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The voltage V that you're computing would be induced between the upper and lower sides of your orange block, and $L$ in your computation is the distance between these two sides. This voltage is induced between any two opposite points on the upper and lower sides provided they are in the area that is covered by the magnetic field. If you connect a wire with ...

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You are taking a shortcut when you say, "The voltage is zero." Voltage is always measured between two points. In electrical engineering, when we say the voltage at point X is V, we actually are measuring the voltage between point X and an implicit other point called "ground". In the electric power grid, "neutral" is ground, by definition. So the voltage ...

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You are having some misconceptions Wire(more pricisely axle) is not rotated between magnetic field in a turbine generator.Wire(more pricisely axle) is used to rotate an coil (present between magnetic fields) present in generator. Magnetic field itself can not force electrons in a conductor to move untill it is at rest. We need an moving conductor so that a ...

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No. The capacitors are in series. This is because one side is at the same potential. But it looks like parallel. If you apply Kirchoff's Loop Law, you will see that they are in series. And in that way too, you will get that answer. Hope that helps.

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NO. The Capacitors are in series as when we go from one capacitor to another we find no junction in between.

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So you have that $\vec{B} = \frac{a}{r^2}\begin{bmatrix}0 & z & -y\end{bmatrix}$ thus the magnitude is $B = \frac{a}{r^2}\sqrt{z^2+(-y)^2},$ where $a$ is unknown. Can you write that as a function of $r?$ Can you investigate what happens as $r$ goes to zero? Are magnetic fields continuous in empty space (a vacuum)? If so, try the next five: What ...

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The magnitude of the B-field is $a/r$ and circulates around the axis. By symmetry, you understand that the magnitude is zero on-axis. But if $a$ is anything but zero, your expression gives an infinite B-field magnitude. Therefore $a$ must be zero and therefore the B-field is also zero everywhere else inside the pipe. The result also follows from Ampere's ...

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If you keep the resistance constant, then $V=IR$ means that voltage is directly proportional to current. If you keep the power constant, then $V=\frac{P}{I}$ means that voltage is inversely proportional to current. However, because $V=IR$, we can write that $P=I^2R$. Therefore, if we say resistance is constant, then power must change with current, which ...

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All AC electric circuits radiate. Indeed, in the United States of America, all consumer electronic items must pass explicit FCC Electromagnetic Interference and FCC Radiation Safety assessment for their sale to be legal, so that one can make sure that the electromagnetic radiations emitted do not interfere with other electronic equipment and are safe for ...

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From classical models, the electron and a proton revolve around their mutual center of mass, which approximately lies on the proton itself, because the proton has a significantly higher mass than the electron. This is why electrons revolve "around" the proton, and hence form the outer layer of an atom. Quantum mechanically, electrons could never form a ...

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If I understand your questions correctly: Yes, it can be somehow the other way around. But we do "know": There are two sort of particles in here, one of them has a certain charge and is light, the other has the opposite charge and is heavy. You can then claim that the heavy ones rather stay in place and the light ones sprint around and make up the current. ...

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Think about what exactly the rate of flow of electrons is? It is the number of electrons per second passing through! The number of electrons is not included in your expression, and that's the problem. Let start over but this time with the number of electrons $n$ included: $$Q=It \Leftrightarrow \\ en=It \Leftrightarrow\\ \frac{n}{t}=\frac{I}{e} ... 1 You have to think more carefully about what exactly Q,I and t signify: In$$ Q = I t$$Q is the charge that is transported by the current I during the time t. If you now write$$ \frac{I}{Q} = \frac{1}{t} then this gives how many times a charge of $Q$ is transported by $I$ during one unit of time (second), since $t$ is the time to transport ...

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