Is there anyway to transfer the electromagnetic energy of a naturally occurring magnet into usable electricity kinda like using the magnet as a battery?

spatularificw2
2022-09-27
Answered

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Domenigmh

Answered 2022-09-28
Author has **7** answers

A magnet does not contain electromagnetic energy, rather, it just has a magnetic field, the force generated by that field on a particle is influenced by the charge and distance. An electromagnet only occurs when there is moving current through a wire.

A magnet can certainly move electrons though a wire, but I believe this would be incredibly inefficient for any real practical application. Just try to imagine a device this would be applied to, a flashlight, for example. Using copper wire (which has electrons that can be moved quite "freely"), the magnet must be moved the distance of the wire quickly back and forth, resulting in even more work than the traditional battery.

A magnet can certainly move electrons though a wire, but I believe this would be incredibly inefficient for any real practical application. Just try to imagine a device this would be applied to, a flashlight, for example. Using copper wire (which has electrons that can be moved quite "freely"), the magnet must be moved the distance of the wire quickly back and forth, resulting in even more work than the traditional battery.

garnirativ8

Answered 2022-09-29
Author has **1** answers

No. The only way I know is to move a wire through the magnetic field thereby creating an electric current. You would need to provide energy to complete the movement of the wires. The electricity generated could then be used to charge a battery. This is what a generator does.

asked 2022-08-12

Imagine that I use a long wire to create an electromagnet. Let's also assume that the current flowing along the wire is constant, and that the wire is winded on the vacuumm.

Is the magnetic field generated by this electromagnet (current flowing over an arbitrary path on a finite volume) proportional to the current?

Is the magnetic field generated by this electromagnet (current flowing over an arbitrary path on a finite volume) proportional to the current?

asked 2022-07-17

How can electromagnets be useful to humans?

asked 2022-07-16

Light can be slowed down to a walking pace and even stopped in a medium of hot rubidium vapour. If an electromagnet was placed in this medium and turned on would the popagation of the magnetic field be slowed down or stopped?

asked 2022-08-11

Let’s say we have two inactive electromagnets spaced one light-minute apart, with north poles pointing toward each other. One of the electromagnets is turned on for ten seconds, and then turned off again (or destroyed). Once the magnetic field from the first electromagnet has propagated to the second, the second electromagnet is turned on. Would the second electromagnet not be repelled away by the magnetic field without exerting a force on the first electromagnet? How is momentum conserved in this scenario?

asked 2022-07-22

Optimal Electromagnet Shape to Attract a Steel Ball

I have an application where I need an electromagnet that generates a pull force on a ~10mm diameter steel ball that is initially ~10mm from the electromagnet.

For a single coil electromagnet with a fixed number of turns and current, is there a core shape that provides the highest attraction force versus distance for a steel ball of a given diameter?

I have an application where I need an electromagnet that generates a pull force on a ~10mm diameter steel ball that is initially ~10mm from the electromagnet.

For a single coil electromagnet with a fixed number of turns and current, is there a core shape that provides the highest attraction force versus distance for a steel ball of a given diameter?

asked 2022-08-31

The core of the problem is "If the magnet is located in this position, what about the attraction of the electromagnet over the magnet?"

What about the dipole-dipole interaction? If suitable, the overall magnetic moment could be given by $\overrightarrow{m}=\overrightarrow{{m}_{coil}}+\overrightarrow{{m}_{core}}$, where $\overrightarrow{{m}_{coil}}=Ni\overrightarrow{S}$ ($i$ = current, $N$ = number of winding, $|\overrightarrow{S}|$ = area of a single loop). Is it correct?

If the dipole - dipole approximation is not suitable, what else?

What about the dipole-dipole interaction? If suitable, the overall magnetic moment could be given by $\overrightarrow{m}=\overrightarrow{{m}_{coil}}+\overrightarrow{{m}_{core}}$, where $\overrightarrow{{m}_{coil}}=Ni\overrightarrow{S}$ ($i$ = current, $N$ = number of winding, $|\overrightarrow{S}|$ = area of a single loop). Is it correct?

If the dipole - dipole approximation is not suitable, what else?

asked 2022-07-16

Temporal distribution of a single photon pulse in an interferometer experiment in vacuum via the Gaussian function $\psi $ :

$\psi (t)={\textstyle \frac{1}{(2\pi {\sigma}^{2}{)}^{1/4}}}{\text{e}}^{-\frac{{t}^{2}}{4{\sigma}^{2}}}{\text{e}}^{\text{i}{\omega}_{0}t}\phantom{\rule{thickmathspace}{0ex}}.$

It is normalised

$\int |\psi (t){|}^{2}\text{d}t=1\phantom{\rule{thickmathspace}{0ex}},$

and the fourier transform is the wave function in the frequency domain,

$\stackrel{~}{\psi}(\omega )=\frac{1}{\sqrt{2\pi}}\int \psi (t){\text{e}}^{-\text{i}\omega t}\text{d}t={\textstyle (}{\textstyle \frac{2{\sigma}^{2}}{\pi}}{{\textstyle )}}^{\frac{1}{4}}{\text{e}}^{-{\sigma}^{2}(\omega -{\omega}_{0}{)}^{2}}\phantom{\rule{thickmathspace}{0ex}},$

such that $|\stackrel{~}{\psi}(\omega ){|}^{2}$ represents the frequency distribution of the photon.

However, since a single photon pulse is still a electromagnetic pulse, is there any link between $\psi (t)$ and the electric field $E(t)$ of this pulse? Like that

$E(t)\sim \text{Re}[\psi (t)]\sim {\text{e}}^{-\frac{{t}^{2}}{4{\sigma}^{2}}}\mathrm{cos}({\omega}_{0}t)\phantom{\rule{thickmathspace}{0ex}}?$

$\psi (t)={\textstyle \frac{1}{(2\pi {\sigma}^{2}{)}^{1/4}}}{\text{e}}^{-\frac{{t}^{2}}{4{\sigma}^{2}}}{\text{e}}^{\text{i}{\omega}_{0}t}\phantom{\rule{thickmathspace}{0ex}}.$

It is normalised

$\int |\psi (t){|}^{2}\text{d}t=1\phantom{\rule{thickmathspace}{0ex}},$

and the fourier transform is the wave function in the frequency domain,

$\stackrel{~}{\psi}(\omega )=\frac{1}{\sqrt{2\pi}}\int \psi (t){\text{e}}^{-\text{i}\omega t}\text{d}t={\textstyle (}{\textstyle \frac{2{\sigma}^{2}}{\pi}}{{\textstyle )}}^{\frac{1}{4}}{\text{e}}^{-{\sigma}^{2}(\omega -{\omega}_{0}{)}^{2}}\phantom{\rule{thickmathspace}{0ex}},$

such that $|\stackrel{~}{\psi}(\omega ){|}^{2}$ represents the frequency distribution of the photon.

However, since a single photon pulse is still a electromagnetic pulse, is there any link between $\psi (t)$ and the electric field $E(t)$ of this pulse? Like that

$E(t)\sim \text{Re}[\psi (t)]\sim {\text{e}}^{-\frac{{t}^{2}}{4{\sigma}^{2}}}\mathrm{cos}({\omega}_{0}t)\phantom{\rule{thickmathspace}{0ex}}?$