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1

We can very well distinguish between magnetic and electric fields by simply observing the movement of a charged particle. As it was said correctly, the magnetic field does not do any work on the particle, meaning it does not accelerate it linearly. It does however apply a force that is perpendicular to the direction of motion. To illustrate, we assume a ...


0

There is only one case where isn't induced a current in the rod. One have to move the rod parallel to the magnetic field. In any other case and especially in any case of rotation the magnetic field will move the electrons inside the rod. For example, if the rotating axis is perpendicular to the rods symmetry axis, there are to pure cases. (1) The rotating ...


0

You can not distinguish the effects of the two forces just by seeing the movement of the charge.You have to measure current,magnetic field and make the appropriate calculations.Even if we talk about energy(no work is done by the magnetic forces,while all the work is done by electric forces)you have to measure it. The only thing you can do without ...


-1

The magnetic field does no work on the particle and hence its energy remains constant. The effect of the electric field changes its energy.


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This equation: $$m_1 v_1 + m_2 v_2 = 0$$ will not work that way. There are multiple ways of looking at a system of moving particles: You can look at it in reference to a random origin, for example one corner of your room. This is what we usually do. And this is where you are trying to apply your equation, but it doesn't work. You can set your origin to the ...


1

In the first case,momentum is conserved because force is applied to each charge from WITHIN the system.So,the center of mass of the system is constant.In the second case,in order for just one charge to move,it has to be put in an EXTERNAL electric field.So,you can see that momentum within the system which consists only of one charge can not be conserved.If ...


2

If one particle is fixed, some force is keeping it fixed, and in the presence of an external force, conservation of momentum doesn't apply. Your second equation is then $v_1 = 0$ (assuming particle #1 is the one that is fixed), not $m_1 v_1 + m_2 v_2 = 0$.


0

To say I am skeptical would be an understatement. I briefly worked on this around 1997 and what I see here, although it matches his description of what he wanted, in no ways matches the effects he claimed for his "original" which exhibited massive positive feedback, ionization, extreme cooling and massive antigravity effects. The over-unity claims were just ...


0

Take it to the limits to examine the problem. Imagine that your weak permanent magnet is really incredibly weak, weaker than the Earth's field, while it's mass is in the kg. When the electromagnet is turned on is the repulsion of that tiny field enough to overcome the attraction of a few kg of "spare" ferromagnetic material to the electromagnet? Well, from ...


0

You might find the Faraday Effect easier. Since it was actually done by Faraday it is quite easy. You need a glass or plastic rod aligned with the magnetic field you want to measure, a monochromatic or coherent light source, a couple of polarizing filters and a photodiode. The magnetic field rotates the plane of polarization proportional to the field in the ...


2

The incorrect part of your statements is that every current element does not have a circular magnetic field around it. Only an infinitely long wire would have perfectly circular magnetic field lines around it. In fact every current element does contribute to the field at point P and has a component both perpendicular and parallel to the wire. However, for ...


0

If you consider a constant current flowing in your coil as you rotate it, you will see it at first rotating clockwise, then, after the 180 degree flip, anticlockwise. If you create the same change in flux, but flip the coil, then you will induce a current in the opposite direction in the frame of the coil. In other words, while it will still look ...


1

To make the correct answer clearer, allow me to introduce the canonical momentum $\vec{p}$, given by: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}$$ This way we can rewrite the Hamiltonian as: $$H=\vec{p}\cdot\vec{\dot{x}}-L$$ Let's start by computing $\vec{p}$: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}=m\vec{\dot{x}}+\dfrac{e}{c}\vec{A}(\vec{x},t)$$ ...


3

The relationship you propose only applies if the current has symmetry with respect to the axis of the wire. Amperes law is that the line integral of the B-field around a closed loop equals the enclosed current times $\mu$. $$\oint \vec{B}\cdot d\vec{l} = \mu I$$ To get your relationship you need to assume that on any circular path around the axis, that ...


0

Ampere's Law. That magnetic field that you calculated depends on $r$ and the enclosed current is $j*ds$ where $j$ is the current density and $ds$ is surface differential. So if you want to know the magnetic field outside of a wire you should take a $r>R$.


0

The magnetic fields for the currents outside exactly cancel each other... That's just how the law of magnetism works.


1

A few points of confirmation/correction: Yes the electrons flow in a direction opposite to the "conventional" current. No there is no "deficiency" if electrons - rather they have a different "potential" which is caused by the chemical reactions in the battery. No you don't have to invoke "surface charges" in the wire in order to understand current - ...


0

In the first diagram, it appears that the rod is moving to the right with uniform speed $v$ and the magnetic field is pointing into the page. In this case, the magnetic force on a positive charge is towards the top of the page. Assuming positive charge is free to move inside the rod, the charge will redistribute until there is no longer a net force on the ...


