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A metal conducts quite well because the there is an electron band that crosses the Fermi level. So, electrons can easily be excited to increase their momentum a bit and consequently move in one direction. Now if you add one electron to the wire, the Fermi level rises. However, you would not be able to see the increase caused by one single electron (or a ...


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Connect the battery to this circuit. At this point, the charge doesn't know that there is a hole in the circuit. Negative charge therefor flows away from the negative battery pole, since it is repelled by this same charge, as far away as it can along the attached wire - this means that the charge will pass through the light bulb, and the light bulb will ...


1

If by our goal is to see the fields due to some charge or current, look at Jefimenko's equations. For electric fields electric field they can be caused by charge density, the rate of which charge density changes or by the rate that current density changes. For magnetic fields they can be caused by current or by the rate at which current changes. Since you ...


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Adding extra electrons isn't going to make any difference, at least as a first approximation. Suppose we have a wire with some applied voltage $V$ and some current flowing $I$. What this means is that the power supply is injecting $I/e$ electrons per second into one end of the wire and extracting $I/e$ electrons from the other end of the wire. The ...


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I think this picture answers the question There battery's terminals are not charged. It's just a chemical reaction that starts the charge. There's a Zn and a CuSO4. When they meet each other via a cable, the Zn gives 2 electrons to ChSO4. Cu gets rid of SO4, leaves 2 extra electrons to SO4 and takes the 2 electrons it got from Zn. Then we have SO4 ...


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I'm not sure about what your problem is. During the travel of the magnet a voltage is induced on the coil because of the changing magnetic field inside the coil. As the magnet closes to the coil, magnetic field increases. When the magnet passes the coil the voltage is reversed because now the magnet field inside the coil is decreasing. ...


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Both good answers. Note that the Earth has a self-capacitance by the same arguments. The Earth carries a net charge and really does have a capacitance. Remember that capacitance is defined in terms of the work that you must do to take a charge from one plate to the other. Two charged objects are involved. Even if the plates are nearly an infinite distance ...


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Clarification from another source: Source: Physics For Scientists And Engineers, Paul A. Tipler and Gene Mosca, Sixth Edition, W. H. Freeman and Company, New York, 2008, p. 971, Fig. 28-20. I maintain that the loop will act the same as the bar. In other words, if you cut a thin slit down the center of the bar and less than the length of the bar (you ...


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What I think is that there is no net emf induced in the first case and hence no current in the loop but there is emf induced in the second case but as the circuit is not complete, there is no current. Here is my reasoning: case 1 in the first case the flux which is (B. A) is not changing and hence emf I. E. d(flux) /dt is 0. One can ...


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Treat the individual electrons in each conductor as if they were in a closed container, and there was otherwise a vacuum in that container. When moving through the magnetic field in the top picture (the ring), I would expect electrons to move to the bottom of the ring, leaving a net positive charge at the top. This will only occur until the electric force ...


2

You are feeling an electric current because the "live" wire of the electrical outlet is connected to the "ground" of your phone. The same effect can be felt in the US (or wherever "back home" is for you) with an improperly wired desk lamp - or one that has an unpolarized plug. The electric field induces a small current in you because your body has a ...


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Let's assume that the power company is supplying a neighborhood with 1000 A of current at 120 V. Since P = IV, the neighborhood is receiving 120 kW of power, which is the "load" seen by the power company. To maximize efficiency, the power company wants to minimize the losses involved with transmitting power to the neighborhood, which occur due to ...


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You are mistaken about your fundamentals. Your initial assumption was that the pd across the resistors was nE not the whole circuit. You implicitly assumed the difference across the whole circui was 0 when you connected the components in a closed circuit. One of Kirchoffs law says the sum of the voltages in a loop in 0.


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In a diode, or PN junction, you have two regions: region P with an excess of holes: the material is doped with atoms having 3 electrons instead of 4 region N with an excess of electron: the material is doped with atoms having 5 electrons instead of 4 Each two regions are neutral but when we place them side by side, some electrons and some holes diffuse ...


