Understanding Helmholtz Coils in CGS Units

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Homework Help Overview

The discussion revolves around the magnetic field produced by Helmholtz coils configured along the Z axis, specifically in the context of CGS units. The original poster seeks to understand the magnetic field at a point along the Z axis and the behavior of odd derivatives of the magnetic field at the origin.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the magnetic field due to the coils and questions the implications of odd derivatives being zero at the origin. Participants explore the symmetry of the magnetic field and its derivatives, discussing the nature of even and odd functions.

Discussion Status

Participants are actively engaging with the concepts, providing insights into the mathematical properties of the magnetic field and its derivatives. There is a focus on clarifying the relationship between the symmetry of the coils and the behavior of the magnetic field at the origin.

Contextual Notes

There is an indication of confusion regarding the application of CGS units and the interpretation of odd and even functions in the context of magnetic fields. The original poster expresses uncertainty about their understanding and seeks further clarification.

stunner5000pt
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Working in CGS units... hence i don't know if my answer is right or wrong anymore... It doesn't 'look' right

Homework Statement


Two coils are placed at l and -l on the Z axis as in the figure. They both carry a current I in the same direction (anticlockwise). Both have a radius of a. (Setup like Helmholtz coils.)

a. What is the field on the z axis due to the two coils?
b. Why are all the odd derivatives of B at the origin equal to zero?

Homework Equations


Due to the loop of current the field a hieght z above the center of the loop is given by
B = \frac{2\pi Ia^2}{c(a^2 + z^2)^{3/2}}

The Attempt at a Solution



a) The field is given by B = \frac{2\pi Ia^2}{c}\left( \frac{1}{(R^2 + (z + l)^2)^{3/2}} + \frac{1}{(R^2 + (z - l)^2)^{3/2}} \right)

b) I know that the first derivative of the B field is zero because there are no magnetic monopoles. But how does that explain why the octopole moment and so forth are zero? I need some kind of hint so if you could please advise... that would be awesome

Thank you for your help
 
Last edited:
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A lot of B fields have first derivatives.
It is wrong to refer to "octupole moments" for the field between the loops.
Odd derivatives at z=0 are zero, because B is an even function of z due to the symmetry of the coils.
 
Last edited:
ok so even functions have odd first derivative and even second derivative odd third derivative and so on. But how does that explain that the odd derivative is zero at the origin? Aren't evne functions symmetric wrt reflection on the Y axis? Why aren't even functions zero at the origin?
 
Mathematically, the derivative of z^(2n) is zero at z=0 for all positive integer n.
Physically, we know the field at the origin is a minimum (it is strongest at +/-L), and is even about the origin. Therefore the tangent is zero.
 
stunner5000pt said:
ok so even functions have odd first derivative and even second derivative odd third derivative and so on. But how does that explain that the odd derivative is zero at the origin? Aren't evne functions symmetric wrt reflection on the Y axis? Why aren't even functions zero at the origin?
An odd function must equal zero at the origin because f(0)=-f(0) for an odd function.
An even function needn't be zero because f(0)=f(0) for an even function.
For instance 2+z^2=2 at the origin.
 
thanks for the help
 

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