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MAGNETISM
AND
ELECTRO-
MAGNETISM
C H A P T E R
Learning Objectives
➣➣➣➣➣ Laws of Magnetic Force ➣➣➣➣➣ Magnetic Field Strength (H) ➣➣➣➣➣ Magnetic Potential ➣➣➣➣➣^ Flux per Unit Pole ➣➣➣➣➣ Flux Density (B) ➣➣➣➣➣ Absolute Parmeability (m) and Relative Permeability (m (^) r ) ➣➣➣➣➣ Intensity of Magnetisation (I) ➣➣➣➣➣^ Susceptibility (K) ➣➣➣➣➣ Relation BetweenB, H, I andK ➣➣➣➣➣ Boundary Conditions ➣➣➣➣➣ Weber and Ewing’s Molecular Theory ➣➣➣➣➣^ Curie Point.^ Force on a^ Current- carrying Conductor Lying in a Magnetic Field ➣➣➣➣➣ Ampere’s Work Law or Ampere’s Circuital Law ➣➣➣➣➣^ Biot-Savart Law ➣➣➣➣➣ Savart Law ➣➣➣➣➣ Force Between two Parallel Conductors ➣➣➣➣➣ Magnitude of Mutual Force ➣➣➣➣➣^ Definition of Ampere ➣➣➣➣➣ Magnetic Circuit ➣➣➣➣➣ Definitions ➣➣➣➣➣ Composite Series Magnetic Circuit ➣➣➣➣➣^ How to Find Ampere-turns? ➣➣➣➣➣ Comparison Between Magnetic and Electric Circuits ➣➣➣➣➣ Parallel Magnetic Circuits ➣➣➣➣➣ Series-Parallel Magnetic Circuits ➣➣➣➣➣^ Leakage Flux and Hopkinson’s Leakage Coefficient ➣➣➣➣➣ Magnetisation Curves ➣➣➣➣➣ Magnetisation curves by Ballistic Galvanometer ➣➣➣➣➣^ Magnetisation Curves by Fluxmete
Designing high speed magnetic levitation trains is one of the many applications of electromagnetism. Electromagnetism defines the relationship between magnetism and electricity
258 Electrical Technology
6.1. Absolute and Relative Permeabilities of a Medium
The phenomena of magnetism and electromagnetism are dependent upon a certain property of the medium called its permeability. Every medium is supposed to possess two permeabilities :
( i ) absolute permeability (μ) and ( ii ) relative permeability (μ r ). For measuring relative permeability, vacuum or free space is chosen as the reference medium. It is allotted an absolute permeability of μ 0 = 4π × 10 −^7 henry/metre. Obviously, relative permeability of vacuum with reference to itself is unity. Hence, for free space,
absolute permeability μ 0 = 4π × 10 −^7 H/m relative permeability μ r = 1. Now, take any medium other than vacuum. If its relative permeability, as compared to vacuum is μ r , then its absolute permeability is μ = μ 0 μ r H/m.
6.2. Laws of Magnetic Force
Coulomb was the first to determine experimentally the quantitative expression for the magnetic force between two isolated point poles. It may be noted here that, in view of the fact that magnetic poles always exist in pairs, it is impossible, in practice, to get an isolated pole. The concept of an isolated pole is purely theoretical. However, poles of a thin but long magnet may be assumed to be point poles for all practical purposes (Fig. 6.1). By using a torsion balance, he found that the force between two magnetic poles placed in a medium is
( i ) directly proportional to their pole strengths ( ii ) inversely proportional to the square of the distance between them and ( iii ) inversely proportional to the absolute permeability of the surrounding medium.
Fig. 6.1 Fig. 6. For example, if m 1 and m 2 represent the magnetic strength of the two poles (its unit as yet being undefined), r the distance between them (Fig. 6.2) and μ the absolute permeability of the surrounding medium, then the force F is given by
F ∝ 1 22 μ
m m r
or F = 1 22 m m k μ r
or (^) F
→
1 2 ^ 2
k m m r μ r
in vector from
where ^ r is a unit vector to indicate direction of r.
or F
→ = 1 3 2 m m k r r
→ where F
→ and r
→ are vectors
In the S.I. system of units, the value of the constant k is = 1/4π.
