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
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