# Tag Info

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Suppose an inductor is connected to a source and then the source is disconnected. The inductor will have energy stored in the form of magnetic field. But there is no way/path to discharge this energy? Short answer: It will find a way/path to discharge this energy. Longer answer: Let's have this simple electric circuit consisting of a battery (...

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It looks like an induction coil with the make and break device at the bottom and a switch right at the bottom. If you connect it up to an accumulator, be very, very careful as the output between the two balls, when separate, could be lethal. Also the electrical insulation elsewhere may be poor and you might get a shock just by touching the switch. Use ...

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It depends. You cannot disconnect an ideal inductor from an ideal voltage source with an ideal switch. These ideal things will break your calculations and you will get an infinite voltage on disconnect. A real inductor has its coil resistance, a capacitance between coils and an insulation between coils that has some great, but pretty much nonlinear ...

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In plain English, it is just Lenz’s law : Lenz's law, named after the physicist Emil Lenz who formulated it in 1834, states that the direction of the electric current which is induced in a conductor by a changing magnetic field is such that the magnetic field created by the induced current opposes changes in the initial magnetic field. It is the basic ...

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The physics you are looking for is electromagnetic induction (https://en.wikipedia.org/wiki/Electromagnetic_induction#Electrical_generator) When you move a permanent magnet relative to a conductor (the copper wire), the magnetic field of the magnet influences the electrons in the copper, creating a current. Really, the energy that you put in to the system by ...

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The fan motor provides a torque $\tau$ which has to accelerate $\alpha$ the fan blades whose moment of inertia is $I$: $$\tau=I\alpha$$ Given how long it takes for the fan blades to stop the frictional torques must be fairly low and so the torque applied by the motor to keep them going must also be low. With the relatively small torque rating, even if the ...

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The minus sign is what makes Maxwell's equations obey causality, so it's a good thing it's there! To see this, you can write out the source-free Maxwell's equations with the sign of $\nabla \times \mathbf{E}$ reversed in Ampère's Law. If you then to follow the standard construction to extract the wave equation from Maxwell's equations, you would obtain ...

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The definition of magnetic flux is $$\Phi = \int_S d\vec{A}\cdot\vec{B},$$ where the integral is not over a closed surface in general. Gauss' Law requires that the integral is over a closed surface, and so there is no contradiction. In particular, look at any basic discussion of Faraday's Law. They always look at simple loops or coils of wire. There are ...

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Koldrakan’s answer explains how the energy is generated. But you might be confused as to why the bulb keeps glowing for some time rather than the light itself fluctuating with the shake. If you didn’t already know this, that's due to the capacitor. A capacitor can store energy in the form of charge. When you shake it the electric energy generated gets ...

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Practically speaking yes, this will almost certainly be a step-down transformer, but I would agree with those other teachers: it can't really be concluded from the wire thickness. It's perfectly possible to build a transformer in which the secondary coil has more windings, but nevertheless use thicker wire. As John Rennie explained, this doesn't normally ...

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An electromotive force doesn't require a conductor --- it doesn't even require matter. The electromagnetic field is a local property of the vacuum, governed by Maxwell's equations. The relevant one in this case is $$\vec\nabla \times \vec E = -\frac\partial{\partial t}\vec B$$ That is, at any point in space, a changing magnitude or direction for the ...

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Duality is actually not $\mathbf{E}\leftrightarrow \mathbf{B}$ (I've used $c=1$), i.e. $(\mathbf{E},\,\mathbf{B})\to(\mathbf{B},\,\mathbf{E})$. It's $(\mathbf{E},\,\mathbf{B})\to(-\mathbf{B},\,\mathbf{E})$. Defining $\mathbf{F}:=\mathbf{E}+i\mathbf{B}$ is a popular way to check this; the above duality is $\mathbf{F}\to i\mathbf{F}$. It's instructive to ...

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The thickness of the wire determines the maximum current the wire can carry without overheating. Thicker wire means a greater current. With a transformer the power coming into the primary, $P_p = V_pI_p$, is the same as the power coming out of the secondary, $P_s = V_sI_s$, (less a few resistive losses) and this means $VI$ is constant. For a step down ...

