Question 1.26: A coil of 800 turns is wound on a closed mild steel core hav......
A coil of 800 turns is wound on a closed mild steel core having a length 600 mm and cross-sectional area 500 mm². Determine the current required to establish a flux of 0.8 mWb in the core.
Step-by-Step
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
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