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MAGNETISM

MAGNETISM. A. Magnetic Field. 1. Magnetic Field Around Electric Current. In Oersted's experiment, a compass is placed directly over a horizontal wire ( h ere viewed from above). When the compass is placed directly under the wire, the compass deflection is r eversed . 2. Biot-Savart Law.

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MAGNETISM

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  1. MAGNETISM

  2. A. Magnetic Field 1. Magnetic Field Around Electric Current • In Oersted's experiment, a compassis placed directly over a horizontal wire (here viewed from above). When the compass is placed directly under the wire, the compass deflection is reversed.

  3. 2. Biot-Savart Law P The magnetic field is dB at some point P r a  i for infinitely long so the limits of is + ~ to - ~ and the limits of is  to 0. a/r = sin  r = a/sin  = a csc 

  4. i a P Biot– Savart Law to define magnetic fieldataround long straight wire. Contoh Soal 1 Suatu kawat lurus panjang dialiri arus sebesar 5 A. Berada di ruang hampa. Tentukan besarnya induksi magnet pada sebuah titik yang berada 10 cm di sebelah kanan kawat, bila kawat tersebut vertikal dan kemana arah induksi magnetnya. Jawab : i = 5 A a = 10 cm = 0,1 m o = 4 10 - 7weber/amp. meter B = 10- 5 weber/m2

  5. x x x Besarnya induksi magnetik di titik P adalah 10- 5 weber/m2. x x x x x x x x x x x x x x x x x Sedangkan bila ditentukan dengan kaidah tangan kanan, arah induksi magnetik adalah: Arah induksi magnet di sebelah kiri kawat adalah keluar bidang kertas, dan di sebelah kanan kawat masuk ke bidang kertas.

  6. 3. Magnetic Field at the center of circular loop of wire a dB cos r dB   dB sin  Vector dB have two component, dB sin and dB cos. The total component vectors dB cos is zero, so

  7. Thus, the total magnetic field at pointP is, If point P is center of circular, than r = a and  = 90o. Magnetic field at center of circular is, For the thin spring with N turns of wire, magnetic field at center point of curcular is,

  8. Contoh Soal 2 Sebuah kawat berupa lingkaran dengan jari-jari 3 cm, dialiri arus listrik sebesar 10 A. Tentukan induksi magnet pada sumbu kawat tersebut yang berjarak 5 cm dari keliling lingkaran kawat. a r  dB sin  Jawab Diketahui a = 3 cm = 0,03 m r = 5 cm = 0,05 m sin = a/r = 3/5 P = 1,44  10- 5 weber/m2

  9. 4. Magnetic Field of a Solenoid X X X X X X X X X X X X X A P C D F dx For example there is solenoid containdN turns of wire The number of turns per unit length is, The radius of solenoid isa Thus, magnetic field at point P in the axis of solenoid by wire element with lengthdx is

  10.  = angle between r and x, whereasa/r = sin  So, r = a/sin  = a cosec . Because x = a cotg, So,dx = - cosec2  d. The Number of magnetic field at pointP by all of length of solenoid wire: If the solinoid have infinetly long, than the limits of the angle become1 = 0oand2 = 180o, If pointP at the center of solenoid, thanmagnetic field at pointP:

  11. If point P at left end of the solenoid,that equation become (for1 = 0oand2 = 90o) : ContohSoal 3 Suatusolenoidapanjangnya 2 meter dengan 800 lilitan. Bilasolenoidaitudialiriarussebesar 0,5 A. Tentukanlahinduksi magnet padaujungsolenoida yang berjari-jari 2 cm. Jawab N = 800 lilitandanl = 2 meter sehingga n = N/l = 800/2 = 400 lilitan/m. 0 = 4  10- 7weber/amp.mdan I = 0,5 A, makadiperolehbesarmedan magnet : B = 0 I n = (4  10- 7)(0,5)(400) = 8 10- 5weber/m2

