Mild steel ring: permeability and inductance

Given: mean circumference $= 30$ cm, $A = 6$ cm$^2 = 6\times10^{-4}$ m$^2$, $N = 500$, $l_g = 1$ mm, $I = 4$ A, $B_g = 1$ T.

(i) Relative permeability of the steel

Total mmf: $F = NI = 500\times4 = 2000$ AT.

Mmf used by the gap:

$$F_g = \frac{B l_g}{\mu_0} = \frac{1\times0.001}{4\pi\times10^{-7}} = 795.8\ \text{AT}$$

Mmf left for the iron: $F_i = 2000 - 795.8 = 1204.2$ AT. The iron length is $l_i = 0.30 - 0.001 = 0.299$ m.

$$\begin{aligned} H_i &= \frac{1204.2}{0.299} = 4027.5\ \text{AT/m} \\ \mu_r &= \frac{B}{\mu_0H_i} = \frac{1}{4\pi\times10^{-7}\times4027.5} = 197.6 \end{aligned}$$

(ii) Inductance

$$\Phi = BA = 1\times6\times10^{-4} = 6\times10^{-4}\ \text{Wb}$$ $$L = \frac{N\Phi}{I} = \frac{500\times6\times10^{-4}}{4} = 0.075\ \text{H}$$

Answer: $\mu_r \approx 198$; $L = 75$ mH