Consider a toroidal solenoid wound on a non-magnetic core, such as shown in Fig. 20. If the flux density is measured on the centre line of the toroid, the relationship between B and H is given. b-y the straight line OA in Fig. 21. Ifnow the space within an unmagnetised ferromagnetic material, the well known magnetisation curve OBCDE, is obtained. The magnetisation has many names being referred to as :
B-H curve, the magnetic saturation curve, the virgin curve or simply the saturation curve.
The difference in flux between the saturation curve and the air line, OA at any magnetising force,is due to the contribution of the magnetic material. This flux is known as the intrinsic flux and gives a true measure of magnetic properties than does the total flux, especially at very high m.m.fs, The point at which the intrinsic flux density curve becomes horizontal gives the intrinsic saturation.
The magnetisation curves can be determined by the following methods provided the materials are in the. form of a ring:
(i) By means of a ballistic galvanometer
(ii) By means of a fluxmeter.
The graphs in Fig. 22 show the relationship between the flux density (B) and the magnetic field obtained for different qualities ofiron. The data for mild steel, wrought iron and sheet steel are so similar that they can be represented by a common graph. Stalloy is an alloy of iron and silicon commonly used in the construction of transformers and A.C. machines.
Example 6. An air cored toroidal coil has 1500 turns and carries a current of 0.2 A The length of the magnetic circuit is 300 mm and cross-sectional area of the coil is 800 mm2. Calculate:
(i) Magnetic field strength,
(ii) Flux density, and
(iii) Total flux within the coil .
Solution. Number of turns, N = 1500
Current, I = 0.2 A
Length of the magnetic circuit, 1 = 300 mm = 0.3 m
Cross-sectional area of the coil, A = 800 mm2 = 800 × 10-6 m2
I. Magnetic field strength, H :
III.Total Flux within the coil, ɸ:
ɸ = B × A
= 1.256 × 10-3 × 800 × 10-6 = 1.005 × 10-6 Wb. (Ans.)
Example 7. A coil of 300 turns and of resistance 10 Ω is wound uniformly over a steel ring of mean circumference 30 cm and cross-sectional area 9 cm2. It is connected to a supply at 20 V.D.C. If rollll’ III’rlllc”ability of the ring is 1500, find:
(i) The magnetising force,
(ii) The reluctance,
(iii) The m.m.f., and
(iv) The flux.
Solution, Given : N = 300 ; R = 10
Example 8. A magnetic circuit consists of an iron ring of mean circumference 80 cm with cross-sectional area of 12 cm2 throughout. A current of 1 A in the magnetising coil of 200 turns produces a total flux of 1.2 m Wb in the iron. Calculate:
I. The flux density in the iron,
II. The absolute and relative permeability of iron,
III. The reluctance of the circuit.
Solution. Given: l = 80 cm = 0.8 m ; a = 12 cm2 = 12 × 10-4 m2 ;
1= 1 A ; N = 200; ɸ = 1.2 mWb == 0.0012 Wb,
Example 9. An iron ring 200 mm mean diameter is made of 30 mm round iron of permeability 900, has an air gap 10 mm wide. It has 800 turns. If the current flowing through this winding is 6.8 amperes, determine:
i. The magnetomotiue force,
ii. The total reluctance of the circuit,
iii. The flux in the ring, and
iv. The flux density in the ring.
Solution. Length of air gap, 19 = 10 mm = 0.01 m
Mean dia. of iron ring = 200 mm = 0.2 m
Mean length of iron path,
li = ( π 0.2 – 0.01)
= 0.618 m
= 7.068 × 10-4 m2
Number of turns, N = 800
Permeability of iron, µr = 900
Current through the winding, I = 68 A.
(i) Magnetomotive force, m.m.f. :
m.m.f. ‘” NT ampere-turns
= 800 × 6.8 = 5440 AT. (Ans.)
Example 10. An iron of 15 cm diameter and 10 cm2 cross-section with a saw cut 2 mm wide is wound with 200 turns of wire. The flux density is 1 Wb/m2. The relative permeability of iron is 600. Find the exciting current. Neglect leakage.