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You are right, $\mathbf{B}$ is a vectorial quantity. Furthermore, it depends on position $\mathbf{r}$. Therefore we need to write $\mathbf{B}(\mathbf{r})$. The magnetic field around two current-carrying wires (with the currents flowing in the same direction) looks like this: (image from schoolphysics - electromagnetism - forces between currents) Notice ...

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If you extend Maxwell's equations to include the possibility of magnetic charge $\rho_M$ and magnetic current $\mathbf J_M$, then Faraday's law and Ampere's law look extremely similar: $$\nabla \times \mathbf E = -\left(\mu_0\mathbf J_M + \frac{\partial \mathbf B}{\partial t}\right)$$ $$\nabla \times \mathbf B = \mu_0\mathbf J_E + \epsilon_0\mu_0\frac{\... 11 But there is no way/path to discharge this energy? What will happen to the stored energy, current and voltage of the inductor in this case? In that case it makes its own circuit with its own path to ground. Often, that is through dielectric breakdown at the switch itself, but the details are highly unpredictable and depend very sharply on environmental ... 10 You might be thinking in comparison to a desk or handheld electric fan. As mentioned by @Farcher, \tau = I\alpha. I, the moment of inertia of a spinning body around a particular axis of rotation, is calculated as follows:$$I = \iiint\rho(x,y,z)||r||^2\ dV$$Or with uniform density,$$I = \rho\iiint||r||^2\ dV$$From this formula, you can see that ... 9 I do not think that I need to draw a closed surface but here is an example of an open surface with a closed loop at its throat. This is likened to a butterfly net. It is often the case that the closed surface is taken to be in the plane of the loop for ease of calculation but this does not always have to be so. The answer to a recent question illustrates ... 9 Usually this extra energy creates a spark due to the high back emf produced. But it is not always possible for a coil to create sparks. It is clear If we try out the experiment. So what happens to the magnetic energy if no sparks are generated? firstly, The sudden switching off would create a potential. difference between the ends of the coil. This means ... 8 Notice, the magnetic magnetic field B at the center of a coil carrying current i, with radius r & having n no. of turns$$B=\frac{\mu_0}{2}\frac{ni}{r}$$hence, magnetic flux \phi linked to the coil is given as$$\Phi=BA=\frac{\mu_0}{2}\frac{ni}{r}\pi r^2=\frac{\mu_0 \pi nir}{2}$$Now, setting the value of \phi, we get$$L=\frac{n\Phi}{i}=...

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It's a bit complicated (Wikipedia). Induction motors work in sync with the AC frequency but have no torque at 0 RPM so they need some arrangement to get them started.

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When we have a DC voltage source with a switch in series with RL and the switch is closed at t=0 then it is said that current is zero initially, but the voltage across inductor is same as that of applied voltage( according to kirchhoff voltage law) so there should be current( according to v=L(di/dt) )but it contradicts the initial statement so how do I ...

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Neither Coulomb nor Biot-Savart are correct equations for the electromagnetic field except in statics. There are time dependent generalizations, such as Jefimenko's equations. \vec E(\vec r,t)=\frac{1}{4\pi\epsilon_0}\int\left[\frac{\rho(\vec r',t_r)}{|\vec r -\vec r'|}+\frac{\partial \rho(\vec r',t_r)}{c\partial t}\right]\frac{\vec r -\vec r'}{|\vec r -\...

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An important point which is somewhat addressed by others but perhaps not clearly enough is (quoting Scotty) "Y' canna break the laws of Physics". You can make everything ideal - semiconductor wire, perfect instantaneously acting switch, infinite insulation - and the basic "rules" governing an inductor still apply. The fact that current flow cannot ...

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If the coil is superconducting the current is still there after you stop moving the magnet. In a normal-metal coil the current dies away due to the coil's resistance, and the extra energy is dissipated as heat.

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Doesn't a battery do this? Also, capacitors. EDIT: With the edit, it looks like the premise of your question could be satisfied by a Van de Graff generator: https://en.wikipedia.org/wiki/Van_de_Graaff_generator which uses friction to strip electrons from a substance, and create an electrostatic potential.

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Piezo electric cells convert mechanical energy to electric energy

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