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1

what will happen to the speed when the load on the motor is fixed and the torque increases? You need to define what you mean by "load" very precisely here to make sense of this question. If by load you mean a constant torque $\tau_L$ opposing the motor's rotation, then if the motor's torque $\tau_M$ output increases we have a nett unbalanced torque on ...


0

If we ignore resistive losses, the electrical power input to the motor must equal the mechanical power generated by the motor. You state that the torque exerted by the motor increases, and the only way to increase the torque is to increase the voltage, which will increase the current and therefore increase the electrical power. This increase in electrical ...


3

Classical you work the intensity of a wave out by integrating the energy arriving over a known area in the course of one cycle, and then divide by the period and the area. The form you exhibit is correct for harmonic waves in SI units if the $E$ is the maximum amplitude of the electric field. But for a general normally-incident plane-wave you are looking ...


0

The force of the changing magnetic field on the electrons in the conductor can be deduced by imagining field lines traversing the conductor (from outside the ring to inside). This results in a force that is not necessarily parallel to the wire - but if the wire is very thin, then a tiny motion of the electrons will cause a polarization of the wire, after ...


0

I think I can answer in the case of the circular and square ring. Firstly, by cylindrical symmetry, we know that the E field is the same all the way around the circle. Why there is no radial component: Imagine rotating the system by 180 degrees about any diameter. At any location, the direction of the magnetic field would reverse, the direction of the ...


0

A high altitude is necessary to get an effective range. An EMP activated at low altitude will only effect electronics close to the source, and if that source is a nuclear bomb there won't really be many electronics left to effect anyway. The point of an EMP is to destroy or damage electronics while leaving the target physically intact; the high altitude of ...


0

Each individual atom is presumably neutral as well. So, we can simplify the model of a neutral object to a neutral atom and then go from there. Each atom has a positive core and a negative cloud of electrons around it as I'm sure you are aware. The electrons on average have a uniform field emitted in all directions and so does the nucleus. Because the ...


0

If you've studied electric and magnetic fields before then you know they are related to forces on charged particles. However, if you look at how the electromagnetic field $F$ interacts with charges to produce forces you see the metric get involved. So the metric is essential to the field as a thing that exert forces, which is the actual definition. There is ...


0

Each of the components of $A^i$ depends on $(x,y,z,t)$ thus you must use the chain rule and evaluate $(dA^i/dt) + (dA^i/dj * dj/dt)$ where the j's are summed over x, y, and z.


4

You usually cannot push your hand through the table, because it's a single solid. The atoms are held together by covalent bonds, which are electromagnetic in nature. Sand on the other hand is grainy - the $SiO_2$ grains do not interact with each other and are only held "in place" because of gravity. You can run your hand through sand similar to driving a ...


0

I do not know whether I understood exactly your question. In any case, I do not think that you need to use Helmholtz decomposition to establish that $E=-\nabla\phi-\partial_t A$ (using $c=1$). Particularly, the fields do not have to fulfill the hypotheses of Helmholtz decomposition in all space for this identity to be true. Maybe we could work in ...


1

What you are trying to do would "break the laws of physics" if it worked. You can't, so it doesn't. As the system rotates there will be losses. Air drag and friction and more. These losses absorb energy and cause things to slow down. For a system to rotate indefinitely it must get energy from somewhere. If there is no input energy it must make its own. A ...


0

See Russell McMahon's answer. Such a device may not work at all, and if it does work it will do so for only a limited time. However, if you want to give it another chance, the magnets you attached to your fan blades may be too thick. The magnet you hold to repel the blades may work for one magnet, but when the next magnet comes too close, the magnetic ...


0

Losses in ferromagnetic materials There are two mechanisms which produce losses in ferromagnetic materials. One of them is hysteresis, the other is eddy currents. Hysteresis Hysteresis losses occur while the magnetic dipoles rotate, meaning while you change the direction of magnetization. You can look at it through single dipoles or through Weiss domains. ...


14

Maxwell's equation can be given in the form $$\text dF = 0$$ $$\text d\star F + J = 0$$ where $F$ is a 2-form and $J$ an $n-1$-form (a current density) which in principle can be generalised to any manifold (for physical reasons one might want to consider pseudo-Riemannian manifolds with signature $(+,-,\cdots,-)$). In the four dimensional theory one usually ...


7

You can generalize Maxwell's equations to an arbitrary number of dimensions by using either the tensor or differential form version, as the vector formalism does not help too much (For instance, in two dimensions, the magnetic field is a (pseudo) scalar field, not a vector field). The equations are then : $\partial_\alpha F^{\alpha\beta} = \mu_0 J^\beta$ ...


2

It's a complicated subject - but very well studied in the context of eddy current brakes, where the retarding force is used to create a braking force without mechanical friction / wear. For me, the starting point for finding out more was this post - in particular the posting by Jim Hardy contained lots of good links. It seems that some of the most ...


