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AntiHelmholtz coil configuration 
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#1
Jun2012, 01:26 PM

P: 1,863

Hi
Say I have to coils in the antiHelmholtz configuration as in the attached picture. It is pretty easy to find the field along the zdirection, as this is introduced in introductory EM. However say that I would like to know the field along the xdirection. This I don't know how to find. What I *do* know is that the Maxwell Equations (div B = 0) tell me that [tex] \frac{dB}{dx} = \frac{dB}{dy} = \frac{1}{2}\frac{dB}{dz} [/tex] But does this imply that the field along x, B(x), is simply B(z), the negated Bfield along the zdirection? Best, Niles. 


#2
Jun2112, 11:44 AM

Mentor
P: 11,580

div B = 0 is equal to [itex]\frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} =  \frac{\partial B_z}{\partial z}[/itex]
Using symmetry, x and y must be the same, therefore [itex]\frac{\partial B_x}{\partial x}=  \frac{1}{2}\frac{\partial B_z}{\partial z}[/itex] 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
Jun2312, 06:08 AM

P: 1,863

Thanks, that was kind of you.



#4
Jun2312, 02:02 PM

Sci Advisor
PF Gold
P: 2,059

AntiHelmholtz coil configuration
The offaxis 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 offaxis 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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