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It is easy to show that the differential and integral forms of Maxwell's equations are equivalent using Gauss's and Stokes's theorems. Correct, they are equivalent (assume no GR, and no QM) in the sense that if the integral versions hold for any surface/loop then the differential versions hold for any point, and if the differential versions hold for every ...

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Yes, in general it affects the current, in particular if that current was going to change, the self inductance $L$ makes it change less quickly than it otherwise would change. Here are some examples: You could put a resistive loop inside a solenoid as a good example. The total $\vec{B}$ field is the $\vec{B}$ field due to the solenoid $\vec{B}_1(t)$, and ...

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If you have current flowing one way through a resistor, then the electrons flow through the other way. Since current flows from the high voltage end of a resistor to the low voltage end, then the electrons come in at the low voltage end and come out at the high voltage end. When electrons (which are negatively charged) go from low voltage to high voltage, ...

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There may be a confusion between enery and power. While the current, which is the number of charges per second, and the energy of the ,electron do'nt change at the output of the resistor, the power does. Power is the amount of energy per unit time, and that does not affect the current which, again, is the amount of charge per unit time.

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A battery works by making charge want to move from one terminal to the other (we call this a potential difference, or voltage). The charge is also moving through the inside of the battery itself (the electrolyte). When you hook up a wire to an anode and cathode of two different batteries, the buildup of charge on the anode of the first battery quickly ...

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A battery is no capacitor, and the actual charge stored in the battery terminals is very low. When you connect the anode of one battery to the cathode of another, that charge is transferred very quickly, and the voltage drops to zero. When you connect anode and cathode of the same battery, a chemical reaction takes place, and charges flow inside the battery ...

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Consider 2D (XY) case for simplicity, $\vec x = x\vec e_x + y\vec e_y$. $\vec e_x$ is unit vector along x-axis. By definition, gradient is the following operator upon scalar: $\nabla=(\frac{\partial}{\partial x}\vec{e_x} + \frac{\partial}{\partial y}\vec{e_y})$ Apply this operator to $\vec p\cdot\vec x$: $\nabla(\vec{p}\cdot \vec ... 1 Take for example a conducting body with an arbitrary shape and volume$V$connected to a battery that will cause a current flow in the body. You don't know the microscopic structure of the body; so in principle for any point$ \mathbf{r} \in V $you can have a different value of$I$: $$I=I(\mathbf{r})=I(x,y,z)$$ Now, take any surface$ \Sigma $in$V$; for ... 1 That is a limiting procedure pretty much similar to the definition of derivative. Fix the point$x$where you want to compute the current density and consider a small disk$D$centred at that point. There will be a certain current$I(D)$flowing across this disk. As this disk shrinks on to the point, the ratio $$\frac{\text dI}{\text d ... 0 according to conversational current, electric current is said to move from the positive terminal to the negative. This is not correct. The conventional current defines positive current in the direction of flow of positive charge, or opposite the direction of flow of negative charge. But conventional current can flow either from a higher potential to ... 0 I see three kinds of lines: On the left I see a vertical line of current, some circular lines (\vec{B} field caused by current?) and some orange lines (rightwards? external \vec{B} field?). The picture on the right might be the exact same thing, look up along the direction of the current. If so, then say \vec{B}_1(x,y,z,t)=B\hat{x}, and ... 6 First, there is nothing wrong with our charge polarity conventions. They are predict electrical phenomena just as accurately as the opposite convention would have done. Did we have a problem if since the begin of their discovery we called them positive particles and negative to protons? We could predict the behavior of electrical phenomena equally ... 0 If you have a thin circuit with a total resistance R, and place it in an external (changing) \vec{B} field, then there is flux through the ring. First, there is flux \Phi_1 from the external, changing \vec{B} field. Since that \vec{B} field is changing, there is an emf due to that. Second, the current from the ring itself produces it's own ... 0 If you have a thin circuit with a total resistance R, and place it in an external (changing) \vec{B} field, then there is flux through the ring. First, there is flux \Phi_1 from the external, changing \vec{B} field. Since that \vec{B} field is changing, there is an emf due to that. Second, the current from the ring itself produces its own ... 0 If you have a stationary ring, with total resistance R, inside an external (changing) magnetic field, then there is flux through the ring. First, there is flux \Phi_1 from the external, changing \vec{B} field. Since that \vec{B} field is changing, there is an emf due to that. Second, the current from the ring itself produces it's own \vec{B} ... 0 The question of what causes what is a good one. It's best to think of fields as changing in time due to the spatial configurations of the fields. An example is:$$ \frac{\partial \vec{B}}{\partial t}=-\vec{\nabla}\times \vec{E}.$$The instantaneous values of \vec{E} not only exert forces on charges (to contribute to their acceleration) but also ... 0 At a fixed time t, the wire is the line (0,vt,z) where z can be any value and s can be your instantaneous distance to this line. Specifically, we can consider \vec{A}(x,y,z,t)=-\frac{\mu_0I}{2\pi}\ln s\hat{z} where s is nothing more than a shorthand for \sqrt{x^2+(y-vt)^2}. Thus, taking the quasistatic approximation,$$\vec{E} = ... 2 I don't really see what the problem is. If$B$is zero on the boundary, your equation shows$0= \mu_0 I(t) + \mu_0 \iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}$, so $$\iint \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}=-I(t)$$ Why is it that you expect that this equation isn't satisfied? I can tell very ... -1 I couldn't put a picture in the comments and I am too lazy to do any drawing without my tablet. I think the problem or at least part of it, is because the magnetic field has discontinuity in space so its curl at those points is not well defined and as a result someone cannot change the surface integral of the curl into a line integral over the perimeter of ... 0 I'll assume that we are neglecting any curvature of space (no GR) and any quantum effects (no QM). The differential and integral forms are entirely equivalent, but the integral forms are not as physically intuitive as you might hope in cases that aren't static or quasistatic. The other factor is that their utility could be questionable. For instance, in a ... 2 There is a mistake in your working. The surface integral of the Poynting vector $$\int \vec{\Pi} \cdot d\vec{S} = \Pi\ 2\pi a h = \frac{a I^2}{2\pi^2 \gamma a^4} 2\pi a h = \frac{I^2 h}{\pi \gamma a^2}$$ As pwf correctly points out, this power matches the power dissipated as Ohmic heating,$VI$, where $$V I = E h I = \frac{I^2 h}{\pi \gamma a^2}$$ This ... 1 This is a great question! I'm not sure where the power is coming from, but I do know that it corresponds to the power flowing out due to Ohmic heating: Note that$R = \frac{h}{\gamma \pi a^2}$and$V = IR$, so$E_z = \frac{V}{h}$(not surprising). Then$\Pi_r|_a = \frac{V I}{2\pi a h}$, and$\oint\Pi\cdot dS = \Pi_r|_a 2\pi a h = VI$, the ohmic dissipated ... 0 Firstly: when the current flows in the positive z-direction and the magnetic field obeys the usual right hand rule then the Poynting vector points into the wire, so there is a minus sign missing in your expression: $$\vec{\Pi}=\begin{cases}\frac{r}{2\pi^2\gamma a^4}I^2(-\vec{e_r})&r< a\\\vec{0}&r> a\end{cases}$$ Secondly: There is actually a ... -2 The Surface integral goes over the surface element of cylinder's lateral Surface, i.e.$d \vec S = r d \phi dz \vec e_r$. Then, you evaluate the integral at$r = a = const.$; you set$r = a$for the Integrand. Integration over$z \in [0,h]$and$\phi \in [0,\pi]$gives the result. 0 Ohmic dissipation is proportional to the square of the current multiplied by a resistance. Yes, the resistance is inversely proportional to the conductivity, but the current induced by a given electric field is also proportional to the conductivity. Hence the Ohmic losses end up being proportional to conductivity. Symbolically:$\$ P_{loss} \propto J^2 R ...