1

You know that $$m\vec a=q\vec v \times \vec B,$$ So in particular, since $\vec B = B\hat z,$ we have $$ma_x=qv_yB,\text{ and } ma_y=-qv_xB. $$ And in our case $B=-\beta x$ so we have $$m\ddot x=-q\dot y\beta x\text{ and } m\ddot y=q\dot x\beta x. $$ You can take the time derivative of the left equation and get $$m\dddot x=-q\ddot y\beta x-q\dot y\beta ...


0

I do some work with magnetic fields in tissue as well for wireless power applications, though we don't typically deal with fields that strong hopefully I can help. First of all human tissue is largely magnetically transparent at low frequencies. While modeling the electromagnetic properties of tissue is very difficult problem (this is why you largely see ...


4

In the sea of electrons picture, the electrons in the conductor are not at rest: they are jiggling about like gas particles, colliding and changing direction constantly. You can think of them as billiard balls at zero gravity, confined in the volume of the piece of conductor at hand. According to thermal physics, their average kinetic energy is related to ...


1

About radiation force in a waveguide: Group velocity conveys energy, momentum and information in a waveguide. Phase velocity is superluminal in a waveguide. See for example "Phase, Group, and Signal Velocity" about the case of a waveguide: A waveguide imposes a "cutoff frequency" $\omega_0$ on any propagating electromagnetic waves based on the ...


6

They do attract one another. Indeed one of the biggest engineering problems in the building of very large scale electrical power generation hardware is the construction of generator rotors and stators to withstand the enormous stresses put on them by the magnetostatic attraction / repulsion between neighboring currents. The yoke of any magnetic system is in ...


2

A very small amount of the energy goes into vibrating the core. This will cause a heavily-loaded transformer to audibly hum. You may find this article http://www.electricaleasy.com/2014/04/transformer-losses-and-efficiency.html useful. Note that, for a well-designed, well-constructed transformer, efficiencies are already very high, and heat losses are not ...


1

Energy is not "lost" - it just changes how it is expressed / stored, and may become less useful (harder to convert to another, more useful form). That's entropy for you. For example, any time you have a source of heat, that source can be used to drive a heat engine and recover some energy. Since motors are usually designed to produce little heating, the ...


2

The "first-quantized setup" is the setup of quantum mechanics, where single particles are considered quantum objects, but fields like the electromagnetic field are still treated classically. However, the derivation of the action in the following is wholly classical (and the first-quantized setup arises when considering $x^\mu$ as quantum fields in the sense ...


3

You need to watch what you mean by the ambiguous term "derive", which can mean either "was derived historically" (i.e. was motivated by or is a derivative of, in the non-mathematical sense) or "is derived logically/mathematically". Historically, I think you are correct that $\boldsymbol{\nabla}\cdot ...


0

Well I dont know if we can prove it but there is a much more elegant way of formulating EM which may be helpful here. As you may know there are two potentials on EM: the scalar potential $\phi$ and the vector potential $\vec{A}$, from which $\vec{E}(t,x)$ and $\vec{B}(t,x)$ are derived. From this two objects and following symmetry considerations you can ...


3

Maxwell derived his equations from 1) charge conservation law; 2) Coulomb's law; 3) Bio--Savart--Laplace law; 4) Faraday's law of induction. The equation $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0}$ was indeed derived from Coulomb's law and in its differential form is written using Gauss--Ostrogradskiy theorem. ...


3

No we cannot prove it; Maxwell postulated that it would hold dynamically because it made the most sense for it to do so as he pondered the displacement current problem. As you likely know, Maxwell pondered the inconsistency between Ampère's law for magnetostatics and the charge continuity equation. Ampère's law for magnetostatics reads $\nabla\times ...


2

What you said is right. The incident field creates the electron-hole pair, then they oscillate normal to the surface. Since the motion is normal to the surface, it radiates in every direction except normal to the surface. (Dipole radiation pattern.) So if you want to avoid that effect, you want the outgoing light you measure to be normal to the surface. If ...


0

Having a magnetic field and moving a conductor non parallel to this field it will be induced a flow of electrons perpendicular to the plane from both the magnetic field vector and the direction of the displacement of the conductor. The induction of electric current takes place because electrons have a magnetic dipole moment and an intrinsic spin. For all ...


0

The electric fields and magnetic fields considered up to now have been produced by stationary charges and moving charges (currents), respectively. Imposing an electric field on a conductor gives rise to a current which in turn generates a magnetic field. One could then inquire whether or not an electric field could be produced by a magnetic field. In 1831, ...


0

In principle, you could use exactly the same steps as are in those notes to derive the full 3D expression. You would have to substitute $\vec{r} = x \hat{\imath} + y \hat{\jmath} + z \hat{k}$ in step 2, and then follow the logic through from there. However, an easier way to find the result for point out of the chosen plane is to exploit the rotational ...


