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54

Most electromagnetic radiation is of very high frequency - the magnetic field changes many times per second. This means that the compass just doesn't have time to "follow" the magnetic field changes. The only thing that does affect a compass is a DC magnetic field - usually this is a large piece of iron etc. that gets magnetized (e.g. by the earth's ...


26

The frequency is one very good argument (and I guess the most important factor) but it might be worth also looking at the magnitudes of the fields. The earth's magnetic field has a strength of roughly $31µT$. The intensity of the sunlight hitting the earth is about $1300W/m^2$. Since the intensity is related to the electric field $E$ of an electromagnetic ...


19

Basically the same reason as what Floris said, but this also has another important aspect: Visible light has a far too small wavelength to affect a compass. Not only does the field oscillate too quickly around an average of zero – even at any single “snapshot” in time of the electromagnetic wave, there would nowhere be a large region where ...


9

One has to make clear that the watches we are using now are no longer using radium , because of radiation danger awareness. Radium dials are watch, clock and other instrument dials painted with radioluminescent paint containing radium-226. The 1900s (decade) were the peak of radium dial production, as radiation poisoning was then unknown; subsequently, ...


8

In microwave ovens what matters is how much energy the radiation carries and how that energy is absorbed by the food. Visible light and IR are rapidly absorbed by most foods, so they would only heat the outer layer of the food. You'd get food with the outside carbonised and the inside raw. Microwaves are far less strongly absorbed by foods, so they ...


7

It is not clear from the question, but let's assume we are in free space/vacuum. Then the Maxwell equations read: $$ \nabla \cdot \mathbf E = 0\\ \nabla \cdot \mathbf B = 0\\ \nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}\\ \nabla \times \mathbf B = \mu_0 \varepsilon_0 \frac{\partial \mathbf E}{\partial t} $$ A plane wave can be written as ...


5

If we assume you are a sphere in space, at the same distance from the sun as Earth, then we can calculate the heat absorbed - and we can calculate how hot you need to be so heat in = heat out (assuming uniform surface temperature, and radiative heat transfer only). For this, we need the Stefan-Boltzmann expression for total emission at a given temperature: ...


5

For light bulbs and other thermal emitters this is definitely true. Their emission follows the black body spectrum (if you neglect absorption due to the glass container). If you want to be picky: Any device, which is operated above 0 K (which applies to all devices) emit thermal radiation according to their temperature. This is not directly related with the ...


4

While for vectors $\vec{B}$ and $\vec{C}$, the cross product $\vec{B}\times\vec{C}$ is indeed perpendicular to both of the vectors, it is simply not the case that the curl of a vector field is orthogonal to the vector field. Do not read too much into the cross product notation. In particular, you can add any constant vector field to $\vec{A}$ without ...


3

In one sense the problem is simple, in that any solution with complex numbers (say with phasors) can be literally translated into a real version (just equate the real and imaginary parts of each complex equation as two real equations). However many phasor type setups are designed to only look for solutions of a particular type, so it might be considered ...


3

In electrodynamics, one uses complex fields only as a calculation trick, since for instance terms like $$ \exp(i \omega t) $$ are usually more easy to handle as $$ \cos(\omega t) $$ . The same trick can be applied for example to the problem of the driven harmonic oszillator: $$ \ddot x + 2\gamma \dot x + \omega_0^2 x = A \cos(\Omega t) $$. Adding $ i \left( ...


3

Project Excalibur The idea of a nuclear pumped X-ray laser was one which was investigated in detail in the Reagan "Star Wars" program of the 1980s, backed by one Edward Teller. Tests were carried out by surrounding the nuke with bundles of rods to create a one-pass laser. Apparently it was nowhere near efficient enough to be used in a military context. ...


3

The measurement of the velocity vector of a nearby, large galaxy is not a simple matter. If you just dropped a spectrograph slit down randomly somewhere in Andromeda then you could get a wide variety of answers, since it rotates with velocities of $\sim +/- 200$ km/s in different parts of its disk. To estimate the centre of mass redshift one must use ...


2

When light hits a barrier, even transparent ones, some light is reflected and some is refracted. This is often described by the transmission coefficient for that material, and at that wavelength. This can happen at the macroscopic barriers and at the smaller barriers between crystals or grains within a material. It is a simple property of waves which does ...


2

Q1: For photons of energies much less gamma rays, the quantum mechanical photon-photon interaction is negligible. This is consistent with the classical electrodynamic description where the principle of superposition holds (electromagnetic waves pass through each other unchanged, as well as through electric/magnetic fields). Q2: in reality, charge is defined ...


