# Anti-Helmholtz coil configuration

by Niles
Tags: antihelmholtz, coil, configuration
P: 1,863
Hi

Say I have to coils in the anti-Helmholtz configuration as in the attached picture. It is pretty easy to find the field along the z-direction, as this is introduced in introductory EM. However say that I would like to know the field along the x-direction. This I don't know how to find.

What I *do* know is that the Maxwell Equations (div B = 0) tell me that
$$\frac{dB}{dx} = \frac{dB}{dy} = -\frac{1}{2}\frac{dB}{dz}$$
But does this imply that the field along x, B(x), is simply -B(z), the negated B-field along the z-direction?

Best,
Niles.
Attached Images
 coils.bmp (248.1 KB, 31 views)
 Mentor P: 11,837 div B = 0 is equal to $\frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} = - \frac{\partial B_z}{\partial z}$ Using symmetry, x and y must be the same, therefore $\frac{\partial B_x}{\partial x}= - \frac{1}{2}\frac{\partial B_z}{\partial z}$ This does not give you any magnetic field! It is just the derivative of the field at some specific point - probably along the central axis. Looking at the (x,y)-plane right in the middle of the coils, you have a field going radially inwards/outwards (depending on the orientation).
 P: 1,863 Thanks, that was kind of you.
 Sci Advisor PF Gold P: 2,080 Anti-Helmholtz coil configuration The off-axis field from a current loop is written as an expansion in elliptic integrals (or sometimes other functions, depending on the coordinate system you choose). Here is a site that came up in a Google search for off-axis field from a loop: http://www.netdenizen.com/emagnettes...s/?offaxisloop but there are many others. The field from a pair of coils is then a sum of the fields from the individual loops. More extensive derivations/explanations are found in advanced E&M texts like those by Jackson, Smythe, and Stratton.

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