Question 1.26: A coil of 800 turns is wound on a closed mild steel core hav...

Now B = Φ /A = (0.8 × 10^{−3}) / (500 × 10^{−6} ) = 1.6 T From Fig. 1.17, a flux density of 1.6 T will occur in mild steel when H = 3,500 A/m. The current can now be determined by re-arranging H = N I / l as follows:

I=\frac{H\times l}{N} =\frac{3,500\times 0.6}{800} =2.625 A