Powering an electromagnet for a long duration

I have just built a powerful electromagnet. What I found is most electromagnet are amp hogs therefore chewing the current out of the battery. What I have since done and ended up doing winding ,8 mm wire around the coil 14000 windings there for given greater resistance. And made a mutiplyer with capacitors, and increased the voltage so I used half the current and saved myself many AH I am still working in the EMF and going to try some how re use it to charge my battery. It’s still an on going project. Good luck . One method of calculating the force produced by a magnetic field involves an understanding of the way in which the energy represented by the field changes. To derive an expression for the field energy we'll look at the behaviour of the field within a simple toroidal inductor. We equate the field energy to the electrical energy needed to establish the coil current.

When the coil current increases so does the magnetic field strength, H. That, in turn, leads to an increase in magnetic flux, greek letter phi. The increase in flux induces a voltage in the coil. It's the power needed to push the current into the coil against this voltage which we now calculate.

Ideal toroid inductor We choose a toroid because over its cross-sectional area, A, the flux density should be approximately uniform (particularly if the core radius is large compared with it's cross section). We let the flux path length around the core be equal to Lf and the cross-sectional area be equal to Ax. We assume that the core is initially unmagnetized and that the electrical energy (W) supplied to the coil will all be converted to magnetic field energy in the core (we ignore eddy currents).

W = time integral v×i dt joules Faraday's law gives the voltage as

v = N×d greek letter phi/dt volts

Substituting -

W = integral N(dgreek letter phi/dt)i dt

W = flux integral N×i dgreek letter phi

Now, N×i = Fm and H = Fm/Lf so N×i = H×Lf. Substituting:

W = flux integral H×Lf dgreek letter phi

Also, from the definition of flux density greek letter phi = Ax× B so d greek letter phi = Ax×dB. Substituting:

W = dens integral H×Lf×AxdB joules

This gives the total energy in the core. If we wish to find the energy density then we divide by the volume of the core material:

Wd= (dens integral H×Lf×Ax dB)/(Lf×Ax)

Wd = dens integral H d B joules m-3 Equation EFH If the magnetization curve is linear (that is we pretend B against H is a straight line, not a curve) then there is a further simplification. Substituting H = B/μ

Wd= dens integral B / µ d B

Wd = B2/(2μ) joules m-3 Equation EFB Compare this result with the better known formula for the energy stored by a given inductance, L:

WL = L×I2/2 joules

A 12 volt car battery might supply about 50 amp.hours of power. They will probably last longer if you don't drain them below half of that. I would suggest that you use two. Use one while you recharge the other. To sustain the field, you must switch the replacement in before taking the other out, but do this quickly. Its not good to connect two batteries in parallel. You might include heavy duty diodes to keep one battery from discharging into the other. (On my boat I use diodes to charge two batteries from one solar panel.) (By the way, 4 W at 12 V draws 0.333 amps.) An alternative would be to use a low voltage DC power supply. A little black box of the type used to power a lap-top computer can easily supply one third of an amp.

Is this correct?

Yes

Is my math correct? Are there batteries this size under 15kg?

You need 2.7 kWh. That's about 10 kilos of high energy density LiIon cells.

So, what you need is a bistable system, that only consumes energy when changing state. You could use a mechanical latch or other mechanism, that uses no energy to hold the weight, and have a solenoid or other actuator flip the latch and release the load.

If you want magnets, then you can use a permanent magnet with a solenoid around it creating a magnetic field opposed to the permanent magnet. When the solenoid is energized, it cancels the permanent magnet's field, and it drops the weight. But the magnet can hold the weight indefinitely. Or you could use a small actuator to pull the permanent magnet away from the object it's stuck on.

There are plenty of other solutions in the same vein that should be much simpler and cheaper than a permanently energized electromagnet.