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There are bound electrons and there are free electrons - the former interact with radiation and electric fields and the like by bouncing around energy levels and emitting photons (dissipating energy, as in a resistive element), the latter do actually move around and are strictly not localized around the nucleus of a few atoms, and largely preserve energy ...

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I need to marry the classical model of the colliding electrons and atoms of the medium with the Quantum principles involving the excitation of charges to higher energy bands and emission of photons. My answer to this question explains something about the concept of the conduction band. Think of it this way. In all the filled shells (including shells ...

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If I had to solve this problem on a paper with a pen but without the Kirchhoff equations then I would do it in four stages in each replacing the voltage sources with a short circuit and the current source with an open circuit. Then one by one putting them back (3 sources - 3 steps). In step 0 without the sources I would calculate the resistance between the ...

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You could use a freeware like pspice student or ltspice which allow you to model electric circuits like the one you descirbed. You can easily calculate currents and voltages in any element in your circuit.

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Higher voltages allow conduction paths that can do less than kill. You can get serious electrical burns and even arcs that pass through the body and leave burn channels, as in lightning strikes, or sitting on a charged laser capacitor (fatal for a Stanford grad student many years ago). Nerve damage and permanent heart arrhythmia can result. Might be good for ...

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Actually it is power that kills. A human body presents variable resistance depending on how good an electrical contact you make with the hot terminals. If you are wearing rubber sneakers, for instance, then you provide a high resistance making it "safe" to touch a hot 120 Vpp wire with one hand. (Don't do that please! This is in no way an invitation or ...

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It may be the current that kills, but it is voltage that is necessary to make the current flow through your body, which is the thing that we usually need warnings about. If there is 1MW of electricity flowing through a superconducting wire at 1V difference from ground, that is a crazy amount of amps, but if you touch it, you will not get 1MA of current ...

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