Magnetic Flux •Unit for flux is weber •The definition of 1 weber is the amount of flux that can produce an induced voltage of 1 V in a one turn coil if the flux reduce to zero with uniform rate. Magnetic Flux density B •Unit for magnetic flux density is Tesla •The definition of 1 tesla is the flux density that can produce a force of 1 Newton per meter acting upon a conductor carrying 1 ampere of current. Magnetic field strength H •Unit for magnetic field strength is Ampere/m •A line of force that produce flux
F = B l I newton Where F = force ; B = magnetic flux density ; l =the length of conductor and I = current in the conductor
Φ = B × A Where Φ = magnetic flux ; B = magnetic flux density and A = area of cross-section
B = µ H Where Η = magnetic field strength ; B = magnetic flux density and µ = permeability of the medium Permeability in free space
µ
o
= B/H = 4π x 10-7 H/m
Relative Permeability (
)
r
Relative permeability is defined as a ratio of flux density produced in a material to the flux density produced in a vacuum for the same magnetic filed strength. str ength. Thus µ
r
= µ /µ = µ r µ
o
∴
µ
or
B = µ r µ oH
o
= B/H
B vs H
Electromagnetic Force (mmf) H turns
mmf
H = NI
Where Η = magnetic field strength ; l =the path length of ; N number of turns and I = current in the conductor
Example 1 A coils of 200 turns is uniformly wound around a wooden ring with a mean circumference of 600 mm and area of cross-section of 500 mm2. If the current flowing flowing into the coil is 4 A, Calculate (a) the magnetic field strength , (b) flux density dan (c) total flux turns
N = 200 turns l = 600 x 10-3 m A = 500 x 10-6 m2 I = 4 A (a )
H = NI/l = 200 x 4 / 600 x 10-3 = 1333 A
(b)
B = µ oH = 4π x 10-7 x 1333 = 0.001675 T = 1675 µ T
(c )
Total Flux Φ = BA = 1675 x 10-6 x 500 x 10-6 = 0.8375 µ Wb
Reluctance ( S ) Ohm‘s law I = V/R [A] Where I =current; V=voltage and R=resistance And the resistance can be relate to physical parameters as R = ρ
l /A
ohm
Where ρ =r =res esis istiv tivit ity y [ohm[ohm-met meter] er],,
l=
length in meter and
A=area of cross-section [meter square] Analogy to the ohm‘s law V=NI=H l
I=Φ and R=S Hl [ ampere / weber ] [ weber ] where S = Φ= µ r µ o A S
Example 2 A mild steel ring, having a cross-section area of 500 mm 2 and a mean circumference of 400 mm is wound uniformly by a coil of 200 turns. Calculate(a) reluctance of the ring and (b) a current required to produce a flux of 800 µ Wb in the ring. (a)
B = µ
S =
Φ
A r
6
=
800 ×10 − 500 ×10
−6
= 1.6T
turns
= 380
µ r µ o A
=
0.4 380 × 4π ×10
−7
× 5 ×10
= 1.667 ×106 [ A / Wb]
−4
(b)
Φ=
mmf
H
H = ΦS
S
H = 800 ×10 −6 ×1.667 ×106 ∴ I =
1342 N
=
1342 200
=
= 1342[ A] = NI
6.7[ A]
Magnetic circuit with different materials l l SA =
1
µ1a1
S = SA For A:
ForB:
SB =
and
+ SB
=
l 1 µ1a1
+
2
µ 2a 2
l 2 µ 2a 2
area of cross-section = a1 mean length = l 1 absolute permeability = µ area of cross-section = a2 mean length = l 2 absolute permeability = µ
1
2
Mmf for many materials in series total mmf = H l + H l A A
B B
HA =magnetic strength in material A l A=mean
length of material A
HB =magnetic strength in material B l B=mean
length of material B
In general Σ (m.m.f) = Σ Hl
Example 3 A magnetic circuit comprises three parts in series, each of uniform cross-section area(c.s.a). They are : (a)A length of 80mm and c.s.a 50 mm 2; (b)A length of 60mm and c.s.a 90mm 2; (c)An airgap of length 0.5mm and c.s.a 150 mm 2. A coil of 4000 turns is wound on part (b), and the flux density in the airgap is 0.3T. Assuming that all the flux passes through the given circuit, and that the relative permeability µ r is 1300, estimate the coil current to produce such a flux density.