F = (^1 ) 4
m m πμ r
N or F = (^1 ) (^4 0) r
m m πμ μ r
N – in a medium
In vector form, F
→ = (^1 ) 4
m m πμ r
r
→ = (^1 ) (^40)
m m πμ r
N
If, in the above equation, m 1 = m 2 = m (say) ; r = 1 metre ; F = 0
1 4 π μ
N
260 Electrical Technology
6.6. Flux Density (B)
It is given by the flux passing per unit area through a plane at right angles to the flux. It is usually designated by the capital letter B and is measured in weber/meter^2. It is a Vector Quantity.
It ΦWb is the total magnetic flux passing normally through an area of A m^2 , then B = Φ/ A Wb/m^2 or tesla (T) Note. Let us find an expression for the flux density at a point distant r metres from a unit N -pole ( i.e. a pole of strength 1 Wb.) Imagine a sphere of radius r metres drawn round the unit pole. The flux of 1 Wb radiated out by the unit pole falls normally on a surface of 4π r^2. m^2. Hence
B = (^12) A (^) 4 r
Φ = π
Wb/m^2
6.7. Absolute Permeability (μμμμμ) and Relative Permeability (μμμμμrr)
In Fig. 6.3 is shown a bar of a magnetic material, say, iron placed in a uniform field of strength H N/Wb. Suppose, a flux density of B Wb/m 2 is developed in the rod.
Fig. 6. Then, the absolute permeability of the material of the rod is defined as μ = B / H henry/metre or B = μ H = μ 0 μr H Wb/m 2 ... ( i ) When H is established in air (or vacuum), then corresponding flux density developed in air is B 0 = μ 0 H Now, when iron rod is placed in the field, it gets magnetised by induction. If induced pole strength in the rod is m Wb, then a flux of m Wb emanates from its N -pole, re-enters its S -pole and continues from S to N -pole within the magnet. If A is the face or pole area of the magentised iron bar, the induction flux density in the rod is B (^) i = m / A Wb/m 2 Hence, total flux density in the iron rod consists of two parts [Fig. 6.3 ( b )]. ( i ) B 0 –flux density in air even when rod is not present ( ii ) B (^) i –induction flux density in the rod B = B 0 + B (^) i = μ 0 H + m /A Eq. ( i ) above may be written as B = μ r. μ 0 H = μ r B 0 ∴ μ r = 0 0
(material) (vacuum)
B^ B B B = ...for same^ H
Hence, relative permeability of a material is equal to the ratio of the flux density produced in that material to the flux density produced in vacuum by the same magnetising force.
6.8. Intensity of Magnetisation (I)
It may be defined as the induced pole strength developed per unit area of the bar. Also, it is the magnetic moment developed per unit volume of the bar.
Let m = pole strength induced in the bar in Wb
Magnetism and Electromagnetism 261
A = face or pole area of the bar in m 2 Then I = m / A Wb/m 2 Hence, it is seen that intensity of magnetisation of a substance may be defined as the flux density produced in it due to its own induced magnetism. If l is the magnetic length of the bar, then the product ( m × l ) is known as its magnetic moment M. ∴ I = m m^ l M A A l V
× = = × = magnetic moment/volume
6.9. Susceptibility (K)
Susceptibility is defined as the ratio of intensity of magnetisation I to the magnetising force H****. ∴ K = I / H henry/metre.
6.10. Relation Between B, H, I and K
It is obvious from the above discussion in Art. 6.7 that flux density B in a material is given by B = B 0 + m / A = B 0 + I ∴ B = μ 0 H + I
Now absolute permeability is μ = 0 μ 0 B H^ I I H H H
μ + = = + (^) ∴ μ = μ 0 + K Also μ = μ 0 μ r ∴ μ 0 μ r = μ 0 + K or μ r = 1 + K /μ 0 For ferro-magnetic and para-magnetic substances, K is positive and for diamagnetic substances, it is negative. For ferro-magnetic substance (like iron, nickel, cobalt and alloys like nickel-iron and
cobalt-iron) μ r is much greater than unity whereas for para-magnetic substances (like aluminium), μr
is slightly greater than unity. For diamagnetic materials (bismuth) μ r < 1.
Example 6.1. The magnetic susceptibility of oxygen gas at 20ºC is 167 × 10^ −^11 H/m. Calculate its absolute and relative permeabilities.
