Answer to Question #289450 in Electrical Engineering for Kunal

Question

A mild steel ring having cross sectional area of 5 cm square and mean circumference of 40 cm has coil of 200 turns wound uniformly around it calculate reluctance at required to produce flux of 800ùrwb Assume ùr for mild steel to be 380 at flux density developed in core

Answer 1

*Assuming 800ùrwb is 800 micro Weber.

$$R=\dfrac{IN}{\Phi}=\dfrac{L}{\mu_0\mu_rA}→I=\dfrac{L\Phi}{\mu_0\mu_rAN}=6.7\text{ A},\\\space\\ R=\dfrac{IN}{\Phi}=1675·10^3\text{ H}^{-1}.$$