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Energy in a Magnetic Field
- A varying potential difference applied to the terminals of an N-turn coil wound on non-magnetic torus produces a varying coil current and hence a varying magnetic field intensity H and a varying flux density Bo.
- The flux linkage of the coil at any instant is given by: λ = N φ = NBo A
- Since the torus is non-magnetic then: Bo = μo H and the B-H curve is a straight line passing through the origin of slope μ (^) o : The permeability of a vacuum
- Since Bo is varying then an emf is induced in the coil and is given by
dt NAdB dt e = dλ = o
- For an instantaneous potential difference of v at the coil terminals: ν = Ri + e
dt ν = Ri + NAdBo The instantaneous power delivered to the core is given by:
dt p = νi = Ri^2 + NAidBo p = pA + p B pA = Ri^2 : Copper losses dissipated as heat
dt AlHdB dt p NAidBo o B =^ = where ( Hl=Ni ) pB = power delivered to the magnetic field Let WB = energy in the magnetic field, where: W p dt AlHdB AlB dB
Bo o
Bo
B =∫ B =∫ 0 =∫ 0 μ
o
WB AlBo 2 μ
2 = J
- positive WB => energy supplied to magnetic field as Bo increases.
- negative WB => energy released from the magnetic field as Bo decreases.
- Al = volume of the space enclosed by the coil
∴ energy density = w (^) B = WAlB
∴ o
Bo o o
wB Bo^ dB B μ μ
2 0
2 2
= =^1
∫ J/m^2
The B-H curve for the non-magnetic material is shown in Figure 14.
Figure 14: B-H Loop for Non-magnetic Material From Figure 14, the energy density for Bo = B ' o is given by the area enclosed by the B-H curve and the Bo axis.
Figure 16: B-H curve for a ferromagnetic material
- If the energy released from the field during the decrease of flux density from B 2 to B 1 is less than the energy supplied to the field for the increase in flux density from B 1 to B 2 , then some of the field energy is lost. This lost energy is dissipated as heat in the magnetic material and is the work done per unit volume in reorienting the magnetic moments of the material. This energy loss is called hysteresis loss.
- For a coil supplied with an alternating current, the magnetic field intensity H is alternating and the steady state B-H loop is shown in Figure 17.
Figure 17:Steady State Loop
- The point ‘ a ’ corresponds to the peak flux density B ˆ^ produced by magnetic field
intensity H ˆ^ , as H ˆ^ is increased from zero to its peak value. The energy per unit volume released from the field as H decreases from H ˆ^ to zero is given by the area amb.
- As H is increased from zero to - H ˆ^ , the energy per unit volume supplied by the source to the field is given by the area bcdn.
- If H is reduced from - H ˆ^ to zero , the energy per unit volume released by the field is given by the area dne.
- Finally, if H is increased from zero to H ˆ^ , the energy per unit volume supplied to the field is given by efam. Thus completing the cycle of H. The loss of energy per unit volume during a cycle at magnetization is given by: bcdn + efam – amb – dne = Area of B-H loop
- This lost energy is dissipated as heat in the magnetic material and represents the work done per unit volume in reorienting the magnetic moments of the material ass it is carried through one cycle of magnetization. This energy loss is called hysteresis loss and is a function of the peak flux density B ˆ^.
- The hysteresis loop may be rescaled to represent φ and F instead of B and H.
This is accomplished by multiplying B and A to produce φ (since: φ = BA ) and multiplying H by l to produce F (since: Hl = Ni = F ).
- The net effect is to multiply any area of the B-H diagram by Al , the volume of the torus.
- Since the area under the B-H loop repeats energy per unit volume, then the area under the φ -F loop represents the energy loos since the φ -F area is now multiplied by volume.
- The area of the φ -F loop represents energy loos per cycle of operation.
- If the material is carried through f cycles of magnetization per second, where f is the frequency of the supplied current, then the hysteresis loss is proportional to the frequency of the supply current i.e.: wB ∝ f
A cross section of the core showing the direction of ie and the flux density B is shown in Figure 19.