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We are not talking about ions here. If you go to the second video you will hear from 7:51 that she talks about electrons and holes. Electrons are free to move around in the n-type material on one side of the junction, while holes, which are simply an empty space missing an electron, are free to move around in the p-type material. n-type: Think of a pure ...


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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 ...


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The short answer This is not 100% true since it assumes DC transmission, but it gives the simplest form of the idea: even if the transmission lines are themselves at high voltages, that doesn't directly mean anything, since voltages are not defined relative to anything special (they're defined relative to some other line which is in parallel with your ...


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There are two different $V$'s here. Suppose the power station outputs at 10,000 V. By the time the wire makes it to your house, this may have dropped to, say, 9,000 V. The $V$ in the first equation refers to the voltage difference you can use, which is 9,000 V (between the wire you receive and ground). The $V$ in the second equation refers to how much ...


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Voltage is a measure of the electric potential difference across two points in a circuit. It may be considered the work done to transport an electric charge. Power lines are made of thick easily conductive material in order to minimize resistance and power loss to heat. But resistance within power lines is fixed, and power is delivered through the line ...


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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.


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The electrons do not even enter the wire, because the redox reaction between the substances in each of the nodes never occurs. Once the wire is connected to each of the nodes, electricity will flow through as electrons will be more attracted to the node with the greater reduction potential.


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In my opinion, the mathematical equation we call Ohm's Law is best taken not as a “law”, a fact about the universe, but as the definition of resistance. $$R \overset{\mathrm{def}}{=} \frac{V}{I}$$ Given this definition of the quantity $R$, we can then make (as other answers have mentioned) the empirical observation that many materials have approximately ...


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To answer your question in one word, "Yes" Now, onto the explanation:- According to Faraday's Law, you will get a current in a conductor when the amount of magnetic flux linked with the conductor changes. Note that it is immaterial whether the source of the magnetic field is a permanent magnet or an electromagnet. All that needs to happen for you to ...


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According to Faraday's law of induction, $$\mathcal{E} = -N {{d\Phi} \over dt}$$ you will need a change in the magnetic flux $\Phi$ in order to get an EMF or an electric field. So if you just put your coil in the magnetic field of the permanent magnet, you will not measure a current. There will only be a current, if you move the coil around so that the flux ...


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Ohm's Law actually follows the definition of power, current and voltage. Let's begin by defining power $P$, current $I$ and voltage $U$ as $P = \displaystyle \lim_{\Delta t \to 0} \frac{E}{\Delta t}$, $I = \displaystyle \lim_{\Delta t \to 0} \frac{Q}{\Delta t}$ and $U = \frac{E}{Q}$. We then find for a constant current $I$ with a constant voltage $U$ that ...


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As 'quasi particles' those newly discovered weyl fermions only exist as specific wave points in a specific material. In that sense, you would not get a 'shocked' as humans are made of 'weyl-fermion-insulated material'


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Ohm's law isn't fundamental and holds true only under certain conditions, like constant temperature for example. However, there is a simple way to think about it. Imagine the flow of massive objects through a wide water pipe. This is like a current. The water pressure causes the objects to flow quickly, that's your voltage. If the pipe is narrow then the ...


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You could start from Drude in zero magnetic field, that equates the derivative of the momentum $\vec p$ by the electrostatic force $\vec F_{el} = q \vec E$ as a product of charge $q$ and electric field $\vec E$ minus a scattering term (with time constant $\tau$; compared to Newtons second law that does not feature the latter, crystal term): $~~~~~~\dot ...


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Ohm's Law is not a construct which can be derived. It is essentially a generalized observation. It is only useful for a few materials (conductors and medium resistivity), and even then virtually all of those materials show deviations from the ideal, such as temperature coefficients and breakdown voltage limits. Rather, Ohm's Law is an idealization of the ...


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There is a sense in which a massive fermion can be thought of as made up from two Weyl fermions of opposite chirality. While I haven't read the paper in question (it's behind a paywall) I get the impression that the quasi-particle detected in these experiments results from electrons behaving as if they were a pair of Weyl fermions, and that the result could ...


3

The ammeter measure the current flowing through itself. If you want to measure the current flowing through another component, then you must make the current through the ammeter equal to the current through the component. If you wire it in series, that's true. If you wired it in parallel, the current would be unevenly divided between the component and the ...