  12. Magnetic Field of a Toroid Circular of a solenoid is a toroid • The number of Magnetic Field of a Toroidis: B = 0 I n

  13. + • 5. The Lorentz Force Law l = conductor length i = current B = uniform magnetic field •  = angle between andB  Becausei = q/t , so B v F = q v B sin  F = q v  B q = charge (coulomb) v = charge velocity (m/s)

  14. Define the direction of Lorentz Force F B  i ContohSoal 4 Sebuahpartikelbermuatan 1 C bergerakdengankelajuan 103 m/s dalammedan magnet homogensebesar 10- 5weber/m2. Arahgerakpartikeltegaklurusterhadaparahmedan magnet. Tentukanbesarnyagaya Lorentz yang dialamipartikeltersebut. Jawab q = 1 C = 10- 6 C  = 90osehingga sin 90o = 1 v = 103 m/s B = 10- 5weber/m2 F = qvb sin  = (10- 6)( 103)( 10- 5) sin 90o = 10- 8newton

  15. + i2 i1 B1 B2 F2 F1 a (1) (2) At each wire length exert Lorentz force. At wire (1) At wire (2) Thus, the magnitudeof F1 = F2is

  16. ContohSoal 5 i1 i2 P (1) (2) • Duabuahkawatpenghantarberaruslistriksangatpanjang, jarakantarakeduanya 20 cm, bila i1 = 2 A dan i2 = 4 A, tentukaninduksi magnet pada P yang beradatepatdiantarakeduakawat. Jawab Olehkawat (1), dengan i1 = 2 A dan r1 = 10 cm = 0,1 m, diperoleh : arahnyategakluruskedalambidangkertas Olehkawat (2), dengan i2 = 4 A dan r2 = 10 cm = 0,1 m, diperoleh : arahnyategakluruskeluarbidangkertas Karena B1dan B2berlawananarah, makaresultankeduannyaadalahB2 - B1 = (8 10- 6 - 4 10- 6) = 4 10- 6weber/m2 denganarahsesuaidengan B2.

  17. B. Electromagnetic induction U G G i U i 1. The induced emf a. Faraday’s law • An induced current is produced by a changing magnetic field • There is an induced emf associated with the induced current

  18. EMF Produced by a Changing Magnetic Field, 1 • A loop of wire is connected to a sensitive ammeter • When a magnet is moved toward the loop, the ammeter deflects • The direction was chosen to be toward the right arbitrarily

  19. EMF Produced by a Changing Magnetic Field, 2 • When the magnet is held stationary, there is no deflection of the ammeter • Therefore, there is no induced current • Even though the magnet is in the loop

  20. EMF Produced by a Changing Magnetic Field, 3 • The magnet is moved away from the loop • The ammeter deflects in the opposite direction

  21. EMF Produced by a Changing Magnetic Field, Summary • The ammeter deflects when the magnet is moving toward or away from the loop • The ammeter also deflects when the loop is moved toward or away from the magnet • Therefore, the loop detects that the magnet is moving relative to it • We relate this detection to a change in the magnetic field • This is the induced current that is produced by an induced emf

  22. b. Motional emf x x x x x x x x x x x x x x x x a a’ d x x x x x x x x x x x x x x x x v x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x c b b’ s A motional emf is one induced in a conductor moving through a constant magnetic field a + Vectorsat rodab v F - b

  23. x x x x x x x x x x x x For example, a conductorabmoving with displacement s,velocity v and time t, thus,the work is W = - F.s. W = i t, so that i  t = - F.s = - B il.s  t = - B l.s  = - B l.(s/ t) = - B l v Is potential differencebetween a and b, can be regarded as an induced emf. ContohSoal 6 x x x x x x x x x x x x P v Q Induksi magnet homogen B = 2 x 10- 2 weber/m2tegaklurusmasukbidangkertas, kawat PQ panjangnya 0,5 meter digerakkansepertipadaGambardengankecepatan 100 m/s. Tentukanbesarnya GGL induksi yang timbulpadakawat PQ.