Solution. μ r =
11 7 0
167 10 1 1 4 10
K
− −
×
= 1.
Now, absolute permeability μ= μ 0 μ r = 4π × 10 −^7 × 1.00133 = 12.59 (^) ××××× 10 −−−−−^7 H /m
6.11. Boundary Conditions
The case of boundary conditions between two materials of different permeabilities is similar to that discussed in Art. 4.19. As before, the two boundary conditions are ( i ) the normal component of flux density is continuous across boundary. B 1 n = B 2 n ...( i ) ( ii ) the tangential component of H is continuous across boundary H 1 t = H 2 t As proved in Art. 4.19, in a similar way, it can be shown
that 1 2
tan tan
θ θ
= 1 2
μ μ This is called the law of magnetic flux refraction.
6.12. Weber and Ewing’s Molecular Theory
This theory was first advanced by Weber in 1852 and was, later on, further developed by Ewing in 1890. The basic assumption of this theory is that molecules of all substances are inherently magnets in themselves, each having a N and S pole. In an unmagnetised state, it is supposed that these small molecular
Fig. 6.
Fig. 6.
Magnetism and Electromagnetism 263
and out of alignment as to reduce the magnetic strength to zero is called Curie point. More accu- rately, it is that critical temperature above which ferromagnetic material becomes paramagnetic.
ELECTROMAGNETISM
6.14. Force on a Current-carrying Conductor Lying in a Magnetic Field
It is found that whenever a current-carrying conductor is placed in magnetic field, it experiences a force which acts in a direction perpendicular both to the direction of the current and the field. In Fig. 6.8 is shown a conductor XY lying at right angles to the uniform horizontal field of flux density B Wb/m^2 produced by two solenoids A and B. If l is the length of the conductor lying within this field and I ampere the current carried by it, then the magnitude of the force experienced by it is F = BIl = μ 0 μ r HIl newton Using vector notation F
→ = (^) I l B
→ → × and^ F^ =^ IlB^ sin^ θ^ where^ θ^ is the angle between^ l
→ and (^) B
→ which is 90º in the present case or F = Il B sin 90º = Il B newtons (∵ sin 90º = 1) The direction of this force may be easily found by Fleming’s left-hand rule.
Fig. 6.8 Fig. 6. Hold out your left hand with forefinger, second finger and thumb at right angles to one another. If the forefinger represents the direction of the field and the second finger that of the current, then thumb gives the direction of the motion. It is illustrated in Fig. 6.9. Fig. 6.10 shows another method of finding the direc- tion of force acting on a current carrying conductor. It is known as Flat Left Hand rule. The force acts in the direc- tion of the thumb obviously, the direction of motor of the conductor is the same as that of the force. It should be noted that no force is exerted on a con- ductor when it lies parallel to the magnetic field. In gen- eral, if the conductor lies at an angle θ with the direction of the field, then B can be resolved into two components, B cos θ parallel to and B sin θ perpendicular to the con- ductor. The former produces no effect whereas the latter is responsible for the motion observed. In that case, F = BIl sin θ newton, which has been expressed as cross product of vector above. *****
***** It is simpler to find direction of Force (Motion) through cross product of given vectors I l
→ and B
→ .
Fig. 6.
264 Electrical Technology
Fig. 6.
6.15. Ampere’s Work Law or Ampere’s Circuital Law
The law states that m.m.f. ***** (magnetomotive force corre- sponding to e.m.f. i. e. electromotive force of electric field) around a closed path is equal to the current enclosed by the path.
Mathematically, H. d s
→ (^) → ∫ =^ I^ amperes where^ H
→ is the vector
representing magnetic field strength in dot product with vector
d s → of the enclosing path S around current I ampere and that is
why line integral (“) of dot product H. d s
→ (^) → is taken. Work law is very comprehensive and is applicable to all magnetic fields whatever the shape of enclosing path e. g. ( a ) and ( b ) in Fig. 6.11. Since path c does not enclose the conductor, the m.m.f. around it is zero.
The above work Law is used for obtaining the value of the magnetomotive force around simple idealized circuits like ( i ) a long straight current-carrying conductor and ( ii ) a long solenoid.
( i ) Magnetomotive Force around a Long Straight Conductor In Fig. 6.12 is shown a straight conductor which is assumed to extend to infinity in either direction. Let it carry a current of I amperes upwards. The magnetic field consists of circular lines of force having their plane perpendicular to the conductor and their centres at the centre of the conductor.
