Current for wrought iron ring with and without air gap

The bar length becomes the mean length of the ring. Area is that of a 2 cm diameter circle.

$$A = \frac{\pi d^2}{4} = \frac{\pi (0.02)^2}{4} = 3.1416\times10^{-4}\ \text{m}^2$$ $$B = \frac{\phi}{A} = \frac{0.05\times10^{-3}}{3.1416\times10^{-4}} = 0.1592\ \text{T}$$

a) No air gap ($l_i = 0.30$ m)

$$\begin{aligned} H_i &= \frac{B}{\mu_0\mu_r} = \frac{0.1592}{4\pi\times10^{-7}\times4000} = 31.66\ \text{AT/m} \\ NI &= H_i l_i = 31.66\times0.30 = 9.50\ \text{AT} \\ I &= \frac{9.50}{600} = 0.0158\ \text{A} \end{aligned}$$

Answer (a): I ≈ 15.8 mA.

b) With a 4 mm air gap

The gap is cut from the same bar, so iron length $l_i = 0.30 - 0.004 = 0.296$ m.

$$\begin{aligned} H_g &= \frac{B}{\mu_0} = \frac{0.1592}{4\pi\times10^{-7}} = 126651\ \text{AT/m} \\ AT_g &= 126651\times0.004 = 506.6\ \text{AT} \\ AT_i &= 31.66\times0.296 = 9.37\ \text{AT} \\ NI &= 506.6 + 9.37 = 516.0\ \text{AT} \\ I &= \frac{516.0}{600} = 0.860\ \text{A} \end{aligned}$$

Answer (b): I ≈ 0.86 A.

A gap of only 4 mm raises the required current about 54 times, because the reluctance of air is 4000 times that of the same length of iron.