# 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 an number of sources listing the formula for the calculation of magnetic field strength -

$$B = permeability * \rho * I= (μ * μ0)*(number of turns/core length)*I$$

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 magnetic field?

• Why would you think the formula would not work? – Jiminion Mar 6 '15 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. – EddieP Mar 6 '15 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! – Floris Mar 6 '15 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? – EddieP Mar 6 '15 at 18:22
• For a pulse, there is also the problem of self-induction, which may limit the maximal current. – user137289 Dec 24 '17 at 19:46

For a normal multipole magnet, the field for a $2n$ multipole is given by: $$B_y+iB_x=n\frac{\mu_0NI}{r_0}(\frac{x+iy}{r_0})^{n-1}$$ 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 $\frac{\partial B_y}{\partial x}=\frac{\partial B_x}{\partial y}=\frac{2\mu_0NI}{r_0^2}$