Suppose that the field strength at point C distant r metres from the centre of the conductor is H. Then, it means that if a unit N -pole is placed at C , it will experience a force of H newtons. The direction of this force would be tangential to the circular line of force passing through C. If this unit N -pole is moved once round the conductor against this force, then work done i.e.
m.m.f. = force × distance = I i. e. I = H × 2 π r joules = Amperes or H = 2
I π r
= H^. d s
→ (^) → ∫ Joules = Amperes =^ I Obviously, if there are N conductors (as shown in Fig. 6.13), then
H = 2
NI π r
A/m or Oersted
and B = μ (^0) 2
NI π r
Wb/m^2 tesla ...in air
= 0 2
r NI r
μ μ π
Wb/m 2 tesla ...in a medium
Fig. 6.
****** M.M.F. is not a force, but is the work done.
Fig. 6.
266 Electrical Technology
Fig. 6.
The magnetic field strength dH
→ due to a small element dl of the wire shown is
dH
→
^
4 2
I d l s s
→ → × π
(By Biot-Savart Law)
or dH
→ = 2 ^
sin 4
Idl u s
θ π ×
(where u ^ is unit vector perpendicular to
plane containing d l
→ and s ^ and into the plane.) or dH
→ = 2 ^ cos 4
Idl u s
φ π
...[∵ θ and φ are complementary angles]
The magnetic field strength due to entire length l :
H
→ = 2 ^ 0
cos 4
l I dl u s
⎡ (^) φ ⎤ ⎢ ⎥ π ⎢ ⎥ ⎣ ⎦
∫
= 2 ^ 0
/ 4
l I r s dl u s
⎡ ⎤ ⎢ ⎥ π ⎢ ⎥ ⎣ ⎦
∫ (^ ) cos = r in Fig. 6. s
φ
= 3 ^^2 2 3/ 2 ^ 0 0
(^4 4) ( )
l l Ir dl (^) u Ir dl u s r l
⎡ ⎤ ⎡ ⎤ ⎢ ⎥ = ⎢ ⎥ π (^) ⎢ ⎥ π⎢ + ⎥ ⎣ ⎦ ⎣ ⎦
∫ ∫
(∵ r is constant) ; s = r^2 + l^2 in Fig. 6.
= 3 2 ^ 4 0 [1 ( / ) ]3 / 2
l Ir dl (^) u r r l
⎡ ⎤ ⎢ ⎥ π ⎢⎣ ∫^ + ⎥⎦
(Taking r^3 out from denominator)
To evaluate the integral most simply, make the following substitution l r = tan φ in Fig. 6. ∴ l = r tan φ ∴ dl = r sec^2 φ d φ and 1 + ( r / l ) 2 = 1 + tan 2 φ = sec^2 φ and limits get transformed i. e. become 0 to φ.
H
→
(^2 ) ^ ^ ^ (^3 2 3 ) 0 0
sec (^1) cos sin 4 sec 4 4
Ir r Ir d u d u u r r r
= sin ^ 4
I (^) u r
φ π N.B. For wire of infinite length extending it at both ends i. e. −∞ to + ∞ the limits of integration would be to , giving 2 ^^ or ^ 2 2 4 2
H I^ u H I u r r
π π^ →^ → − + = × = π π
.
( ii ) Magnetic Field Strength along the Axis of a Square Coil
This is similar to ( i ) above except that there are four conductors each of length say, 2 a metres and carrying a current of I amperes as shown in Fig. 6.17. The Mag- netic Field Strengths at the axial point P due to the op- posite sides ab and cd are Hab and Hcd directed at right angles to the planes containing P and ab and P and cd respectively. Now, Hab and H (^) cd are numerically equal,
Fig. 6.
Magnetism and Electromagnetism 267
hence their components at right angles to the axis of the coil will cancel out, but the axial components will add together. Similarly, the other two sides da and bc will also give a resultant axial component only.
