Looking to power an electromagnet requiring 12 DC volts at 4 Watts, and wondering what size battery I would require to do this for 4 weeks?

My maths so far is, and I am not sure this is correct:

12 V / 4 W = 0.375 Amps
0.375 Amps * 24 * (7 * 4) = 252 AmpHours or 252,000 mAh

Is this correct? The magnet I am looking for can hold 20kg, and the magnet weighs 100g, so would it even be able to carry anything other than just the battery for such a duration?

4 weeks is with added duration to what it would usually take, but I have no idea where, if my maths is correct, I would even be able to find a battery of this size that isn't super heavy. Is my math correct? Are there batteries this size under 15kg?

  • $\begingroup$ How did you divide 12 by 4 and get 0.375? You should be dividing 4 by 12 (i.e., watts by volts, not volts by watts) but that doesn’t give 0.375 either, although it gives something close. $\endgroup$ – G. Smith May 28 at 5:24
  • $\begingroup$ With correct maths it's 224Ah, but that is a large battery. I doubt that you'll find one that weighs less than 50 kg. Do a search for "12V 250Ah battery" and check the specs. $\endgroup$ – hdhondt May 28 at 5:29
  • $\begingroup$ @G.Smith Thank you. I am not sure how I got 12/4 = 0.375 but I messed up P=IV. Classic. With such a large size this isn't looking reasonable. I will have to look at electromagnets with more pull. $\endgroup$ – Jack May 28 at 6:07
  • $\begingroup$ @hdhondt Thanks for this. Having a look now, but I will likely have to move to another magnet, to see if at any size it is reasonable. Thanks for the help! $\endgroup$ – Jack May 28 at 6:08
  • $\begingroup$ @hdhondt Looking currently at a 12V with 7.8W that has 1000lb of force (or 450kg). This might be possible because my calculations at 7.8/12 = 0.65 which would give 436.8Ah. They are easily within the measurements, so it is theoretically possible. Now just a matter of maximising cost for what I am hoping to do! $\endgroup$ – Jack May 28 at 6:14

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

  • $\begingroup$ Welcome, Craig! This site supports the use of Latex for writing equations. You can find some info here and at the pages it links to. The main tutorial is here. $\endgroup$ – PM 2Ring Jun 14 at 7:05

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