1

To start with a particle loses kinetic energy, and therefore momentum, when radiating electromagnetic energy in some electric field, it is the basic reason why the planetary model of an atom cannot work. Brehmstrahlung, "braking radiation" or "deceleration radiation") is electromagnetic radiation produced by the deceleration of a charged particle when ...


1

I can give a back-of-the-envelope derivation of a drag force that ignores fringe effects and other complications. Say the conductor is a plate of thickness $\Delta z$ traveling with velocity $v$ in the x direction. Take the magnetic field to be constant in a rectangular area, with the the $\Delta y$ side perpendicular to the velocity much longer than $\Delta ...


1

I don't have the expertise you're looking for, but here's a crack at it: This is a really hard problem whose answer depends on the material properties. In particular, the answer depends on how big the 'eddies' are: if the current moves in roughly a big circle then the effect is large, but if there are lots of little eddies the effect is small. However, ...


0

Charges in motion relative to a reference frame produce a magnetic field in that reference frame. The direction of the field at a point is perpendicular to the velocity of the charge AND perpendicular to an imaginary line segment drawn between the charge and the point. You really have to think 3D, and it helps to get a partner to help hold some balls (for ...


0

Possibly a simpler answer that you may be looking for, its along a different axis... If you lay the loop on the table, the poles are Up and Down. Think about the right hand grab rule... if you were to wrap your right hand around the magnet - if your fingers are pointing the direction of the current flow, the thumb of your right hand points north


0

There is a difference between your two cases. When you are talking about a charge passing between magnets you are thinking of it as a uniform magnetic field. But it is not uniform, it gets stronger as you approach the magnets. if it were a uniform field the magnetic dipole you put in the middle would not feel a force along the line of the magnets, although ...


0

What you're asking for would be a finite-element model of the room, and there's a number of interesting papers that use it to solve the wave equation, usually for sound waves in a solid. You could definitely co-opt these to deal with a complex geometry, but it may take a very long to simulate. The problem is going to be dealing with the boundary conditions. ...


1

The derivation assumes the wire is a perfect conductor, and also that it is negligibly thin. If it had some resistivity, then you're right, there would be an electric field in the wire, but even in that case the electric flux $\int \vec{E}\cdot\text{d}\vec{a}$ would be negligible, and so would its time derivative.


0

The physics creating eddy currents and EMFs in inductors is the same: Faraday's law of induction. $ \oint_C {E \cdot d\ell = - \frac{d}{{dt}}} \int_S {B_n dA} $ The strength of any induced current and voltage is dependent on: 1) The amount of magnetic flux ($\int_S {B_n dA}$) 2) The rate at which the flux is changing So for the loop in your first ...


2

There is never actually an electric field in a conductor in the electrostatic sense. An E field is always generated perpendicular to a charged surface (the wire). For any wire carrying current, the electric field tends to radiate outward from the wire. The magnetic field will be circulating around the wire such that the Poynting vector, $ \vec S = \vec E ...


0

For the flat Amperian Loop, The current flowing through the wire that pierces the surface of the loop) is I. However, there is no field piercing the surface of the loop. Now, why is there is no field piercing the loop:- Of course the field between the plates of the capacitor no way pierces the surface of the loop "Isn't there a field inside the wire, ...


0

Note that light (in one of its interpretations) is a Electromagnetic Wave. Now, there is a huge difference between the oscillating magnetic field in an EM wave and a magnetic field generated by a permanent magnet or an electromagnet.\ One of the differences is that magnitude of the magnetic field in an EM wave happens to be very small. In fact, the ...


2

yes! it is. It is without doubt that the radiation of an antenna works in the way you described in your question. But what is a radio wave? It is important to differ between radio waves and electromagnetic waves. Radio waves are pulsed EM radiations. In each pulse electrons inside the antenna rod get accelerated and this led to photon emission. All this ...


0

For slowly moving charges, magnetism is just a relativistic correction, so the relative size of its effect is $O(v^2/c^2)$. Since $v$ is very small for charges in a wire (less than 1 cm/s), the effect will be insignificant. Since parallel currents attract, the current will be attracted to the center of the wire a tiny, tiny bit.


0

The emf of an inductor (solenoid) is directly proportional to the change in current per time. When you connect a solenoid to a voltage source, the back-emf is initially exactly equal to the voltage applied. So, if you apply a perfect voltage source to a perfect inductor, the current rises linearly forever. In the real world, after a while the resistance of ...


-1

Yes it will rotate due to magnets pushing and so providing motive forces, however not for too long at all this is because if you take the friction in account for the object that is generated between the pivot and the magnets. Furthermore, the air-resistance might not be too much but it will definitely extrapolate in results over few tens or minutes and will ...


1

The answer to your question is Maxwell's equations for Electromagnetics. Of course the variation of the electric field creates a magnetic field and vice versa, but we don't use the term induction in this case. \begin{align} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{E} & \:=\:-\:\dfrac{\partial \mathbf{B}}{\partial t} \tag{01}\\ ...



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