Series-Parallel Magnetic Circuit:

Figure shows an electromagnet made of cast steel contain a coil of 500 turns wound on the central limb. The cross-sectional area of the outer limbs and central limb are 2.5 cm² and 6 cm² respectively. Find out the current that the coil should carry to generate a flux of 0.6 m Wb in the air-gap. The magnetization curve for cast steel is as specified below :

Figure

Solution

There are two equal parallel paths ACDE and AGE.

Flux density in either parallel path is half of that in the central path as flux divides into two equal parts at point A.

Total m.m.f needed for the whole electromagnet = m.m.f required for path

EF + m.m.f required for the air-gap + m.m.f required for either of the two parallel paths, say path ACDE.

Flux density in the central limb and air-gap = 0.6 × 10/6 × 10^{- 4} = 1 Wb/m² .

Corresponding value of H found from the given data = 900 AT/m.

So, AT for central limb = 900 × 25 ×10^{- 2} = 225 AT

H in the air-gap =  B /μ₀

=0.6 × 10^{- 3} / 4π× 10^{- 7}

= 477.46 AT/m

So, AT required for air-gap = 477.46 × 0.8 ×10^{- 2} = 3.82 AT.

Flux density in the path ACDE is 0.5 Wb/m² and corresponding H is 540 AT/m.

So, AT required for path ACDE = 540 × 100 × 10^{- 2} = 540 AT

Total AT required = 225 + 3.82 + 540 = 768.82

Current required = 768.82 /500 = 1.54 Amp .