As seen from Eq. ( ii ) above,
Hab = 4
I π r
[cos θ − cos (180º − θ)] =
. 2 cos cos 4 2
I I r r
θ θ
π π Now r = (^) a^2^ + x^2 ∴ H (^) ab = 2 2
. cos 2
I a x
θ π +
Its axial components is Hab ′ = Hab. sin α = 2 2
cos
. sin 2
I a x
θ α π + All the four sides of the rectangular coil will contribute an equal amount to the resultant magnetic field at P. Hence, resultant magnetising force at P is
H = 2 2
cos
- sin 2
I a x
θ × α π +
,
Now cos θ = (2 2 2 )
a a + x
and sin α = 2 2
a a + x
∴ H =
2 2 2 2 2
( ). 2
a I π a + x x + a
AT/m.
In case, value of H is required at the centre O of the coil, then putting x = 0 in the above expres- sion,
we get H =
2 2
. 2.
a I I a a a
= π π
AT/m
Note. The last result can be found directly as under. As seen from Fig. 6.18, the field at point O due to any side is, as given by Eq. ( i )
=
/ 4 45º 45º / 4
sin. cos. 2cos 45º.^2 (^4 4 4 4 )
I (^) d I I I a a a a
−π −
π
θ θ = − θ = = π ∫ π π π Resultant magnetising force due to all sides is H = 4 1.^22 (^4 )
I a a × = π π AT/m ...as found above
( iii ) Magnetising Force on the Axis of a Circular Coil In Fig. 6.19 is shown a circular one-turn coil carrying a current of I amperes. The magnetising force at the axial point P due to a small element ‘ dl ’ as given by Laplace’s Law is
dH
→ = (^2 ) 4 ( )
Idl π r + x The direction of dH is at right angles to the line AP join- ing point P to the element ‘ dl ’. Now, dH can be resolved into two components :
( a ) the axial component dH ′ = dH sin θ ( b ) the vertical component dH ″ = dH cos θ Now, the vertical component dH cos θ will be cancelled by an equal and opposite vertical com- ponent of dH due to element ‘ dl ’ at point B. The same applies to all other diametrically opposite pairs of dl ’s taken around the coil. Hence, the resultant magnetising force at P will be equal to the sum of all the axial components.
Fig. 6.
Fig. 6.
Magnetism and Electromagnetism 269
∴ H = (^2 ) 1 1
sin. cos 2 2
NI (^) d NI l l
θ (^) θ ∫θ θ^ θ =^ −^ θθ
= 2
NI l (cos θ 1 −cos θ 2 ) ... ( iv )
The above expression may be used to find the value of H at any point of the axis, either inside or outside the solenoid. ( i ) At mid-point, θ 2 = (π − θ 1 ), hence cos θ 2 = −cos θ 1 ∴ H = 2 2
NI l
cos θ 1 = NI l
cos θ 1
Obviously, when the solenoid is very long, cos θ 1 becomes nearly unity. In that case, H = NI l
AT/m –Art. 6.15 ( ii ) ( ii ) At any point on the axis inside a very long solenoid but not too close to either end, θ 1 ≅ 0 and θ 2 ≅ π so that cos θ 1 ≅ 1 and cos θ 2 = −1. Then, putting these values in Eq. ( iv ) above, we have
H ≅ 2
NI l
× 2 = NI l It proves that inside a very long solenoid, H is practically constant at all axial points excepts those lying too close to either end of the solenoid. ( iii ) Towards either end of the solenoid, H decreases and exactly at the ends, θ 1 = π/2 and θ 2 ≅ π, so that cos θ 1 = 0 and cos θ 2 = −1. Hence, from Eq. ( iv ) above, we get
H = 2
NI l In other words, value of H is decreased to half the normal value well inside the solenoid. Example 6.2. Calculate the magnetising force and flux density at a distance of 5 cm from a long straight circular conductor carrying a current of 250 A and placed in air. Draw a curve show- ing the variation of B from the conductor surface outwards if its diameter is 2 mm.
Solution. As seen from Art. 6.15 ( i ) , H = 250 2 2 0.
