Question

An iron-ring of mean diameter 19.1 cm and having a cross-sectional area of 4 sq. cm is required to produce a flux of 0.44 mWb. Find the coil-mmf required. 

If a saw-cut 1 mm wide is made in the ring, how many extra amp-turns are required to maintain the same flux ? 

B - μ_r curve is as follows :

B (Wb/m2)	0.8	1.0	1.2	1.4
μr	2300	2000	1600	1100

Answer 1

For a mean-diameter of 19.1 cm, Length of mean path, l_m = π × 0.191 = 0.6 m

Cross-sectional area = 4 sq.cm = 4 × 10⁻⁴m² 

Flux, φ = 0.44 mWb = 0.44 × 10⁻³ Wb 

Flux density, B = 0.44 × 10⁻³/(4 × 10⁻⁴) = 1.1 Wb/m² 

For this flux density, μ_r = 1800, by simple interpolation.

H = B/(μ_oμ_r) = 1.1 × 10⁷ /(4π × 1800) = 486.5 amp-turns/m.

m.m.f required = H × l m = 486.5 × 0.60 = 292

A saw-cut of 1 mm, will need extra mmf.

H_g = B_g/μ_o = 1.1 × 10⁷/(4π) = 875796

AT_g = H_g × l_g = 875796 × 1.0 × 10⁻³ = 876

Thus, additional mmf required due to air-gap = 876 amp-turns