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If your power supply is sourcing a positive current toward the ground, that corresponds to a flow of positive charge from the supply to ground. This is equivalent to a flow of negative charge from the ground to the power supply. In a real wire, only negative charges can flow, so the second thing will happen: electrons (which have a negative charge) will ...

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The external force on the whole current-carrying wire in region $V$ can be expressed as $$\mathbf F = \int_V \rho\mathbf E_{ext} + \mathbf j \times \mathbf B_{ext}\,d^3\mathbf x.$$ where $\mathbf E_{ext}, \mathbf B_{ext}$ is the external electric and magnetic field and $\rho,\mathbf j$ is the charge and current density. Stationary and moving wire will ...

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The current flows through both resistors. What works for me is to use the analogy of fluid through pipes under pressure. Imagine two huge tanks, connected by two small pipes of different sizes. The pressures in each of the two tanks are analogous to the two voltages. The pipes are analogous to the resistors. The fluid is analogous to the electrons. The ...

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I know I'm a little late, but I'll take a shot at answering this for you. I'm actually very much a beginner at understanding electronics myself, so everyone: please keep me honest! There has been some criticism of your question, as it does not show a complete circuit. I need to agree with this, as any reliable calculations within a circuit require ...

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Theoretically, these requirements arise from the way you connect the measurement devices to the rest of the circuit. A voltmeter is connected in parallel, as you said. Say that you are trying to measure the voltage drop across a resistor $R$ through which passes a current $i$. If the internal resistance of the voltmeter is comparable to $R$, then the ...

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In analyzing circuits one must always consider the possibility that things you've ignored in some situations cannot be ignored in others. For instance, there is always the possibility of an internal series resistance in a voltage supply or a parallel resistance in a current supply. Usually we ignore those if the external resistance is much higher than the ...

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According to my understanding, indeed you could define a physical quantity like $$\vec{I} = I\;\vec{n_d}$$ where $\vec{n_d}$ is the unitary drift direction. There is no problem with that. But what is the most important is to understand the harmony between different quantities. I mean that there is some little subtleties between $I$ and $\vec{j}$. The ...

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A general solution is to Assume a potential difference between the electrodes (the two balls in the first problem, or the ball and the plane in the second problem). Calculate the electric field present in the medium surrounding the electrodes. Use the microscopic form of Ohm's Law: $$\vec{J}=\sigma\vec{E}$$ and integrate over a closed surface ...

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First let's examine what corrosion is. "In the most common use of the word, this means electrochemical oxidation of metal in reaction with an oxidant such as oxygen. Rusting, the formation of iron oxides, is a well-known example of electrochemical corrosion." from wiki Keeping it simple, when you put a DC charge across the cathode and anode you are ...

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It appears to me that you are slightly confused with regards to the concept of current in conductors. Now, if I only choose one side of this rectangle, and apply external electrical field ∑ only to it, what EMF would I create on the conductor? I would simply say ∑, however then I had the following idea, and I started to doubt if I create 2∑ instead ...

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EMF is actually the potential difference across the conductor. In your case, since $V = \int \Sigma .dr$, therefore the resulting EMF is $\Sigma$ multiplied by the length of the conductor in direction of applied electric field.

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Yes. There are ammeters which are usually used in AC circuits which measure current on the basis of Joule heating of the ammeter wire. You can use a simple compass needle. If the deflection of the needle is appreciable, say about 3 degrees, it will probably interfere with the reading of a shunted-galvanometer ammeter.

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Any ammeter which uses a shunt resistor will ignore magnetic fields (if it is well-designed), and measure only the voltage across the shunt. Since the shunt is a known resistance, $$i = \frac{V}{R}$$ For instance, this is the standard way a DMM measures current.

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