I r
= = π π ×
795.6 AT/m
B = μ 0 H = 4π × 10 −^7 × 795.6 = 10 −−−−−^3 Wb/m^2
In general, B = 0 2
I r
μ π Now, at the conductor surface, r = 1 mm = 10−^3 m
∴ B =
7 3
4 10 250 2 10
− −
π × × π ×
= 0.05 Wb/m^2
The variation of B outside the conductor is shown in Fig. 6.22. Example 6.3. A wire 2.5 m long is bent (i) into a square and (ii) into a circle. If the current flowing through the wire is 100 A, find the magnetising force at the centre of the square and the centre of the circle. (Elec. Measurements; Nagpur Univ. 1992)
Solution. ( i ) Each side of the square is 2 a = 2.5/4 = 0.625 m Value of H at the centre of the square is [Art 6.17 ( ii ) ]
= 2 2 100
I a
×
π π ×
= 144 AT/m ( ii ) 2 π r = 2.5 ; r = 0.398 m
Value of H at the centre is = I /2 r = 100/2 × 0.398 = 125.6 AT/m
Fig. 6.
270 Electrical Technology
Fig. 6.
Example 6.4. A current of 15 A is passing along a straight wire. Calculate the force on a unit magnetic pole placed 0.15 metre from the wire. If the wire is bent to form into a loop, calculate the diameter of the loop so as to produce the same force at the centre of the coil upon a unit magnetic pole when carrying a current of 15 A. (Elect. Engg. Calcutta Univ.)
Solution. By the force on a unit magnetic pole is meant the magnetising force H. For a straight conductor [Art 6.15 ( i )] H = I /2 π r = 15/2π × 0.15 = 50/π AT/m Now, the magnetising force at the centre of a loop of wire is [Art. 6.17 ( iii ) ] = I / 2 r = I / D = 15/ D AT/m Since the two magnetising forces are equal ∴ 50/π = 15/ D ; D = 15 π/50 = 0.9426 m = 94.26 cm. Example. 6.5. A single-turn circular coil of 50 m. diameter carries a direct current of 28 × 104 A. Assuming Laplace’s expression for the magnetising force due to a current element, determine the magnetising force at a point on the axis of the coil and 100 m. from the coil. The relative permeabil- ity of the space surrounding the coil is unity.
Solution. As seen from Art 6.17 ( iii ), H = 2
I r
. sin 3 θ AT/m
Here sin θ = 2 2 2 2
(^25) 0. 25 100
r r x
= =
sin 3 θ = (0.2425) 3 = 0.01426 ∴ H =
28 10 4
2 25 76.8 AT/m
6.18. Force Between Two Parallel Conductors
( i ) Currents in the same direction. In Fig. 6.23 are shown two parallel conductors P and Q carrying currents I 1 and I 2 amperes in the same direction i. e. upwards. The field strength in the space between the two conductors is decreased due to the two fields there being in opposition to each other. Hence, the resultant field is as shown in the figure. Obviously, the two conductors are attracted towards each other.
( ii ) Currents in opposite directions. If, as shown in Fig. 6.24, the parallel conductors carry currents in opposite directions, then field strength is increased in the space between the two conduc- tors due to the two fields being in the same direc- tion there. Because of the lateral repulsion of the lines of the force, the two conductors experience a mutual force of repulsion as shown separately in Fig. 6.24 ( b ).
6.19. Magnitude of Mutual Force
It is obvious that each of the two parallel conductors lies in the magnetic field of the other conductor. For example, conductor P lies in the magnetic field of Q and Q lies in the field of P. If ‘ d ’ metres is the distance between them, then flux density at Q due to P is [Art. 6.15 ( i ) ]
B = 0 1 2
I d
μ π Wb/m
2
272 Electrical Technology
***** Strictly speaking, it should be only ‘ampere’ because turns have no unit.
Fig. 6.
Also, it is seen from Fig. 6.23 that when the two currents flow in the same direction, net field strength midway between the two conductors is the difference of the two field strengths.
Now, H 1 = I 1 /2π and H 2 = I 2 /2π because r = 2/1 = 2 metre
∴ 1 2 2 2
I I − π π
= 7.95 ∴ I 1 − I 2 = 50 ... ( i )
Force per unit length of the conductors is F = 2 × 10 −^7 I 1 I 2 / d newton ∴ 2.4 × 10 −^4 = 2 × 10 −^7 I 1 I 2 /2 ∴ I 1 I 2 = 2400 ... ( ii ) Substituting the value of I 1 from ( i ) in ( ii ) , we get (50 + I 2 ) I 2 = 2400 or I 22 + 50 I 2 −2400 = 0 or ( I 2 + 80) ( I 2 −30) = 0 ∴ I 2 = 30 A and I 1 = 50 + 30 = 80 A
Tutorial Problems No. 6.