0

The whole point of a 'density' is that it is independent of volume. For example, if you have simple mass density, the equation is $$rho=\frac{mass}{volume}$$ Which produced the same answer regardless of volume (assuming uniform density). The same is true for magnetic density. When you get a bigger box and integrate around the boundary, you will get a ...


1

Let us assume that the magnetic field we're talking about is homogeneous in the relevant regions of space. To create such magnetic field in a smaller box requires less energy than to create the same magnetic field in a larger (that is with a larger support as a vector field) box. This is why you clearly have more energy in the larger box than the smaller box ...


-2

I truly feels that the equations of 1. Magnetic energy density(Ub=0.5B^2/mu);Ub=mag.energy/volume 2. Electircal energy density(Ue=0.5epsilon*E^2) are godsends. To understand this question answer,you must review certain fundamental things like physical meaning of energy density:it tells that energy density is same everywhere in the space,note that not the ...


0

When one has a distribution of charges $q_1,\dots,q_n$ at points $\vec{r}_1,\dots,\vec{r}_n$, the energy of the system is given by the sum of the energy of each particle due to its interaction with the others divided by two since each interaction is counted twice i.e. $$U=\sum_{i=1}^n\sum_{\substack{j=1 \\ j \neq i ...


1

The dielectric constant, or more appropriately, the dielectric function, can be thought of as a measure of screening. A simple relation for which to picture this is: $V_{eff} = V_{ext}/\epsilon$ Therefore, in TMDs, since the electrons are more mobile in the planes, they tend to screen potentials with a greater efficiency. This gives a higher dielectric ...


0

Given a certain four-current $J^\mu = (c \varrho, \vec{j})$, that is a charge density $\varrho$ and current density $\vec{j}$. the four-potential $A^\mu = (\Phi / c, \vec{A})$ is given by: $$ A^\mu(\vec{r},t) \propto \int \frac{j^\mu(\vec{r}\ ', t_r)}{|\vec{r}-\vec{r}\ '|} d^3r'$$ with $t_r = t - \frac{|\vec{r}-\vec{r}\ '|}{c}$ if one takes the retarded ...


0

Your question consists actually of two parts, I will answer them one-by-one: Why is the magnetic field circular? Any vector field $\vec F$ can be decomposed into a rotational part and a divergent part, according to the Helmholtz decomposition theorem $$\vec F = - \vec \nabla \Phi + \vec \nabla \times \vec A $$ This is a purely mathematical statement and ...


1

You have forgotten that the vector $\vec{{S}}$ is actually a function of $\phi$. If you quickly switch into cartesian coordinates $\vec{S}=\cos\left(\phi\right)\vec{i}+\sin\left(\phi\right)\vec{j}$ you can see that when you integrate the $\vec{S}$ component around $2\pi$ the sine and cosine give zero. You have to be quite careful when using curvelinear ...


1

My question is that if we repeat the above procedure in vacuum, would the sparks be produced? Not in perfect vacuum, because there is no gas to ionize. An understanding of this behavior can be attained by studying Paschen's Law (see http://en.wikipedia.org/wiki/Paschen%27s_law), which describes the breakdown voltage for given distance and pressure.


2

There can be in principle both magnetic sources and electric sources. Including these, one may write the full Maxwell equations as $d \star F = \star j_E,$ $d F = \star j_M,$ where $j_{E,M}$ are the electric/magnetic currents, respectively (as I've written them they are 1-forms). Since there are no experimentally observed magnetic charges (magnetic ...


0

Without being sure of my answer,i would say that this difference in magnitude has to do with the charges themselves. The first time you did it,you pushed negative charges in one direction.When you reversed the polarity,then you already had a situation which was not in equilibrium as the first time.The first time you had a flow in which electrons and protons ...


1

No it is not correct because your Amperian loop is wrong.The circle has a more difficult integral.You can not get B out of the integral in a circular Amperian loop because in each point there is a different angle between B and dl,so the dot product between them changes. By saying that it is a long thin sheet i assume that it has infinite dimensions and zero ...


1

but I don't understand how he used this method to calculate potential difference. To be sure, this isn't a calculation of potential difference (since the value is path dependent) but is instead the work associated with moving a unit test charge along the path. The flux rule is simply that the line integral of the electric field along a closed path ...


0

From the point of view of electromagnetism I can see it the following non-mathematical way. $curl$ or $rot$ differential operators calculates magnitude of "vorticity" of the field, i.e. how much it is "spinning" or changing in rotation. Now, "vorticity" of the "vortex" is how much this "vortex" spinning again. Take, for example, long spring and connect its ...


-1

We have to add both of them. It is no more a choice between two of them.