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This problem can be solved using a simple integral. First, we take the formula for resistance and rewrite it small lengths of wire. I also break up the cross sectional area in terms of thickness $t$ and circumference $2\pi r $ $$dR = \frac{\rho dl}{2\pi rt}$$ Think of this as the resistance of a single ring of plastic sheet. Now we integrate over the ...


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Since Michael has already pointed out that the problem as stated has no answer, I will answer a different question instead: if we have a resistive spherical shell with inner radius $a$, outer radius $b$, and bulk resistivity $\rho$, and the surfaces of this shell are coated with a conductive layer, what is the resistance between the inner and outer surface? ...


3

If you really mean "points", see the answer to this question. Basically, the logic is as follows: If you try to inject a finite current at a "point" in a bulk, it will necessarily lead to a divergent current density $\vec{J}$ in a neighborhood of that point, proportional to $r^{-2}$ (where $r$ is the distance from the injection point.) A divergent current ...


3

Basically, you want to find the proportionality between the total current and the voltage difference between cathode and anode. Let's assume that the current flow is radial under steady-state conditions, which basically allows me to ignore the $z$-direction throughout. In a steady-state solution, we will have $\nabla \cdot \vec{J} = 0$; moreover, if we ...


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During summer due to heat, the intra molecular space increases and the substance expands. As the substance expands, the weight of the conductor increases resulting in the increased sag.


0

The amount of work done by unit charge between any two nodes of current carrying circuit is called the potential difference between those nodes. The amount of work done against the electric field by displacing (without acceleration) a unit test charge from one terminal to other terminal in an open circuit is called the electromotive force. Obviously when ...


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First of all power will alway be same i.e VI (primary side of amplifier) = VI (secondary side of amplifier). However V and I individually can change. So in your case: Case 1: Voltage= 20 current will become I = 0.5 amp so net power = 10 watt Case 2: If current I = 8 amps V will be = 10/8 Volts so net power = 10 watts


1

I disagree with the other two answers, or at least believe they are not very clear. The important point is that, as you suggested, a induction cooker only consumes significant amounts of energy when a pot is actually on top of it. There are, of course other losses, but without any metal object in the vicinity, the cooktop is like a transformer without a ...


0

I believe an induction cooktop works by creating an alternating magnetic field, which heats any ferromagnetic materials close to the inductor. If you have an inductor switched on, but you're not heating anything, the inductor is still producing a magnetic field. It takes energy to sustain such a field, even if there's nothing being heated. Also, even if ...


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Induction cooktops contain electromagnets below each pot or pan station. When a station is switched on, electric current flows through wire wrapped around an iron core. In order for magnetic flux to be induced in the iron core, the electric current must constantly change, so the current must alternate. The iron core concentrates the magnetic flux ...


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Its simple. The electric line and the hair carry same charges, and since like Charges repel each other,this charges tends to push very far away from each other, and so this causes the hair to rise. The principle is 'the more the quantity of charges the more the hair tends to rise straight'.


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Just like in the case of the van da graaf generator, when a person touches the charged metal ball (dangerous) the excess stored electron on the ball travel into the person's body, now, the charges can't get to the ground(since the person's standing on a rubber). You are now filled up with electrons. The electrons don't like each other and are trying to get ...


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Related question on EE: Does perfect insulation exist? (especially the part about vacuum) Insulators and conductors The property of a material to carry charges from one point to another is what electric current is. The difference between insulators and conductors lies in the electron band structure they posses. In conductors the Fermi-level ...


0

Power(P) = Voltage(V) * Current(I) That law describes the relationship between power, voltage, and current in a conductor. It means that, if you measure the current flowing in the conductor, and you measure the voltage difference from one end of the conductor to the other at the same instant, then the product of voltage and current will be the rate at ...


0

The conditions needed to produce lightning have been known for some time. However, exactly how lightning forms has never been verified so there is room for debate. Leading theories focus around separation of electric charge and generation of an electric field within a thunderstorm. Recent studies also indicate that ice, hail, and semi-frozen water drops ...



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