# Calculating magnetic field strength for a very small electromagnet

I am trying to calculate the magnetic field (in tesla/gauss) of an electromagnet that is very small and has very few windings. For example 12 windings over 0.003 meters. I know this is not going to produce a very strong field, but I would like to pulse a strong current through the coil very briefly to make it stronger. I have found a number of sources listing the formula for the calculation of magnetic field strength -

$$B = \mu \rho I= μ μ_0\frac{N}{L}I$$ where $$N$$ is the number of turns and $$L$$ the length of the core, e.g. as listed here.

My question is

• can this formula be applied to my electromagnet design?

• Does the size and low number of windings on my electromagnet mean this formula is not valid?

• Is there any other way I can calculate/estimate the magnetic field?

• Why would you think the formula would not work? Mar 6, 2015 at 17:18
• Another person told me it was not valid for small electromagnets after I compared the results I got from that formula, and one for a U shaped electromagnet that I found on en.wikipedia.org/wiki/Electromagnet which was turns*current = B((Lcore/μ)+(Lgap/μ0)) The problem was, I got a smaller result in Teslas for the U shaped electromagnet than the straight core electromagnet. I know something is wrong with this, but I am not sure what. Mar 6, 2015 at 17:42
• The field depends on the shape. You can't apply the formula for a straight magnet and expect it to give the right result for a U shaped magnet, or vice versa! Mar 6, 2015 at 18:10
• The formula I was using for the U shaped magnet was taken from en.wikipedia.org/wiki/… Where it said "For an electromagnet with a single magnetic circuit, of which length Lcore of the magnetic field path is in the core material and length Lgap is in air gaps" the formula is turns*current = B((Lcore/μ)+(Lgap/μ0)) I am not sure if this is the correct formula for a U shaped electromagnet or not. Does anyone know the formula for the U shaped magnet? Am I using the wrong one here? Mar 6, 2015 at 18:22
• This formula is technically for an infinite solenoid. However should still give good results. Apr 14, 2022 at 23:04

For a normal multipole magnet, the field for a $$2n$$ multipole is given by: $$B_y+iB_x=n\frac{\mu_0NI}{r_0}\left(\frac{x+iy}{r_0}\right)^{n-1}$$ where $$r_0$$ means the distance between the pole and the origin of the coordinate system. $$NI$$ means each pole consists of $$N$$ turns of wire carrying current $$I$$.
So, $$n=1$$ is a dipole with gap $$2r_0$$; and $$n=2$$ is a quadrupole with field gradient $$\frac{\partial B_y}{\partial x}=\frac{\partial B_x}{\partial y}=\frac{2\mu_0NI}{r_0^2}.$$