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13

Your intuition about the meaning of the divergence operator is wrong. In physics it's easiest to think intuitively about divergence by using the divergence theorem which states $$\int_V dV \ \nabla \cdot \mathbf{B} = \int_{\partial V} \mathbf{B} \cdot d\mathbf{S}$$ where $\partial V$ is the surface area surrounding the volume $V$. The magnetic field has ...


6

Divergence means the field is either converging to a point/source or diverging from it. Divergence of magnetic field is zero everywhere because if it is not it would mean that a monopole is there since field can converge to or diverge from monopole. But magnetic monopole doesn't exist in space. So its divergence is zero everywhere. Mathematically, we get ...


4

There are 2 ways of looking at the speed of an electric current. The electrons themselves move quite slowly: of the order of 1.0 m/hour. In other words, if you have a 1m long cable, an electron will take a whole hour to get from one end to the other. For your example, in a copper wire of radius 1 mm carrying a steady current of 10 Amps, the drift velocity ...


3

No, that is not enough to say that $B=0$. You must also consider that $$\nabla\times E=-\frac{\partial B}{\partial t}$$ which means that for a magnetic field that is constant spatially but not in time, your conditions would be true but your $B$ field would not be $0$ If, however, we had a case where $\nabla\times E=0$ as well, then (aside from being a very ...


3

Well, a large number of countries, after the break even ( actually 60% of output over input energy) in energy of the prototype tokamak in JET joined into creating ITER, a prototype Tokamak design designed to have output energy in megawats. If interested you should go to the FAQ of the link given for ITER . There exist alternate projects: Of the ...


2

To get a lot of energy from a fusion reactor you need lots of D-T fusion events per second, and this means a combination of reasonably high density and very high temperature. This is extraordinarily difficult to achieve. In particular as you try and increase the plasma density it gets increasingly difficult to maintain the plasma in a stable state. There ...


2

Probably not, but it depends on the geometry of your coil. For a couple of dollars at the hardware store you can get a big stack of those coin magnets. If the answer to your question were yes in general, it'd be harder to break apart the big stack of magnets that to separate two of them. That's not consistent with my experience. In general for a dipole ...


2

Faraday's law $\mathcal{E}=-d\Phi/dt$ can be used in a variety of situations, including ones where the phrase "motional EMF" is appropriate. Your particular expression $-vBL$ is applicable only for a very particular situation. Probably a sliding bar, which is part of a circuit, in a uniform magnetic field. That expression can be derived using Faraday's law, ...


2

See this explanation from here: Right at the heart of the Earth is a solid inner core, two thirds of the size of the Moon and composed primarily of iron. At a hellish 5,700°C, this iron is as hot as the Sun’s surface, but the crushing pressure caused by gravity prevents it from becoming liquid. Surrounding this is the outer core, a 2,000 km thick layer of ...


2

Can we conclude that B=0? For a general field it is wrong because every constant vector will satisfy those conditions. But for the magnetic field is it enough? It depends on what facts about magnetic field you want to admit into your hypothetical situation. If you assume the Maxwell equations with vanishing sources and the condition $\nabla \times ...


2

John Rennie's answer is correct for a DC series connected motor and, almost certainly, this is the kind of motor you (the OP) are talking about. An interesting way of writing John's answer "backwards" is that you have just observed the reason why the most powerful traction motors are exactly this kind of motor - almost all DC train and tram motors are ...


2

When a motor is turning it acts as a generator and produces a back EMF that opposes the applied EMF. See my answer to Top angular speed of electric motor for more on this. A frictionless motor would draw no current when not under load, though obviously real motors do draw some current because of frictional losses. If you load the motor you reduce the back ...


1

The Verdet constant is a coefficient which sums up the magneto-optical properties of the medium. So, the temperature and wavelength dependence are wrapped up in it. Fundamentals of Photonics by B.E.A. Saleh expresses the Verdet constant in terms of the wavelength as $$ V\simeq-\frac{\pi\gamma}{\lambda n} $$ where $\lambda$ is the wavelength of the light ...


1

The term "Rosenberg-Coleman effect" originates from the article Heliographic latitude dependence of the dominant polarity of the interplanetary magnetic field. It is also referred to as the "dominant polarity effect". As the Earth orbits the Sun, the Earth travels above and below the equator of the Sun. According to Rosenberg and Coleman, the polarity ...


1

from a, b and c: An electromagnetic field is propagating by changing the field that's generated by electrically charged particles pass through the air and the space that is devoid of particles of space. An electromagnetic wave propagates, not an electromagnetic field. An EM wave is a propagating disturbance in the existing electromagnetic ...


1

Your intuition is correct. The ring develops a current, and thus a magnetic field only when the external magnetic field changes. A steady DC current would behave like a regular magnet: the ring would not move, and in fact resist motion (since that would entail changing magnetic field). However, it's takes a lot of current, which causes two problems: Power ...


1

As Draksis said, the condition is that the integral over any volume has to be zero. If you want a formal proof, here you go: Let's call $f(\mathbf{x}) = \nabla \cdot \mathbf{B}$, and assume it is continuous. Suppose there's some $\mathbf{x}_0 \in \mathbb{R}^3$ with $f(\mathbf{x}_0) \neq 0$, and let's say that $f(\mathbf{x}_0) > 0$ (the proof for $f < ...


1

Intuitively, if the volume integral of a function is 0 over any arbitrary volume, the function itself must be 0 at all points in space. More concretely, consider a function for which $\int_V \, f \, \mathrm{d}x = 0$ for any volume $V$. Then, $\int_{V+dV} \, f \, \mathrm{d}x = 0$ for any infinitesimal addition to V. $$\int_{V+dV} \, f \, \mathrm{d}x - ...


1

In lamination, sheets of metal are separated by insulators. Charge can't flow between plates (they're insulated from each other), so there can't be large eddy currents perpendicular to the plate. Obviously large eddy currents can still flow on the plane of the metal sheets. If you properly align the source of the emf and the plate, you can rid of most of ...


1

A magnetic field exerts a force on a moving charge. Given a magnetic field, $\vec{B}$, and a charge, $q$, moving with velocity, $\vec{v}$, the magnetic force, $\vec{F}$, on the charge is:$$\vec{F}= q(\vec{v} \times \vec{B})$$ The directions of these with respect to one another can be found using the right hand rule. See the picture below. The magnetic ...


1

Electromagnetic fields propagate at the speed of light, however in a real circuit the rate at which the electromagntic field grows is controlled by the inductance of the solenoid. You probably know that a changing magnetic field through a conductor creates a current - this is after all the way all our electricity is generated. When you apply a voltage to a ...


1

Water and air are both weakly magnetic, with magnetic permeabilities on the order of $10^{-6}$ H/m whereas iron, a strongly magnetic material, has a permeability of about 0.25 H/m. Thus, if you could create the supercavitation in water, a magnetic field would do just about nothing to maintain the cavity.


1

Water is actually not a non-magnetic material that many people seem to think it is, it is actually diamagnetic. This means that it will reduce the external magnetic field by aligning anti-parallel to the field. However, water has a susceptibility on the order of $\chi_v\simeq-10^{-5}$. This will change the strength of the magnetic attraction by about ...



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