2

Theoretically yes, the laser principle does not consume any material. There is a light source that excites the electrons in the material to higher levels, they deexcite to some intermediate one, here the avalanche of photons appears producing the laser light and leaving the electrons in the ground state. And you can repeat the process without a loss.


2

The radiation produced by a mobile phone does have the same heating effect (dielectric heating) as a microwave oven. However, as you suspected, the difference is power. Power is a measure of energy per unit time which, for a given frequency, equates to the number of photons per unit time (not 'power per photon', as your title says). As listed here, ...


2

Your argument that the energy should radiate away would be true if your inductor were a good antenna, in which case it would be a bad inductor! The problem is an impedance mismatch: The inductor produces a magnetic field (which stores the energy you inquire about), but little electric field. That is the wrong ratio, or impedance, to couple to the vacuum ...


2

(This type of question has been asked by 4 users but in those questions they either gave an example of a wooden box or a room and they got answers that the light is absorbed by the wood or the walls of the room. But in my question its the case of mirrors.) In this case, the light would be absorbed by de "viewer". You would need some type of device ...


2

In light propagation, oscillation does not mean any movement in space. It is the value of the electromagnetic field, at one given point in space, that oscillates. The picture that you quote does not represent the movement in space, but the electromagnetic field value as a function of time. Compare to waves in water: if you put a little boat on the water, ...


1

We say the electromagnetic wave is oscillating because something waves as the wave passes by. Light does propagate as per the above image, but it isn't the full story. For a bit more, have a look at the Wikipedia electromagnetic radiation article and note this: "Also, E and B far-fields in free space, which as wave solutions depend primarily on these two ...


1

light is supposed to possess relativistic moving mass even though it does not possess any rest mass. m2c2=M2c2-M2v2 where m is rest mass and M is relativistic mass and v=c. this gives m = 0 , but M is not zero the value of M can be calculated from the experimental data on radiation pressure.


1

A photon has a rest mass of nought (where the rest mass $m$ is the Lorentz-invariant quantity in the four-momentum's Minkowski norm squared $E^2/c^2 - p^2 = m^2 c^2$). However, a lightfield of energy $E$ gravitates and itself has a gravitational source equivalent to a mass $E/c^2$. Also, a system of photons has a nonzero rest mass (see reference), as does ...


1

The problem with this question (although your question is still a natural one for those thinking about light to ask) is that it mixes the ideal and the real. You describe an ideal situation with your mirrors, but then ask for what would happen in real life. No actual mirror has reflection coefficient of 1 (which would represent 100% reflectivity) and so any ...


1

We know that light is massless so why does a black hole's gravity attract light? Because gravity doesn't just attract objects with mass. It alters the path of light too. Because gravity is caused by a concentration of energy which "conditions" the surrounding space, altering its metrical properties, whereupon we talk about spacetime curvature. But note that ...


1

I don't really see a difference between the microwave measurement and the definitions of the SI units. We have Unit of time: second: The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom. Unit of length: meter: The ...


1

Now, what I'm wondering is this- do we have a single model/theory whose equations accurately describe the behavior of light in all scenarios? Yes, for all intents and purposes: QED. Clearly, we can not say that QED works at all scenarios (e.g., very tiny length scales and very high energy scales), since we can not test it in all scenarios. But, as ...


1

The coil is an inductor, which stores energy in a magnetic field. The coil & capacitor together form a basic electronic oscillator. If you start out with a charged capacitor, it will start to discharge, forcing a current through the coil. This will set up a magnetic field in the coil, which takes up some of the energy that used to be stored in the ...


1

Let us work in the Coulomb gauge, i.e. $\phi=0$ and $\nabla\cdot\mathbf{A}=0$. Then the electric and magnetic fields are defined as $$\mathbf{E}=-\dot{\mathbf{A}},\quad \mathbf{B}=\nabla\times\mathbf{A}$$ Now one solves the wave equation for $\mathbf{A}$. The constraint $\nabla\cdot \mathbf{A}=0$ tells us that $\mathbf{A}$ is transverse to the wave vector. ...


1

The vector potential is transverse to the direction of motion. To see this, perform a gauge transformation into the Coulomb gauge, such that $\nabla\cdot\mathbf{A}=0$. Using a plane wave solution $\mathbf{A}_0\cos(\mathbf{k}\cdot\mathbf{r}-\omega t)$ we then find $\mathbf{k}\cdot\mathbf{A}_0=0$, i.e. the potential is perpendicular to the direction of motion. ...



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