1. The force between two long parallel conductors is 15 kg/metre. The conductor spacing is 10 cm. If one conductor carries twice the current of the other, calculate the current in each conductor. [6,060 A; 12,120 A] 2. A wire is bent into a plane to form a square of 30 cm side and a current of 100 A is passed through it. Calculate the field strength set up at the centre of the square. [300 AT/m] (Electrotechnics - I, M.S. Univ. Baroda )
MAGNETIC CIRCUIT
6.21. Magnetic Circuit
It may be defined as the route or path which is followed by magnetic flux. The law of magnetic circuit are quite similar to (but not the same as) those of the electric circuit. Consider a solenoid or a toroidal iron ring having a magnetic path of l metre, area of cross section A m^2 and a coil of N turns carrying I amperes wound anywhere on it as in Fig. 6.25.
Then, as seen from Art. 6.15, field strength inside the solenoid is H = NI l
AT/m
Now B = μ 0 μr H =^0 r^ NI l
μ μ Wb/m^2
Total flux produce Φ = B × A = 0 r^
A NI l
μ μ Wb
∴ Φ = / (^0) r
NI l μ μ A
Wb
The numerator ‘ Nl ’ which produces magnetization in the magnetic circuit is known as magnetomotive force (m.m.f.). Obviously, its unit is ampere-turn (AT)*. It is analogous to e.m.f. in an electric circuit.
The denominator 0 r
l μ μ A
is called the reluctance of the circuit and is analogous to resistance in
electric circuits.
∴ flux = m.m.f. reluctance
or Φ = F S Sometimes, the above equation is called the “Ohm’s Law of Magnetic Circuit” because it resembles a similar expression in electric circuits i. e.
Magnetism and Electromagnetism 273
current = e.m.f. resistance
or I = V R
6.22. Definitions Concerning Magnetic Circuit
1. Magnetomotive force (m.m.f.). It drives or tends to drive flux through a magnetic circuit and corresponds to electromotive force (e.m.f.) in an electric circuit.
M.M.F. is equal to the work done in joules in carrying a unit magnetic pole once through the entire magnetic circuit. It is measured in ampere-turns.
In fact, as p.d. between any two points is measured by the work done in carrying a unit charge from one points to another, similarly, m.m.f. between two points is measured by the work done in joules in carrying a unit magnetic pole from one point to another.
2. Ampere-turns (AT). It is the unit of magnetometre force (m.m.f.) and is given by the product of number of turns of a magnetic circuit and the current in amperes in those turns. 3. Reluctance. It is the name given to that property of a material which opposes the creation of magnetic flux in it. It, in fact, measures the opposition offered to the passage of magnetic flux through a material and is analogous to resistance in an electric circuit even in form. Its units is AT/Wb.*
reluctance = 0 r
l l A A
= μ μ μ
; resistance =
l l
A A
In other words, the reluctance of a magnetic circuit is the number of amp-turns required per weber of magnetic flux in the circuit. Since 1 AT/Wb = 1/henry, the unit of reluctance is “reciprocal henry.”
4. Permeance. It is reciprocal of reluctance and implies the case or readiness with which magnetic flux is developed. It is analogous to conductance in electric circuits. It is measured in terms of Wb/AT or henry. 5. Reluctivity. It is specific reluctance and corresponds to resistivity which is ‘specific resistance’.
6.23. Composite Series Magnetic Circuit
In Fig. 6.26 is shown a composite series magnetic circuit consisting of three different magnetic materials of different permeabilities and lengths and one air gap (μ r = 1). Each path will have its own reluctance. The total reluctance is the sum of individual reluctances as they are joined in series.
∴ total reluctance = 0 r
l μ μ A
= 1 2 3
1 2 3 0 1 0 2 0 3 0
a r r r g
l l l l A A A A
μ μ μ μ μ μ μ ∴ flux Φ =
0
m.m.f.
r
l μ μ A
6.24. How to Find Ampere-turns?
It has been shown in Art. 6.15 that H = NI / l AT/m or NI = H × l ∴ ampere-turns AT = H × l Hence, following procedure should be adopted for calculating the total ampere turns of a composite magnetic path.