Figure 19:Eddy Currents in a Torus – ie decreasing The cross section shows a circular path of elementary width that is concentric with the boundary of the cross section. This path may be considered to extend around the entire torus. For a very low frequency current ie , as ie changes then the flux density B in the elementary section and an emf is induced in the ferromagnetic core, resulting in a circulating current i in the core, whose direction produces a flux to oppose the changing flux that produces the circulating current. These circulating currents are called eddy currents and results in i^2 R losses in the core. At low supply current frequencies, the induced emf and circulating eddy currents are very small and are negligible hence the λ − ie loop at low frequencies reflect hysteresis
losses in the core with the eddy current losses being negligible. At high supply current frequencies, the flux density at any point in the core depends on the coil current ie and the circulating currents i. At the surface of the core, B depends on ie alone since this is the only current that encircles the path of B. The flux density of B at the centre of the core is dependent on the circulatory core eddy currents, i. If ie is varying rapidly, then a rapid decrease in ie results in a large decrease in flux linkage and a large induced emf in the elementary section towards the centre of the core. This large induced emf results in a large circulatory eddy current whose direction produces a flux to maintain the
original flux density value at the centre of the core. Hence the effect of these eddy currents is to prevent a change in flux density towards the centre of the core. At the instant when ie is zero, B at the centre of the core is zero and at very high frequencies the flux density at the centre of the core is prevented from changing due to the inhibiting effect of the large circulatory eddy currents. The centre of the core is virtually unused under these conditions and the phenomenon is known as the magnetic skin effect. Under these conditions, the flux in the core is concentrated on the surface of the core. At very low frequencies an instantaneous rising coil current ie , would produce a flux linkage λ , in the core. But at larger frequencies, coil current of ie is unable to produce flux linkage of λ , since the circulatory eddy currents has the effect of reducing the flux linkage in the core. A higher volume of coil current of value ie2 is necessary to produce flux linkage of λ , at these higher frequencies.
The result is the broadening of the λ − ie loop towards the right and is shown by
the movement of a to a’. At very low frequencies, when the coil current is derceasing to zero, an instantaneous current of ie3 produces a flux linkage of λ 3 in the core. At high frequencies, the same amount of current of ie3 would produce eddy currents in the core which would tend to produce the same flux linkage of λ 3 , the coil current would have to be decreased to a value of ie4. This has the effect of broadening the λ − i e loop towards the left as shown by the movement from b to b’. The net effect
is a broader λ − ie loop due to eddy currents in the core at high frequency
operation. The crosshatched area of the λ − ie loop is therefore responsible for hysteresis
losses, while the shaded area is due to eddy current losses.
The broadening effect of the λ − ie loop due to eddy currents is increased as the
frequency of the coil supply current increases.
dt
dB n
ha dt e = dφ =
The resulting eddy current flow through the lamination of resistivity ρ and of path length approximated to 2h. The side area of each lamination is given by:
n A al L =
The resistance of the lamination to eddy current flow is given by:
( / )
al n
k h A R = ρl^ = ρ
The power loss in a lamination is given by:
k hn
al dt
dB n
ha R p e 2 ρ
2 2
2 2 2 = =
Power loss in n laminations is given by:
2 (^ )
2 2
2 2 p (^) core _ eddy = neR = kaρn dBdt lah
where lah = volume of core
21 (^ )
2 2 p (^) core _ eddy = (^) na kρ dBdt lah where a / n = lamination thickness For an alternating flux density given by: B = Bm sin ω t dBdt (^) = ω Bm cos ωt = 2 π. fBm cos ω t
∴ p core eddy na kρ dBdt lah π^2 f^2 Bm^22 ωt
2 2 _ = 21 ( ).^4. cos
The core loss due to eddy currents is proportional to: a. The square of the lamination thickness a / n and b. The square of the supply frequency f. Electrical steel sheets, manufactured for the purpose of laminations are covered with a thin surface layer of oxide and then a coat of vanish or higher resistivity inorganic material. These substances are used to insulate the laminates when they are stacked to form a core.
MAGNETIC SYSTEMS
The coil in the torus in Figure 21 extends around the entire circumference resulting in zero magnetic flux density outside the torus.
Figure 21:Coil wound on Torus A ferromagnetic torus with a coil around part of the circumference is shown in Figure 22.
Figure 22: Partially wound iron torus
MAGNETIC CIRCUITS WITH DIFFERENT MATERIALS
Resistive circuits can be made up of many resistive elements. In the same way magnetic circuits can be made up different ferromagnetic materials of different dimensions along the flux path. An example is shown in Figure 24.
