Solenoids and Toroids — Question 7

Question 7

The flux through a toroid cross section uses a logarithmic average field

An ideal air-core toroid has rectangular cross section of height h and radii a and b. Its field is $B(r)=\mu_0NI/(2\pi r)$. Define flux through a radial cross section by $\Phi=\int_a^bB(r)h\,dr$. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Find the flux and the cross-section average magnetic field.

2. Find the effective radius whose local field equals that average; evaluate it for a=.0800 m and b=.120 m.

Question 7 - Solution

The flux through a toroid cross section uses a logarithmic average field

(a) Method and calculation

$$\Phi=\frac{\mu_0NIh}{2\pi}\int_a^b\frac{dr}{r}=\boxed{\frac{\mu_0NIh}{2\pi}\ln(b/a)}.$$$$\langle B\rangle_A=\frac{\Phi}{h(b-a)}=\frac{\mu_0NI}{2\pi}\frac{\ln(b/a)}{b-a}.$$

(b) Interpretation and further calculation

$$B(r_{\mathrm{eff}})=\langle B\rangle_A\Rightarrow\boxed{r_{\mathrm{eff}}=\frac{b-a}{\ln(b/a)}}.$$$$r_{\mathrm{eff}}=\frac{.0400}{\ln1.5}=\boxed{.0986521\,\mathrm m}<\frac{a+b}{2}.$$

Physical check

The average exceeds the field at the arithmetic mean radius because inverse radius is a convex function. Flux is defined here without invoking induction.