2

When power is consumed, the voltage and current rises and fall together (with some variation due to power factor). As you consume energy from the grid the voltage generally stays the same but the current rises and falls depending on your consumption. As you reduce your demand to zero, the current falls to zero, then when you start feeding into the grid, ...


0

If particle A is attracted towards particle B because of an electric charge or gravity, then both will have a momentum towards or away from each other. Thus, momentum is conserved. Grab some paper and a pencil, go on, right now, I'll wait. Make a number line. Put A on -1 and B on 1. They will meet at 0. -1 + 1 = 0, and 1 - 1 = 0. Thus, A's velocity is ...


-1

I think as you are not a physicist it is quite hard to answer this without going into the details of quantum field theories. In the case of the electromagnetic force the theory which describes the interaction is Quantum Electrodynamics or QED for short. This theory describes physics on scales which are much smaller than things we encounter in our everyday ...


2

According to relativity, If magnetic field is just an electric field viewed from a different frame of reference It is true that a pure electrostatic field in an inertial reference frame (IRF) will be observed as a mix of electric and magnetic fields in some relatively moving IRFs. However, in the general (time varying) case, it is not possible to ...


5

Your statement is not really true, since if you only have a magnetic field in one frame of reference, then it can never be viewed as just an electric field in another frame of reference. And vice-versa. As described here, the magnetic field can be defined as (e.g. in Jackson's Classical Electrodynamics) the field that is responsible for the Lorentz force ...


1

The best rubber material to use is pure gum rubber. Rubber from an inner tube (tire) contains carbon and balloons may have other colorants which are conductive and will not work well. A wider belt will carry more charge. The 'combs' or charge pick-offs at the base and collecting sphere should not touch the belt but come very close - about 1/16". The tube ...


1

It's funny that all the answers so far forgot the simple and elegant following criterion : quantum mechanics appears when $$\dfrac{\hbar\omega}{k_{B}T} > 1$$ with $\hbar\approx6,63.10^{-34}\text{J}\cdot\text{s}$ the Planck constant and $k_{B}\approx1,38.10^{-23}\text{J}\cdot\text{K}^{-1}$ the Boltzmann constant, $\omega$ and $T$ being the (angular) ...


2

Yes, there must be. The solenoidal law tells us that the normal component of the B-field must be continuous across any interface. There are no sources or sinks of B-field. Therefore the lines of magnetic field are continuous through your bar magnet - they come in through one end (very roughly speaking) and out through the other.


0

It seems to me that, unless I am mistaken, the fields defined above do not satisfy $$\nabla \times \textbf{B} = \epsilon_0 \mu_0\frac{\partial \textbf{E}}{\partial t}.$$ As long as $$ \omega^2\epsilon_0\mu_0=k^2+\frac{n^2\pi^2}{a^2} $$ then they do satisfy the above equation.


0

Electromagnetic Radiation falls under the umbrella of Electromagnetic Fields. There are all kinds of Electromagnetic fields. For example stationary Electric fields between fully charged parallel plates. Some EM fields propagate. They travel in space. These fields are periodic, they are traveling EM waves. They are a solution to the wave equation derived ...


2

The distance on top, $R$ or $r$ effectively cancels the $r^3$ on the bottom to give overall a $r^{-2}$ factor, i.e. inverse square. Often you'll find expressions like this in EM, with things like: $$ {\vec r \over r} = \hat{r} $$ so that $$ {\vec r \over r^3} = {\hat{r} \over r^2} $$


3

The Hall voltage can indeed be equal to zero if the electrons and holes balance out. You can find the formula in these Hall Effect lab notes by Pengra, Stoltenberg, Dyck, Vilches, eq. 16: $$R_H = \frac{1}{|q|}\frac{n_h \mu_h^2 - n_e \mu_e^2}{(n_h \mu_h + n_e \mu_e)^2}$$ where $\mu$ is mobility and $n$ is density and $e$ means electron and $h$ means hole ...


3

In real-life conductors you always have some of both kind of charge carriers (electrons and holes), so, the main question is, which of those is more abundant? In fact materials with positive and negative hall coefficients can be found.


-3

to me magnetism seems to work with super small temperature changes occuring thrue two locked in densities in two compartments locked against each other, the compartment with the higher density is the compartment with more resistance,creating heat on one side, making a flow between them possible. we seem unable to measure these small differences, but i can ...


0

A permanent magnet owes its magnetic field to microscopic currents. If you imagine each tiny current as a circular one laying in a plane such as in the picture below, you will see that the currents immediately facing each other are going in opposite directions and therefore cancel out. What is left is just the surface current that seems to flow on the ...


0

As far as I understand the question. The question is when EMF source is attached and detached. EMF Source Attached. Consider the EMF source to be a battery in a circuit.A battery maintains a potential difference between its ends and hence provides an electrostatic force for the charges in the circuit. EMF Source Unattached. When the EMF source (battery) ...



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