medan elektromagnetik 2-9
TRANSCRIPT
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ME-2-9 1
9. Gaya Magnetik, Bahan Magnetik dan Induktansi
Dr. Achmad [email protected] Antena & PropagasiRuang B-306
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ME-2-9 2
Pustaka Elektromagnetika Edisi 7 oleh William H.Hayt
dan John A.Buck, Penerbit Erlangga, 2006 Fundamentals of Applied Electromagnetics
by Fawwaz T. Ulaby, Prentice Hall International Inc., 2001
Electomagnetics by Joseph A.Edminister, Schaum’s ouline Series
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outline Gaya pd muatan bergerak Gaya pd unsur arus diferensial Gaya antara unsur arus diferensial Gaya dan torka Magnetisasi Syarat batas magnetik Rangkaian magnetik Induktansi & induktansi bersama
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Gaya pd partikel
Persamaan Gaya Lorentz
IdLxBdFKxBdSdFJxBdvdF
Gaya pada unsur arus diferensial •Gaya pada partikel bermuatan bergerak melalui medan magnetik•dF = dQ v x B
IdLxBF
KxBdSF
JxBdvF
s
vol
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Arus yg sama dlm arah ke dalam bahan disediakan oleh(a) Muatan positif(b) Muatan negatif
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Gaya Lorentz
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Right Hand Rule
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Gaya magnetik (Magnetic Force) Gaya pada unsur arus diferensial Notasi F Satuan Newton
LBBL dIIdF
ILxBF
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Gaya antara unsur-unsur arus diferensial
212
12112
4 R
adLIdH R
BIdLdF
2222 dBdLIdFd
1212212
21024
RadLdLR
IIdFd
2212
1122102
12
1212
2102 44
dLR
dLaIIRadL
dLII
F RR
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contoh
z
zz
ax
HHB
mAax
axIH
67
0103
104
/215
2
dLBIF
pNaF x8
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Gaya antar arus arah sama
mNdI
F /2
20
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Gaya antar arus berlawanan
mNdI
F /2
20
Dua utas filamen sejajar berjarak d yg dialiri arus sama besar ttp berlawanan arahgaya total tolak menolak
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Torka/Torsi
Atau momen gaya Arah torka normal terhadap keduanya (gaya F dan
lengan pengumpil R) dan dalam arah maju sekerup putar kanan
Notasi T, satuan Nm (Newton meter)m = momen dwikutub magnetik (A.m2)S = luas (m2)
FRT BmBST I
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Gaya & Torka
121121
21
2211
0FRFRRT
FFFRFRT
FRT
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contoh yzzy
x
aBaBIdxdFBIdxadF
sisipadavektorGaya
001
01
1____
xy
yzzyy
y
adxdyIB
aBaBIdxdyadFRdT
dyaR
0
00111
1
21
21
21
yx
xy
adxdyIBdTdT
adxdyIBdTdTdTdT
042
031
13
BIdSdT
BaIdxdyaBaBIdxdydT zxyyx
000
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dT=IdSXBdS : luas vektor dari sosok diferensial
dm = I dSdm : momen dwikutub magnetikdT = dm X B
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contoh
mNaF
mNaaaaBILF
mNaaF
mNaaaaaBILF
x
xzyy
zy
zyzyx
4,6
4,68,06,02104
4,22,3
4,22,38,06,01104
4
3022
3
3011
mmNaaaTFRFRFRFRT
xxx .8,44,24,244332211
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Torsi pada loop berarus
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Sifat Bahan Magnetik
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Magnetics Properties
Materialm = R-1(x 10-5)ParamagneticIron oxide (FeO)720Iron amonium alum66Uranium40Platinum26Tungsten6.8Cesium5.1Aluminum2.2Lithium1.4Magnesium1.2Sodium0.72Oxygen gas0.19
DiamagneticAmmonia-.26Bismuth-16.6Mercury-2.9Silver-2.6Carbon (diamond)-2.1Carbon (graphite)-1.6Lead-1.8Sodium chloride-1.4Copper-1.0Water-0.91
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Magnetisasi dan Permeabilitas
I = arus bebas, IT = arus total, Ib = arus terikat
M=magnetisasi
IdLMB
dLMI
IdLB
b
T
.
.
.