***** From the ratio Φ = m.m.f. reluctance
, it is obvious that reluctance = m.m.f./Φ. Since m.m.f. is in ampere- turns and flux in webers, unit of reluctance is ampere-turn/weber (AT/Wb) or A/Wb.
Fig. 6.
Magnetism and Electromagnetism 275
Fig. 6. It should be noted that reluctance offered by the central core AB has been neglected in the above treatment.
6.27. Series-Parallel Magnetic Circuits
Such a circuit is shown in Fig. 6.30 ( a ). It shows two parallel mag- netic circuits ACB and ACD con- nected across the common magnetic path AB which contains an air-gap of length l (^) g. As usual, the flux Φ in the common core is divided equally at point A between the two parallel paths which have equal reluctance. The reluctance of the path AB con- sists of ( i ) air gap reluctance and ( ii ) the reluctance of the central core which comparatively negligible. Hence, the reluctance of the cen- tral core AB equals only the air-gap reluctance across which are connected two equal parallel reluc- tances. Hence, the m.m.f. required for this circuit would be the sum of ( i ) that required for the air-gap and ( ii ) that required for either of two paths (not both) as illustrated in Ex. 6.19, 6.20 and 6.21.
The equivalent electrical circuit is shown in Fig. 6.30 ( b ) where the total resistance offered to the voltage source is = R 1 + R æ R = R 1 + R /2.
6.28. Leakage Flux and Hopkinson’s Leakage Coefficient
Leakage flux is the flux which follows a path not intended for it. In Fig. 6.31 is shown an iron ring wound with a coil and having an air- gap. The flux in the air-gap is known as the useful flux because it is only this flux which can be utilized for various useful purposes. It is found that it is impossible to confine all the flux to the iron path only, although it is usually possible to confine most of the electric current to a definite path, say a wire, by surrounding it with insulation. Unfortunately, there is no known insulator for magnetic flux. Air, which is a splendid insulator of electricity, is unluckily a fairly good magnetic conductor. Hence, as shown, some of the flux leaks through air surrounding the iron ring. The presence of leakage flux can be detected by a compass. Even in the best designed dynamos, it is found
Fig. 6.
Fig. 6.
276 Electrical Technology
that 15 to 20% of the total flux produced leaks away without being utilised usefully. If, Φ t = total flux produced ; Φ = useful flux available in the air-gap, then
leakage coefficient λ = total flux useful flux
or λ = t
Φ Φ In electric machines like motors and generators, magnetic leakage is undesirable, because, al- though it does not lower their power efficiency, yet it leads to their increased weight and cost of manufacture. Magnetic leakage can be minimised by placing the exciting coils or windings as close as possible to the air-gap or to the points in the magnetic circuit where flux is to be utilized for useful purposes. It is also seen from Fig. 6.31 that there is fringing or spreading of lines of flux at the edges of the air-gap. This fringing increases the effective area of the air-gap. The value of λ for modern electric machines varies between 1.1 and 1.25.
6.29. Magnetisation Curves
The approximate magnetisation curves of a few magnetic materials are shown in Fig. 6.32. These curves can be determined by the following methods provided the materials are in the form of a ring : ( a ) By means of a ballistic galvanometer and ( b ) By means of a fluxmeter.
6.30. Magnetisation Curves by Ballistic Galvanometer
In Fig. 6.33 shown the specimen ring of uniform cross-section wound uniformly with a coil P which is con- nected to a battery B through a revers- ing switch RS , a variable resistance R 1 and an ammeter. Another secondary coil S also wound over a small portion of the ring and is connected through a resistance R to a ballistic galvanometer BG. The current through the primary P can be adjusted with the help of R 1. Sup- pose the primary current is I. When the primary current is reversed by means of RS , then flux is reversed through S , hence an induced e.m.f. is produced in it which sends a current through BG. This current is of very short duration. The first deflection or ‘throw’ of the BG is proportional to the quantity of elec- tricity or charge passing through it so long as the time taken for this charge to flow is short as compared with the time of one oscillation. If θ = first deflection or ‘throw’ of the galvanometer when primary current I is reversed. k = ballistic constant of the galvanometer i. e. charge per unit deflection.
then, charge passing through BG is = k θ coulombs ... ( i ) Let Φ = flux in Wb produced by primary current of I amperes ; t = time of reversal of flux ; then
rate of change of flux = 2 t
Φ (^) Wb/s
Fig. 6.
