Hot answers tagged electromagnetism
9
This is a good question with a lot of deep math and physics behind it (information theory). I will try to give you a casual answer.
Signal to noise ratio:
First, you should ask yourself what a "signal" is. For example, when you listen to the radio, especially AM radio, you hear the sounds / music / voices just fine even though there is static / noise in ...
8
If you are not well-acquainted with special relativity, there is no way to truly explain this phenomenon. The best one could do is give you rules steeped in esoteric ideas like "electromagnetic field" and "Lorentz invariance." Of course, this is not what you're after, and rightly so, since physics should never be about accepting rules handed down from on ...
5
Not really. A magnetic field alone doesn't create electricity. A changing magnetic field does. The Earth's magnetic field does change a tiny bit but not enough to really generate much.
The other option is to move the inductor in the magnetic field. The Earth's magnetic field is quite homogeneous over short distances though so the coil would need to move ...
4
Moving charge always produces a magnetic field. If you have a non-zero current then you have non-zero moving charge and a magnetic field will be produced.
You can achieve essentially no magnetic field though by using two wires right next to each other each carrying current in the opposite directions. As long as the wires are very close and the amount of ...
3
J. D. Jackson in the introductory remarks of his chapter on 'Radiation Damping, Classical Models of Charged Particles' (3rd edition), says that the problem of radiation reaction on motion of charged particles is not yet solved. He says that we know how to find motion of charged particles in given configuration of EM fields and also how to calculate EM fields ...
2
In the passage you quoted, McDonald is describing the very edge of the thin conducting disk. Recall that although the disk is thin, it does have a finite thickness. Therefore, its edge looks like this, with the conductor drawn in grey:
One way to describe this geometry is to say the internal angle is $\pi/2$. McDonald chooses to instead say that the ...
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In ferromagnetic materials there is an unpaired electron in the outermost orbital, giving an overall magnetic moment equal to one electron spin to the atom. In a ferromagnetic bulk crystal, these orbitals can overlap between neighbouring atoms which causes the spontaneous magnetisation through the exchange interaction. This interaction is incredibly short ...
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The question you have formulated is not an easy one to answer (correctly). But the question you've formulated isn't quite the question that I see. The good news is that the text of the question you've posted implies a much simpler question; it's just asking for the energy change.
You can probably assume that the acceleration is dominated by the circular ...
2
Electric and magnetic fields are what the electromagnetic field 'looks like' from a particular (inertial) frame of reference.
Take a charged particle: In its rest frame, it appears to generate an electric field only and no magnetic field at all. From a different frame of reference (in particular one in relative motion), we'll see the charge moving, thus a ...
2
For this particular circuit, the voltage across the R1/C1 branch #1 is fixed by V1, and that across R2 (branch #2) is also fixed, again by V1. That is, the fixed V1 decouples the two branches, so they can be solved separately (circuit #1 = voltage source V1 across branch #1, and circuit #2 = V1 across branch #2). Once these circuits are solved, the current ...
2
A diagram may help:
Here, the charged particle was initially stationary, uniformly accelerated for a short period of time, and then stopped accelerating.
The electric field outside the imaginary outer ring is still in the configuration of the stationary charge.
The electric field inside the imaginary inner ring is in the configuration of the uniformly ...
2
The force law you show gives us the total force on the wire. This force comes from the sum total of forces on all the electrons moving through the wire. So imagine that your wire is supported at either end, and the magnetic field is strictly between the two supports, so the total force on the wire is – in some sense – between the two supports. Then, if ...
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The second problem is quite tough. J. D. Jackson comments, in the introductory remarks of his chapter on 'Radiation Damping, Classical Models of Charged Particles', that we know how to solve classical electrodynamics problems in two ideal conditions - a) given charge and current densities, how to compute the fields and b) given the fields, how to find the ...
1
I'm guessing you're thinking of using disk magnets like these ones. The force between this type of magnet is somewhat complicated to calculate, but at distances larger than the depth of the disk the force falls away as distance squared i.e.:
$$ F \propto \frac{1}{d^2} $$
This means the force falls very rapidly with distance. In the bed that Dan mentioned ...
1
Your calculations are correct, provided the cylinder is indeed ohmic. The constant $E$ you're getting is the difference in electric field between both terminals.
As for the current flowing from inside to outside, as you said the cross sectional area will be different, and so will the length. The length $L=r_b-r_a$, but the cross sectional area is not ...
1
http://www.antenna-theory.com/antennas/shortdipole.php is a website with useful info., including formulas.
To oversimplify, it seems to say that once the antenna is a tenth or less of the wavelength, the exact ratios don't matter so much. The antenna is inefficient, but it works for both sending and receiving. If you can detect the signal, of course you can ...
1
Many ham antennas include coils that help the antenna appear to be the right length for the frequency in use, there are also trapped antennas there the coil will block frequencies above a specific point and the frequency drops the coil will allow the energy to pass to the antenna element on the other side so at high freq you have a shorter antenna and as you ...
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The main reason is that in Ham radio you care about transmitting, in that case you need to make sure the antenna is in the right length so you get a standing wave inside the antenna. You can read about standing wave ratio here.
If you are just receiving then you could use any wire, loop antennas are practical for low frequency transmissions in AM, where you ...
1
Firstly, long wavelengths are used in the carrier waves. These are affected less by everyday matter, and are good at spreading out. They can reflect, but they aren't distorted or diffracted much. In contrast, light waves are absorbed everywhere, and X rays and higher are very directional.
That's all the physics involved. It's good enough for a radio to ...
1
What causes these constants to have the values they do is simply our choice of a system of units.
When you have a unitless constant, it makes sense to ask why it has the value it does. For example, two of the lines in the visible spectrum of hydrogen have wavelengths in the exact integer proportion of 28/25. When this was first discovered, it made sense to ...
1
In the static case you can solve Maxwell equations using a vector potential via the poisson equatuion for the magnetic potential.
$\Delta \vec A(\vec r)=-\mu_0 \vec J(\vec r)$
Using the Greens function for the Laplace operator yields the solution of this differential equation.
$\vec A(\vec r)=\frac{\mu_0}{4 \pi}\int d^3r' \frac{\vec J(\vec r')}{|{\vec ...
1
There are two phenomena in your question.
(1) Let us first understand how magnetic field can be considered to "arise" because of relativity. Imagine a frame of reference in which a charge $Q$ is at rest. If another charge $q$ is brought in its vicinity, it will experience only an electrostatic force. Now get on to another inertial frame of reference moving ...
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There are three different measures of an object's mass: its inertial mass $m_i$ (defined by Newton's second law), its passive gravitational mass $m_p$ (defined by how much force it feels in a gravitational field), and its active gravitational mass $m_a$ (defined by the strength of the gravitational fields it makes). You get qualitatively different ...
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