Figure 24: Magnetic system with two different materials
Here the magnetic flux is produced in the cast iron section. The flux and flux density in the cast steel section can be computed along the parameters of the two materials. Applying the continuity law for magnetic flux which states:
∫ A B^. d^^ A =^0
r r
for a closed surface in a magnetic field.
i.e.:
Bi Ai = BsA s
or Bi Ai − BsAs = 0
∴ (^) i s s iB A
B = A
Applying Ampere’s Circular Law:
∫ H^ dl =∫ AJd^ A
r (^). r r. r
∫ H^ dl =^ Hi^ li + Hsls
r (^). r
∫ A J^ dA = Ni =^ F
r r .
The equivalent magnetic circuit is shown in Figure 25.
Figure 25: Equivalent Magnetic Circuit for Fig. 24
Ri φ = Fi = Hil i Rs φ = Fs = Hsl s F = Fi + F s
ri o i
i i i i ii ii A
l BA R Hl Hl = φ = = μ μ
rs o s
s s s s ss ss A
l BA R Hl Hl = φ = = μ μ
φ
R R R^ F
= i + s =
SOLUTION:
The equivalent circuit of the magnetic system is shown in Figure 25. Assuming that the flux densities Bi and Bs , and the magnetic field intensities Hi and Hs are uniform throughout the cross sections, applying Ampere’s Law:
a. ∫ =∫
A
H dl Jd A
r r r r
..
∫ H^ dl =^ Hi^ li + Hsls
r r .
Hi and Hs can be determined for each material since the flux through them is given as 0.25 x 10-3^ Wb. Ai = 25 x 25 x 10-6^ m^2 As = 12.5 x 25 x 10-6^ m^2
6250.^25101060.^4
3 = = ×× − =
− i (^) A i B φ^ T
Hi from the B-H curves = 710 A/m
3120.^25. 5 10106 0.^8
3 = = ×× − =
− s (^) A s B φ^ T
Hs from the B-H curves = 480 A/m
2
( 80 25 ) 2100 25 12.^5 + ×
li = − + ^ − +
li = 0.2425 m ls = 30 x 10-3^ m H (^) il i + Hsls = Ni
∴ 710 0.^24255004803100. 373
3 = + = × × × × =
− N i Hil^ i Hsls A
b. R = (^) φF =^ Ni φ =^5000. 25 ××^010.^373 − 3 = 746 × 103 A/Wb
c. B = μr μoH
Bi = μriμoH i
∴ μ = (^) μ = 4 π × 100 .−^47 × 710 = 448 o i ri i H
B
μ = (^) μ = 4 π × 100.^ −^87 × 480 = 1330 o s rs s H
B
d. = (^) μ μ = 448 × 4 π ×^010.^2425 − (^7) × 252 × 10 − 6 = 690 × 103 ri o i i i A R l A/Wb
7 2 6 57.^7103
3 10 = ×
× × × × ×
= = ×
− −
− μ (^) rsμo s π s s A R l A/Wb
The equivalent magnetic circuit for the system is shown in Figure 28.
Figure 28: Equivalent Magnetic Circuit for System
SOLUTION:
1 3 7 3 6 0.^531106
300 10 = ×
× × × ×
= = = ×
− −
− μA π R R l A/Wb
2 7 3 6 0.^148106
100 10 = ×
× × × ×
= ×
− −
− R (^) π A/Wb
The mmf of the two loops are: F = R 1 φ 1 + R 2 φ 2 Eqn. R 2 (^) φ (^) 2 = R 3 φ 3 Eqn.
and φ 1 (^) = φ 2 + φ 3 Eqn.
Equations 1, 2 and 3 are three equations with three solutions. 25 × 10 −^6 = 0. 531 φ 1 + 0. 148 φ 2 0 = 0. 148 φ 2 (^) + 0. 531 φ 3 0 =− φ 1 + φ 2 + φ 3
which yields: φ 1 = 19.3 x 10-6^ Wb φ 2 = 15.1 x 10-6^ Wb φ 3 = 4.21 x 10-6^ Wb
- (^3101060). 0965 6 1
×
= = ×
−
− B (^) A φ (^) T
6 2
×
= = ×
−
− B (^) A φ (^) T
- (^21101060). 0210 6 3
×
= = ×
−
− B (^) A φ (^) T
