# Anti-Helmholtz coil configuration

1. Jun 20, 2012

### Niles

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.

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2. Jun 21, 2012

### Staff: Mentor

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).

3. Jun 23, 2012

### Niles

Thanks, that was kind of you.

4. Jun 23, 2012

### marcusl

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/emagnettest/offaxis/?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.