# Tag Info

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Inductor opposes any "change" in the current in the circuit. As you know, $$V_L = -E = L\frac{di}{dt}$$. The more the change in current the greater the opposition. But this does not mean that if, for instance, the current is rising then inductor would start reducing the current. The presence of inductor only affects how fast the change is happening,...

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You are right, there is no back EMF without current. BUT: the electrons making the current are the same, whose motion generates the magnetic field that creates back EMF. Therefore, it is not like you first briefly accelerate, then note the back EMF, decelerate, accelerate again etc. It is a smooth process described by Maxwell's differential equations of ...

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I view that there is "some" posts and quasi same questions. I will answer for now briefly. A "complete" simplified answer for later. Your trouble understanding is justified. It is a theoritical "simplified" case. For good "understanding", one may apply the" propagating wave theory". Remark .This theory ...

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the i vs t graph above only reveals the current that has already been slowed down by the back emf but not the current before the drop. Stop thinking that there has to be some initial current to cause the back-emf. There was no current 'before the drop'. It started at zero exactly as the equation says it must, because that is the definition of inductance (if ...

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For $t < 0$ we have $i(t) = 0$. The inductor feels no pressure from the voltage source and so no back-emf is needed to keep the current at zero. At $t = 0$ we also have zero inductor current but the inductor is now feeling the pressure from the voltage source. The back-emf in this case exactly cancels out the voltage source emf to make the current ...

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Did you mean "without permanent magnet" in the last sentence? The excitation field MUST come from somewhere. Choices: Permanent magnets / Fed back from alternator output / From independent source eg battery. I cannot think of any other proper alternatives (but they may exist :-). Some few machines have an internal alternator to provide field energy ...

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Is there a way to derive theoretically the proportion of the current (or of the power) produced by the alternator which is lost (has to be re-injected) in order to supply the electromagnets of the device with power? The proportion is not a fixed relationship. There are a lot of details of the machine design that determine how much power is required to ...

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It's amazing that the technology is quite consolidated and dates back to the end of the World War II (https://en.wikipedia.org/wiki/Microwave_oven). A magnetron used for radars straightforwardly reused for something to last more that WW-II: desire for peace? big market move ensuring a long-lasting production after the boom for war reason? The typical ...

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The basic reason is that the water molecules in the food resonate at the 2.45 GHz frequency that the microwave oven uses. This is what allows the microwaves to couple to the water molecules, causing them to vibrate and thus heat up the food. It's also the reason why a microwave over doesn't affect, very much, objects that do not have water in them (metal ...

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You can cook food with any wavelength of EM radiation that is sufficiently absorbed by the food. Of course if you have a sufficiently strong source, you could even heat food with very penetrating radiation like x-rays. Ionizing radiation such as UV and x-rays will destroy proteins and vitamins though. Very low frequency radiation is also very penetrating and ...

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Sort of. The question is not whether they can be used, but whether they can be used efficiently. It sounds a bit strange to talk about "antenna matching" when one of the antennae is a lasagne, but that's what's going on. The target will absorb EM radiation of different wavelengths with different levels of efficiency. Other wavelengths will be ...

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You certainly can cook food with other wavelength like solar oven. Actually most kitchen oven use heating elements that produces infra-red (light), which is the primary heating factor. The wavelength also has to be such that it can produce heat on organic materials, radiowave, which are long waves mostly goes through. Microwave, of shorter wavelength, will ...

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As others have pointed out you can try some degaussing or reversing the current flow in your electromagnet. Both solutions will be difficult to implement and you may have to deal with BOTH the electromagnet residual and the attracted object residual fields. The force provided by your electromagnet is proportional to distance between the attracted object and ...

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I would suggest two possible solutions: First, make sure you are using a "soft" magnetic material. 1045 steel is somewhat "hard" meaning that it can be slightly permanently magnetized. Even "soft" materials may require heat treating to remove remanence. I would try a lower carbon steel like 1018 or 1010. Second, when the ...

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The most efficient and clean way to erase remanent magnetization is to heat the core to above its Curie temperature. But this is often impractical. When using AC fields to demagnetize, it is very important that the current profile also has a decay term. If you only run an AC current, and then switch it off, the magnet will be stuck with whatever field ...

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You can reduce magnetism in magnetic objects by degaussing. This usually involves running an AC or random magnetic field through the object, which can be done with a coil and alternating current. I can't see the coil construction within the magnet but you also may be able to briefly reverse the current to remove the object.

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It seems like your electromagnet is made from a material that has some residual magnetism. There is nothing that can be done about it, at least directly. You can use a reverse polarity switch, which will reverse the polarity momentarily and the attached object would then definitely fall. Try this principle first without knocking the steel object down ...

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Visible light is a wide-band while specialized cameras are a narrow-band sometimes with a variations as thin as 3 nanometer in lambda deviation. Such nanocomposites are indeed harder to produce and demand appropiate band filtering which furthers the whole costs of the technology itself,

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