the integrals over the charge distribution you will get the following fields ...


7

Suppose you start with a linear solenoid. Due to the Lorentz force charge particles travel in circles (or helices) inside the solenoid so they can't reach the walls of the solenoid. But obviously the trouble is that they will leak out of the ends. Now we curve the solenoid round and join its ends together to make a torus so now the particles can't leak out ...


3

The words ultimately refer to the same "objects" but they describe different aspects of them. "Permanent magnets" are objects and they're defined by the external property (what they look like from outside) that the magnetic field remains nonzero around these objects without any activity. On the other hand, "ferromagnets" are materials and the focus is on ...


1

Do any physical real entities exist? All the particles we consider elementary (the most well-known being the electron) do not occupy any finite amount of space, i.e. are pointlike, at least according to our present understanding of the standard model. However, that understanding could change as new theories emerge or as experimental evidence is found. As ...


1

I'm going to say no, because I interpret a "physical entity" as something which we can observe, and therefore confirm it's existence. For instance: Particles (0 dimensional): Mathematically are points, but when we observe them we observe them to have sizes because of the observation process (bouncing photons off atoms, electrons off other electrons, even at ...


0

In a perfect transformer, if we run a current through either the primary or secondary coils, we are guaranteed in the quasistatic case that $\Phi_1=\Phi_2$ for each individual turn of the coils. Now suppose we ran a current $I_1$ through the primary coil, and waited long enough so that $I_1$ was steady and $I_2$ (current induced on second coil) was zero. we ...


0

Matter is full of positively charged protons, and those positively charged protons attract the negatively charged electrons. In general, if you take a chunk of neutral matter and add one extra electron then the attractive force due to the protons in the matter outweighs the repulsive force due to the electrons in the matter, and the extra electron is bound ...


0

EMF $\mathcal{E}$ is defined as the work done in taking unit positive charge around a complete circuit. So to take a charge of $dq$ around a complete circuit is $\mathcal{E} dq$. $dq = Idt$ So the work done is $\mathcal{E}I dt$


0

Generators often have two sets of windings - one arranged on the stator (fixed) and the other on the rotor (spinning). One is chosen/designed to be the excitor (could be either the rotor or the stator); the excitor is fed a small amount of electrical power to maintain a magnetic field. The other winding generates the output as it moves through the field ...


0

To answer your question it takes a little bit longer as usual. Let us start with permanent magnets. Where the magnetic field of permanent magnets come from? The process of its production is the next: Some material is milled to powder, then it will be pressed into its form under the influence of a strong magnetic field and sintered. Why this we do? The best ...


3

First of all, don't think of multipole moments as separate things that have their own individual meaning. Instead, think of them as parts of one thing. Once we have all the parts written down, we can start naming and organizing each one to determine its contribution to the whole. Now, for your question Is there a physical interpretation for multipole ...


3

A permanent magnet has a fixed north/south polarity - in this example, lets say north is facing up and south is facing down. This magnet has a membrane of some kind attached to its north face. An electromagnet beneath the permanent magnet can switch the direction of its north/south polarities by changing the direction of the electric current running ...



Top 50 recent answers are included