Helmholtz Coils

A system for producing a small region of space with nearly uniform magnetic field, called Helmholtz Coils, consists of two identical coils of radius R on the same (z) axis as shown here:  

							
The z axis is the symmetry axis, P is the point of interest (we take it to be the origin ({0,0,0})); so the centers of the two coils are at {0,0, -d/2} and {0,0,+d/2}.  [Clearly, we are taking nudges from Helmholtz to derive a useful arrangement for creating a desired B at P.]  We are interested in B at P but realize that for small changes in position about P the field will not change dramatically.

We will explore the behavior of B as a function of z only, striving for uniformity.

(a)  Derive B along the z axis for identical currents I running through the two coils 

If you want B to be ‘additive’ for the two coils, and pointing in the +z direction, what directions should I flow in each coil?

THIS FIGURE MAY BE OF USE:
						

ENTER YOUR DERIVATION (OR JUST YOUR RESULTS) HERE:

(b)  It’s desirable to have  (OverVector[B])[z] be uniform over at least a small range of z around the origin.   We can use a Taylor Series for the above expression for  (OverVector[B])[z] to explore this.  First, Use the Series Function to generate a few terms.  Here is a Quicky Tutorial on Series:

M’s Series function can  be used to form a Taylor series of a function about a given point.  For a one D function f[x]:

Series [f,{x,(Subscript[x, 0]),n}]
generates a power series expansion for f about the point x=(Subscript[x, 0]) to order (((x-(Subscript[x, 0])))^n).  

Normal[Series [f,{x,(Subscript[x, 0]),n}]]
generates a power series expansion for f about the point x=(Subscript[x, 0]) to order (((x-(Subscript[x, 0])))^n) and drops the ((O[x])^n) (which cannot be evaluated).

 Consider xo = 0 and n = 6

ClearAll["`*"]
Series[f[x],{x,0,6}]
Normal[Series[f[x],{x,0,6}]]  (*  this drops the ((O[x])^n) Results in the desired Taylor series out to Order 6 *)

Now apply the above to form the Taylor Series for B[z], the two coil B-Field along the z axis.

B[z_] = (1/2) II (R^2) ((1/((((R^2)+((((d/2)-z))^2)))^(3/2)))+(1/((((R^2)+((((d/2)+z))^2)))^(3/2)))) μzero
Series[B[z],{z,0,6}]
BSeries[z_] = Normal[Series[B[z],{z,0,6}]]  (*  this drops the ((O[x])^n)  *)

Examine carefully the ODD TERMS (HINT:  they are missing!).

(c)  Take these odd derivatives separately using M’s D (derivative) function  and show that at z = 0, they disappear  (seems to hold for all odd n).  Reminder:   D[f, {x,n}] gives the multiple derivative   (∂^n)f/∂(x^n)

(*  Input Cell  *)

(d)  The second derivative term in the Taylor Series, ((192 II (R^2) ((d^2)-(R^2)) (z^2) μzero)/((((d^2)+4 (R^2)))^(7/2)))  , looks like an easy one to make vanish if we pick a certain d.  Assign d to it’s obvious value to make the second term in the Taylor Series vanish and re-write the series (BSeries[z]) with this particular d:

(*  Input Cell  *)

(e) Use Manipulate to vary d and examine the changes in the Plots of B[z] (I assumed II, R, and μzero to be all equal to 1)  Take note of the shapes of the curves near d = R.    OK;  I’ll do it for you.  Look how I defined B[z_,d_] := ......    and wrote Manipulate with the Manipulate[d;.......].  This is a trick that makes Manipulate accept the function written for B.

Remove["`*"]
B[z_,d_] := (1/2) II (R^2) ((1/((((R^2)+((((d/2)-z))^2)))^(3/2)))+(1/((((R^2)+((((d/2)+z))^2)))^(3/2)))) μzero
II = 1; R = 1; μzero = 1;
B[z,d]
Manipulate[d;Plot[B[z,d], {z,-1.5R, 1.5 R},PlotRange-> {0,1}],{d, 0, 2 R, Appearance -> "Open"}]

(f)  Sticking with two coils separated by d = R,  reverse the current in ONE of the coils. Derive a new B[z] and again use Manipulate to vary d and examine the changes in the Plots of B[z] (again, I assumed II, R, and μzero to be all equal to 1).