Φ = BC AC = 0.3 × 1.5 × 10
−4
−4
= 0.45 × 10 Wb
Mmf = Φ S = H l = N I ΦS a = Φ
Material a
ΦS b = Φ
Material b
airgap
ΦS c = Φ
Total mmf
and
a
=
µ r µ o Aa b
µ r µ o Ab
c µ r µ o Ac
=
=
0.45 ×10
−4
× 80 × 10
1300 × 4π ×10 0.45 × 10
−4
−7
× 50 × 10
× 60 ×10
1300 × 4π ×10
−7
−3 −6
44.1 At
−3
× 90 × 10
−6
= 18.4 At
−6
= 119.3 At
0.45 ×10 −4 × 0.5 ×10 −3 −7
=
1300 × 4π ×10 ×150 ×10
NI = ΦS a + ΦS b + ΦS c = 44.1 + 18.4 + 119.3 = 181.8 At I =
181.8 4000
−3
= 45.4 ×10 A = 45.4mA
Leakages and fringing of flux leakage fringing
Magnetic circuit with air-gap
Leakages and fringing of flux
Some fluxes are leakage via paths a, b and c . Path d is shown to be expanded due to fringing. Thus the usable flux is less than the total flux produced, pr oduced, hence
Leakage factor =
total flux usable flux
Example 4 A magnetic circuit as in Figure is made from a laminated steel. The breadth of the steel core is 40 mm and the depth is 50 mm, 8% of it is an insulator between the laminatings. The length and the area of the airgap are 2 mm and 2500 mm2 respectively. A coil is wound 800 turns. If the leakage factor is 1.2, calculate the current required to magnetize the steel core in order to produce flux of 0.0025 Wb across the airgap.
Ba = H a =
Φ
Aa Ba µ o
= =
2.5 × 10
−3
2500 × 10 1 4π ×10
−7
−6
= 1T
= 796000[ AT / m]
mmf = H = 796000 × 0.002 = 1594[ AT ] Total flux = ΦT
=
flux in airgap × leakage factor
=
0.0025 ×1.2 = 0.003Wb
92% of the depth is laminated steel, thus the area of cross section is ∴
AS = 40 x 50 x 0.92 = 1840 mm2=0.00184m
BS
=
Φ T AS
=
3 ×10 −3 0.00184
= 1.63T
From the B-H graph, at B=1.63T, H=4000AT/m ∴ mmf in the steel core = Hl = 4000 x 0.6 = 2400 AT
Total mmf. = 1592 + 2400 = 3992 AT ∴ NI = 3992
I = 3992/800 = 5 A
D
Magnetic circuit applying voltage law
Analogy to electrical circuit
applying voltage Kirchoff’s law Mmf in loop C = NI = HLl L + HMl M outside loop NI= HLl L + H Nl N And in loop D 0 = HMl M + H Nl N In general Σ (m.m.f) = Σ Hl
P IL
L
IM
IN
M
N
E Q
At node P we can also applying current Kirchoff’s law Φ L = Φ M + Φ N
Or
Φ L - Φ M - Φ N = 0
In general:
Σ Φ = 0
Example 5 A magnetic circuit made of silicon steel is arranged as in the Figure. The center limb has a cross-section area of 800mm 2 and each of the side limbs has a cross-sectional area of 500mm 2. Calculate the m.m.f required to produce a flux of 1mWb in the center limb, assuming the magnetic leakage to be negligible.
3 4 0 m m
1m m 1 50 m m
34 0 m m
Φ = B × A
B =
Φ
A
3
1×10 −
=
800 ×10
Looking at graph at B=1.25T µ r =34000 Apply voltage law in loops A and B
−6
3 4 0 m m
m.m. f = Φ A S 1 + ( Φ A + Φ B ) ( S 2 + S a ) S 1
=
1 µ r µ o A1
S 2 =
S a
=
340 ×10 34000 × 4π ×10
150 ×10
−7
−3
× 500 ×10
34000 × 4π × 10 × 800 ×10 =
4π ×10
−7
= 15915
−3
−7
1× 10
−6
−6
= 4388
−3
× 800 × 10
−6
= 1.25T
= 994718
1 m m
A
1 5 0 m m
3 4 0 m m
B
Since the circuit is symmetry Φ A =Φ B
m.m. f = ΦS 1 + ( 2Φ ) ( S 2 + S a ) In the center limb , the flux is 1mWb which is equal to 2 Φ Therefore Φ =0.5mWb
m.m. f = 0.5 ×10
−3
(15915) + (1×10 )( 4388 + 994718)
= 8 + 999 = 1007
−3
A
Hysteresis loss
Materials before applying m.m.f (H), the polarity of the molecules or structures are in random.
After applying m.m.f (H) , the polarity of the molecules or structures are in one direction, thus the materials become magnetized. The more H applied the more magnetic flux (B )will be produced
When we plot the mmf (H) versus the magnetic flux will produce a curve so called Hysteresis loop loop 1. OAC OAC – whe when n mor more e H app appli lied ed,, B increased until saturated. At this point no increment of B when we increase the H. 2. CD- when when we reduc reduce e the the H the the B also reduce but will not go to zero. 3. DEDE- a neg negat ativ ive e val value ue of H has has to applied in order to reduce B to zero. 4. EF – when when app apply lyin ing g more more H in the the negative direction will increase B in the reverse direction. 5. FGCFGC- whe when n redu reduce ce H will will red reduc uce eB but it will not go to zero. Then by increasing positively the also decrease and certain point it again change the polarity to negative until it reach C.
Eddy current When a sinusoidal current enter the coil, the flux Φ also varies sinusoidally according to I. The induced current will flow in the magnetic core. This current is called eddy current. This current introduce the eddy current loss. The losses due to hysteresis and eddy-core totally called core loss. To reduce eddy current we use laminated core metal
insulator
