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 **mm ^{2}. 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 mm^{2} = 800 × 10^{-6} m^{2}

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 *cm ^{2}. 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.*

**(VTU, 1999)**

**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 *cm ^{2} 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 cm^{2} = 12 × 10^{-4} m^{2} ;

*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,

*l _{i} *= ( π 0.2 – 0.01)

= 0.618 m

= 7.068 × 10^{-4} m^{2}

Number of turns, N = 800

Permeability of iron, µ_{r} = 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 cm ^{2} cross-section with a saw cut *

*2*

*mm wide is wound with 200 turns of wire. The flux density is*

*1*

*Wb/m*

^{2}. The relative permeability of iron*is 600. Find the exciting*

*current.*

*Neglect leakage.*

**(K.U.) **

**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.*

**(Indore University)**

**Solution**. Mean length of iron ring (or iron path), l_{i} = 1 m (neglecting air gap length)

Length of air gap. l_{g} = 2 mm = 0.002 m

Number of turns, N = 400

Permeability of iron, µ_{r} = 400

Current through the coil, *I *= 2.5 A

**Flux density, B **

Total AT required = AT required for iron ring (AT_{i})+AT required for air gap (AT_{g})

i.e. AT = AT_{i} + AT_{g}

= H_{i}l_{i} + H_{g}l_{g}

**Example 12**. *An iron ring of mean circumference of 900 mm and cross-sectional **area: **of 250 mm ^{2} 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.*

**(Roorkee University) **

**Solution**. *Given: l _{i} *= 900 – 2 = 898 = 0.898 m ;

*A*= 250 mm

^{2}= 250 × 10

^{-6}m

^{2 }

ɸ = 500 µ Wb = 500 x 10^{-6} Wb; 19 = 2 mm = 0.002 m; µ_{r} = 1200.

**(i) ****Before the air gap is cut:**

**Example 13**.

*A ring his a diameter of 24 cm and a cross-sectional area of*

*1000*

*mm*

^{2}. The ring*is made up*

*of semicircular sections of cast iron and cast steel with each joint having a reluctant*

*equal to an air*

*gap of 0.2 mm. Find the ampere-turns required to produce a flux of*8 ×

*10-*

^{4}Wb.*The*relative permeability

*of caststeel and cast iron are 900 and 170 respectively. Neglect*

*fringing*

*and leakage effects.*

**(Gujarat University)**

**Solution**. Given: ɸ = 8 × 10^{-4} Wb

A = 1000 mm^{2} = 1000 × 10^{-6} = 0.001 m^{3}

Total length of air gap, l_{g} = 2 × 0.2 = 0.4 mm = 0.0004 m

(µ_{r})_{c.s}. = 170

**Example 14***. A magnetic ring has a mean circumference of *1.8 *m and is of 0.012 **m ^{2} 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, l_{g} = 4.8 mm = 0.0048 m

Length of Iron path = 1.8 *– l _{g} *= 1.8 – 0.0048 = 1.795 m

Area of cross-section, *A *= 0.012 m^{2 }

Flux in the air gap, ɸ_{g} = 0.8 m Wb

Number of turns, *N *= 210

Permeability of Iron, µ_{r}, = 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 = AT_{i} + AT_{g}

**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 N _{1} *=

*350, N*=

_{2}*500 and N*=

_{3}*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.*

**(Pune University)**

**Solution. ***Given: l _{i} *= 1.2 – 0.004 = 1.196 m ;

µ_{r} = 1400 ;

l_{g} = 4 mm = 0.004 m ;

*A *= 3 × 3 = 9 cm^{2} = 9 × 10^{-4} m^{2 }

N_{1} = 350 ;

N_{2} = 500 ;

N_{3} = 500 ;

I_{1} = 1.5 A ;

I_{2} = 3.5 A ;

I_{3} = 2.5 A.

**Flux in the air gap, **ɸ_{g}:

By applying the right-hand thumb rule, it is found that fluxes produced by the currents I_{1} and I_{2} are directed in the clockwise direction through the iron core whereas that produced by current I_{3} 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.*

**(Allahabad University) **

**Solution. **Cross-section of the bar = 2 cm × 2 cm

Number of turns, N = 600

Flux, ɸ = 2 mWb = 0.002 Wb

Relative permeability, µ_{r} = 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

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