Solenoids and Toroids — Question 2

Question 2

Integrating circular turns gives different finite-solenoid fields at the center and end

A thin solenoid of radius R and length ell is centered at z=0, with uniform turn density n and current I. Each turn contributes $dB=\mu_0nIR^2\,dz\prime/[2(R^2+(z-z\prime)^2)^{3/2}]$ along +z. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Integrate to obtain the on-axis field B(z).

2. Find the center and end fields for ell=4R, expressed as multiples of $\mu_0nI$, and find their ratio.

Question 2 - Solution

Integrating circular turns gives different finite-solenoid fields at the center and end

(a) Method and calculation

$$\int\frac{R^2\,du}{(R^2+u^2)^{3/2}}=\frac{u}{\sqrt{R^2+u^2}},\quad u=z-z\prime.$$$$\boxed{B(z)=\frac{\mu_0nI}{2}\left[\frac{z+\ell/2}{\sqrt{R^2+(z+\ell/2)^2}}-\frac{z-\ell/2}{\sqrt{R^2+(z-\ell/2)^2}}\right]}.$$

(b) Interpretation and further calculation

$$B(0)=\mu_0nI\frac{\ell/2}{\sqrt{R^2+\ell^2/4}}=\frac2{\sqrt5}\mu_0nI.$$$$B(\ell/2)=\frac2{\sqrt{17}}\mu_0nI,\quad\boxed{B_{\mathrm{end}}/B_{\mathrm{center}}=\sqrt{5/17}=.542326}.$$

Physical check

Only in the very long limit is the end field one-half of the center field; finite geometry changes both values.