Question

A magnetic circuit made of mild steel is arranged as shown in Fig. . The central limb is wound with 500 turns and has a cross-sectional area of 800 mm² . Each of the outer limbs has a cross-sectional area of 500 mm² . The air-gap has a length of 1 mm. Calculate the current rquired to set up a flux of 1.3 mWb in the central limb assuming no magnetic leakage and fringing. Mild steel required 3800 AT/m to produce flux density of 1.625 T and 850 AT/m to produce flux density of 1.3 T

Answer 1

Flux in the central limb is = 1.3 mWb = 1.3 × 10⁻³ Wb 

Cross section A = 800 mm² =800 × 10⁻⁶ m² 

∴ B = Φ/A = 1.3 × 10⁻⁶ /800 × 10⁻⁶ 

= 1.625 T

Corresponding value of H for this flux density is given as 3800 AT/m.

Since the length of the central limb is 120 mm. m.m.f. required is 

= H× l = 3800 × (120 × 10⁻³ ) = 456 AT/m. 

Air-gap

Flux density in the air-gap is the same as that in the central limb. 

H = B/μ₀ = 1.625/4π × 10⁻⁷ = 0.1293 × 10⁻⁷ AT/m 

Length of the air-gap = 1 mm = 10⁻³ m 

m.m.f. reqd. for the air-gap = H × l = 0.1293 × 10⁷ × 10⁻³ 

= 1293 AT.

The flux of the central limb divides equally at point A in figure along the two parallel path ABCD and AFED. We may consider either path, say ABCD and calculate the m.m.f. required for it. The same m.m.f. will also send the flux through the other parallel path AFED. 

Flux through ABCD = 1.3 × 10⁻³ /2 = 0.65 × 10⁻³ Wb 

Flux density B = 0.65 × 10⁻³ /500 × 10⁻⁶ = 1.3 T 

The corresponding value of H for this value of B is given at 850 AT/m. 

∴ m.m.f. reqd. for path ABCD = H × l = 850 × (300 × 10⁻³ ) 

= 255 AT 

As said above, this, m.m.f. will also send the flux in the parallel path AFED. 

Total m.m.f. reqd. = 456 + 1293 + 255 = 2004 AT 

Since the number of turns is 500, I = 2004/500 = 4A