I need to calculate the magnetic field strength at 30 cm distance of a coil with a diameter of 17 cm. Directly at the coil surface I measure 100 Gauss.

Please advise.

4 answers

I do not understand what you mean by the "surface" of a coil. Is it the open circular area pi*D^2/4?

In what direction from the coil is the 30 cm distance measured? Is it along a perpendicular to the coil through the center?

There is a formula for the magnetic field of a circular coil at various distances

B(z) = (muo*I/2)*a^2/(z^2+a^2)^(3/2)

The ratio B(z)/B(z=0) is
[a^2/(z^2+a^2)^(3/2)]/(1/a)
= a^3/(z^2+a^2)^(3/2)
= (8.5)^3/(30^2 + 8.5^2)^3/2
= 61.41/3.032*10^4 = 2.03*10^-3

Multiply that by the B field at the center of the coil for your answer.

I get 0.203 Gauss. You are in the far field z>a region where B falls with the cube of distance, z.

If these are measurements you made, what value did you get at z = 30 cm, on axis?
This is a circular coil and indeed the open area is 227 cm
No not perpendicular but parallel from the coil (measured 30 cm away from above the coil).
My Gauss meter is not sensitive enough to pick up the signal at this distance.
Can you somewhat elaborate more on the various terms you use in your formula?
I obtained the formula from the book "Foundations of Electromagnetic Theory" by Reitz and Milford. You need to use the Biot-Savart law. That book only does the case of the field normal to the coil, along the axis. a is the coil radius and z is the dstance away along the axis. "muo" is mu-zero, the permeability of free space.

I suspect that you will also have a 1/z^3 falloff at large distances, and proportionality with muo*I, but do not know what the exact formula.

Googling "magnetic dipole field" may help find you the answer.
Great, thanks a lot!