A Current-Carrying Coil as a Magnetic Dipole — Question 6

Question 6

The Helmholtz spacing removes the quadratic axial field variation

Two identical thin N-turn coils of radius R lie at z=+s and z=-s. Equal currents I produce moments along +z. Write $B(z)=C[f(z-s)+f(z+s)]$, where $C=\mu_0NIR^2/2$ and $f(u)=(R^2+u^2)^{-3/2}$. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the coil separation 2s that makes the second derivative of B vanish at the midpoint.

2. Find the midpoint field at that spacing and identify the first possible order of axial field variation near the midpoint.

Question 6 - Solution

The Helmholtz spacing removes the quadratic axial field variation

(a) Method and calculation

$$f\prime\prime(u)=\frac{3(4u^2-R^2)}{(R^2+u^2)^{7/2}},\quad B\prime\prime(0)=2Cf\prime\prime(s).$$$$B\prime\prime(0)=0\Rightarrow4s^2=R^2\Rightarrow\boxed{2s=R}.$$

(b) Interpretation and further calculation

$$B(0)=\frac{\mu_0NIR^2}{(R^2+R^2/4)^{3/2}}=\boxed{\left(\frac45\right)^{3/2}\frac{\mu_0NI}{R}}.$$$$B(z)=B(-z),\quad B\prime\prime(0)=0\Rightarrow\boxed{B(z)-B(0)=O(z^4)}.$$

Physical check

Equal same-sense coils automatically remove odd axial powers; the Helmholtz spacing additionally removes the quadratic term.