Circular ring with radial air gap, DC current

The ring is a toroid with a circular cross-section. Its mean diameter and section size follow from the outer and inner diameters.

Given: outer diameter $= 5.5$ cm, inner diameter $= 3.5$ cm, $\mu_r = 2000$, $l_g = 2$ mm, $N = 1000$, $\Phi = 0.3$ mWb.

Geometry

The radial thickness is $5.5 - 3.5 = 2$ cm, which is the diameter of the circular section. So

$$\begin{aligned} D_{mean} &= \frac{5.5 + 3.5}{2} = 4.5\ \text{cm} \\ A &= \frac{\pi}{4}(0.02)^2 = 3.142\times10^{-4}\ \text{m}^2 \\ l_i &= \pi\times0.045 - 0.002 = 0.13937\ \text{m} \end{aligned}$$

Flux density

$$B = \frac{\Phi}{A} = \frac{0.3\times10^{-3}}{3.142\times10^{-4}} = 0.9549\ \text{T}$$

Mmf

$$\begin{aligned} F_i &= \frac{B l_i}{\mu_0\mu_r} = \frac{0.9549\times0.13937}{4\pi\times10^{-7}\times2000} = 52.95\ \text{AT} \\ F_g &= \frac{B l_g}{\mu_0} = \frac{0.9549\times0.002}{4\pi\times10^{-7}} = 1519.8\ \text{AT} \\ F &= 1572.8\ \text{AT} \end{aligned}$$

Current

$$I = \frac{F}{N} = \frac{1572.8}{1000} = 1.573\ \text{A}$$

Answer: $I \approx 1.57$ A