0
0
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bJM
TJB
0
JH
Kerapatan arus
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Magnetisasi & Permeabilitas
0
00
1)(
R
mR
Rm
m
HHHBHM
Dimana : M = magnetisasim = suseptibilitas magnetik (kerentanan magnetik)R = permeabilitas relatif
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contoh Bahan ferit dg R=50 dan bekerja dg
kerapatan fluks rendahhubungan linier dpt dipakai
B=0,05 Tm= R-1= 50-1=49
B= R 0HH=0,05/50x4x10-7=796A/m
Magnetisasi : M=mH=39000A/mB= 0 (H+M)
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Syarat batas magnetik
121211 NNN aaBB
KaHHBB
Ntt
NN
1221
12
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BN1S-BN2 S=0BN1- BN2 = 0
Ht1L – Ht2L = KLHt1– Ht2= K(H1 – H2) x aN12 = K
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contoh Daerah 1 1=4 H/m, z>0 Daerah 2 2=7 H/m, z<0 K=80 ax A/m pada z=0 B1=2ax-3ay+az mT di daerah 1 Tentukan B2
BN1=(B1-aN12)aN12=az mTBN2=BN1=az
Bt1=B1-BN1=2ax - 3ay Ht1=Bt1/1=500ax – 750ay A/mHt2=Ht1-aN12xK=500ax -670ay A/m Bt2= 2Ht2=3,5ax-4,69ay mTB2=BN2 + Bt2 =3,5ax – 4,69ay + az mT
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Rangkaian Magnetik Magnetomotansi (arus magnetomotoris) : Vm
Satuan : ampere-lilit : Aturn (A.t)
Reluctance Rasio magnetomotansi dengan fluks total Satuan : ampere-lilit perweber (A.t/Wb) S
dVm
LH dVm
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contoh Toroida berteras udara dg N=500 lilitan Luas penampang S=6 cm2
Jejari r=15 cm Arus kumparan I=4 A Vm,sumber=2000A.t =d/S =1,25x109 A.t/Wb =Vm,s/ = 1,6x10-6 Wb B=/S=2,67x10-3 T H=B/=2120 A/m Atau H2r=NI H=NI/2r=2120 A/m
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Aplikasi Feromagnetik : Transformers
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Aplikasi Feromagnetik : Electromagnets
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Aplikasi Feromagnetik : Magnetic Tape recording
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Transformator
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Motor DC
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Induktansi (induktansi diri) Hasil bagi pertautan fluks total (Flux Linkage)
dengan arus, satuan : H (Henry)
N = jumlah lilitan
INL
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Konduktor sesumbu dengan jari-jari konduktor dalam a, konduktor luar b<r<c
abId
ln20
abd
L ln20
abL ln
20
•Indukatansi untuk panjang d
•Induktansi permeter
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•Induktansi timbal balikPertautan fluks timbal balikM12=M21
1
12212 I
NM
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Contoh : Induktansi-diri dan timbal balik pd 2 buah solenoida sesumbu berjejari R1 dan R2 (R2>R1) dialiri arus I1 dan I2 dengan n1 dan n2 lilitan/m
)0()0(
2222
1111
RaInHRaInH
z
z
2121012
2111012
RnnM
RIn
122121021
2122021
MRnnM
RIn
Dg. Cara yg sama 2
2122
1
1211
In
L
In
L
Induktansi :
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Latihan-1Rangkaian Magnetik spt gambar. Anggap B=0,8 Wb/m2 pada titik di tengah kaki kirinya. Hitung (a) Vm,udara (b)Vm,baja (c) Arus yang diperlukan dlm kumparan 1.500 lilitan yang melilit kaki kirinya
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Latihan-2
Dua buah kumparan toroida berpenampang bujur sangkar spt gambar. Jika kumparan dalamnya 500 lilitan dan kumparan luarnya 4.000 lilitan, tentukan (a) Ldalam (b) Lluar (c) M
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Latihan-3
Sebuah Solenoida panjang berjejari 3 cm dan mempunyai 5.000 lilitan/m, masing-masing lilitan dialiri arus 0,25 A. Dalam daerah 0<<a dlm solenoida R=5, sedangkan utk a<<3 cm R=1. Tentukan a shg (a) terdapat fluks total sebesar 10 Wb (b) fluks tsb terbagi sama antara kedua daerah