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46

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 ...


25

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 $31┬ÁT$. 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 ...


22

Yes, there are an uncountable infinity of possible wavelengths of light. In general the frequency spectrum for Electromagnetic (e.g light, radio, etc) is continuous and thus between any two frequencies there are an uncountable infinity of possible frequencies (just as there are an uncountable number of numbers between 1 and 2). Two things to consider in ...


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 ...


16

Formally there are an infinite number of different wavelenghts. However, any given physical system can only be found in a finite number of distinct physical states. To create a light source with a wavelength $\lambda$ that is well defined up to some resolution $\delta\lambda$, requires observing it within a system of size of the order of ...


14

The wavelengths that stimulate vitamin D production are between 280nm and 320nm, which is called UVB. You would need to use a detector capable of measuring light in this wavelength. However there is no need, because normal windows are made from soda-lime glass and this transmits no wavelengths shorter than about 350nm. Some Googling will find you the ...


11

Sir Elderberry, Punk_Physicist and the Count Iblis have all given correct answers in principle. There are two phenomena (really thought experiment, rather than practical, devices) that one needs to heed. A finite measuring time $T$ can only resolve frequencies to within an uncertainty of the order of $1/T$. This is the reciprocal relationship between the ...


8

The reason has to do with time dilation, and specifically, with the resulting red shift. A black hole forms from a collapsing star, which is of course made of brightly glowing matter. The event horizon forms in the centre and moves outwards while the star-matter falls towards it. Because of gravitational time dilation, the infalling matter never crosses the ...


7

I believe that currently, light in free space (i.e., not in a waveguide or a crystal or anything tricky) is believed to be able to have all values of frequency/wavelength/energy. As you say, it is continuous. There are highly speculative theories that perhaps this is not true "all the way down" but thus far we have no evidence that the wavelength spectrum is ...


7

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 ...


6

This is a curious question. Many black holes are detected and identified due to light emitted from material falling into them. When black holes accrete matter, conservation of angular momentum would usually lead to the formation of an accretion disk. The release of gravitational potential energy as material falls means that this disk can be hot, and it is ...


4

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 ...


3

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 ...


3

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

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

We can represent a monochromatic electromagnetic wave by one of its fields, the $\vec E$ or $\vec B$ field (or $\vec H$ in the case of the diagram further down). For example, we can write $$\vec E (\vec r, t) = \vec E_0 e^{i(\vec k \cdot \vec r - \omega t)}$$ where the relevant part for this question is $\omega$, the angular velocity. When the frequency of a ...


2

For electrodynamics simulations I have found the following applets particularly useful. www.falstad.com Below is a screenshot of the TE electrodynamics simulation showing a still frame of the electromagnetic fields of an oscillating electric dipole. The arrows represent the strength and direction of the electric field. The magnetic field is represented by ...


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

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( ...


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

In general, $$\mathbf E = -(\nabla \phi + \frac{\partial \mathbf A}{\partial t})$$ For the source free case, and in the transverse gauge with appropriate boundary conditions on $\phi$, $$\mathbf E = -\frac{\partial \mathbf A}{\partial t}$$ Which can be checked by taking the curl of both sides $$\nabla \times \mathbf E = -\nabla \times \frac{\partial ...


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. ...


1

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 ...



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