Example 11. An iron ring of mean length 1 meter has an air gap of 2 mm and a winding 400 turns. If the permeability of the iron is 400 when a current of 2.5 amperes flow through flow through the coil, find the flux density.
Solution. Mean length of iron ring (or iron path), li = 1 m (neglecting air gap length)
Length of air gap. lg = 2 mm = 0.002 m
Number of turns, N = 400
Permeability of iron, µr = 400
Current through the coil, I = 2.5 A
Flux density, B
Total AT required = AT required for iron ring (ATi)+AT required for air gap (ATg)
i.e. AT = ATi + ATg
= Hili + Hglg
Example 12. An iron ring of mean circumference of 900 mm and cross-sectional area: of 250 mm2 has flux of 500 µWb. Calculate the m.m.f. acting on the iron ring. An air gap of 2 mm is now cut in the ring. Find the flux in the ring if the m.m.f. remains same. A relative permeability of 1200 may be assumed.
Solution. Given: li = 900 – 2 = 898 = 0.898 m ; A = 250 mm2 = 250 × 10-6 m2
ɸ = 500 µ Wb = 500 x 10-6 Wb; 19 = 2 mm = 0.002 m; µr = 1200.
(i) Before the air gap is cut:
Solution. Given: ɸ = 8 × 10-4 Wb
A = 1000 mm2 = 1000 × 10-6 = 0.001 m3
Total length of air gap, lg = 2 × 0.2 = 0.4 mm = 0.0004 m
(µr)c.s. = 170
Example 14. A magnetic ring has a mean circumference of 1.8 m and is of 0.012 m2 cross-section, and is wound with 210 turns. A saw cut of 4.8 mm width is made in the ring. Calculate the magnetising required to produce a flux of 0.8 milliweber in the air gap.
Assume leakage factor as 1.25 and permeability of iron as 400.
Solution. Mean circumference of the ring = 1.8 m
Length of air gap, lg = 4.8 mm = 0.0048 m
Length of Iron path = 1.8 - lg = 1.8 – 0.0048 = 1.795 m
Area of cross-section, A = 0.012 m2
Flux in the air gap, ɸg = 0.8 m Wb
Number of turns, N = 210
Permeability of Iron, µr, = 400
Leakage factor, = 1.25
Magnetising current I:
Total ampere-turns required (AT) = AT required for iron ring + AT required for air gap
i.e., AT = ATi + ATg
Example 15. A rectangular core shown in Fig. 25 has a mean length of magnetic path of 120 cm, cross-section of (3 cm × 3 cm), relative permeability of 1400 and an air gap of 4 mm cut in the core. The three coils carried by the core have number of turns N1 = 350, N2 = 500 and N3 = 500; and the respective currents are 1.5 A, 3.5 A and 2.5 A. The directions of the currents are as shown. Find the flux in the air gap.
Solution. Given: li = 1.2 – 0.004 = 1.196 m ;
µr = 1400 ;
lg = 4 mm = 0.004 m ;
A = 3 × 3 = 9 cm2 = 9 × 10-4 m2
N1 = 350 ;
N2 = 500 ;
N3 = 500 ;
I1 = 1.5 A ;
I2 = 3.5 A ;
I3 = 2.5 A.
Flux in the air gap, ɸg:
By applying the right-hand thumb rule, it is found that fluxes produced by the currents I1 and I2 are directed in the clockwise direction through the iron core whereas that produced by current I3 is directed in the anti-clockwise direction through the core.
Example 16. A cast steel magnetic structure made for a bar of section 2 cm × 2 cm is shown in Fig. 29. Determine the current that the 600 turn magnetising coil on the left limb should carry so that a flux of 2 mWb is produced in the right limb. Take u, = 600 and neglect leakage.
Solution. Cross-section of the bar = 2 cm × 2 cm
Number of turns, N = 600
Flux, ɸ = 2 mWb = 0.002 Wb
Relative permeability, µr = 600
Since paths C and D are in parallel with each other w.r.t. path E, the m.m.f. (NI) across the two is the same