  24. Jawab Denganmenggunakanpersamaan  = - B l v, dapatdihitung:  = - (2 x 10- 2 )(0,5)(100) = - 1 volt Jadi GGL induksipadakawat PQ sebesar 1 volt denganarahdari Q ke P. 2. Laws of Electromagnetic induction a. Magnetic Flux Magnetic Flux is number of magnetic force lines on certain region area on perpendicular direction. B  = B A = magnetic flux(weber), B = Dencity of magnetic force lines/magnetic field (weber/m2), A = area of region to be coveredby B (m2) A

  25. Faraday’s law of induction states that “the emf induced in a circuit is directly proportional to the time rate of change of the magnetic flux through the circuit”  =- /t  =- d/dt For the instantaneous induced is: If the circuit consists of N loops,than an emf is induced in every loop and Faraday’s law becomes Negative sign indicate direction Lenz,s law.

  26. ContohSoal 7 Suatukumparandengan 500 lilitandiberikanmedan magnet. Apabilaterjadiperubahanfluks magnet sebesar 2 10- 3weberdalamwaktu 1 detik, tentukanbesarnyagayageraklistrikinduksi yang timbulpadaujung-ujungkumparanitu. Jawab  = -500 2.10- 3/1 = - 1 volt

  27. 3. Implementations ofMagnetic Induction primary (input) secondary (output) a. Transformer • V1 : V2 = N1 : N2 V1= Potential at primary coils, • V2= Potential at secondary coils, • N1= Quantities of primary coils, • N2 = Quantities of secondary coils Ideal transformer V1 i1 =V2 i2 • i1 = Current at primary coils, • i2 = Current at secondary coils

  28. Actuality there is no ideal transformer, there is dissipation energy. Dissipation energy coused by (1) Joule warming, and (2) warming of eddy current. • Magnitude oftransformerefficiencyis: ContohSoal 8 Sebuahtransformatorstep downmempunyaiefisiensi 80%, jumlahlilitan primer 1000 lilitan, sedangkansekundernya 500 lilitan, apabiladaya yang diberikanpadaprimernya 2000 watt dengankuatarus 4 ampere. Tentukan (a) dayapadasekundernya, dan (b) Kuataruspadasekundernya.

  29. Jawab • a) 80% = P2/2000 x 100%, sehinggadiperoleh P2 = 1600 watt (b) Tegangan primer: P1 = V1 i1 2000 = V1 4, sehinggadiperoleh V1 = 500 volt Tegangansekunder: V1 : V2 = N1 : N2 500 : V2 = 1000 : 500, sehinggadiperoleh V2 = 250 volt Jadikuataruspadasekunder i2 = P2/V2 = 1600/250 = 6,4 ampere

  30. b. Dynamoand Alternator B A C K D S  • Dynamo is device changing agent mechanical energy to electrical energy. • Generally dynamo named generator. • Alternator is dynamoproduced alternating current.

  31. E Thus,induced emf of ac generator is sinus function. A B max t max t C D Work principle of dc generator is,

  32. 4. Self-Inductance P P L L S S When closed circuit, like figure, at first the lamp P to light. Then turn switch S of, butthe lamp P still to lihgt in a moment,because arise self-induction currentcoused by change of magnetic flux at coil L, from there is current became there is not current. The lamp light when the circuit is closed. The lamp light when the circuit start opened

  33. Self-Induction current appearance at a coil produced self-induced emf. L = self-induced (henry), • di/dt = The rates of current change(A/s), •  = self-induced emf (dalam volt)

  34. Definition: Self-inductanceis 1 henry, if at coils appear self-induced emf 1 volt withrates of current change 1 ampere persecond. L = Self-induced of coil (henry), • N = Quantities of coils, = Magnetic flux on coils, i= Current on coils (ampere) ContohSoal 9 Padasebuahkumparan yang mempunyai 500 lilitan, terjadiperubahancepatfluksmagnetnya 0,05 weber/s danperubahancepatkuatarusnya 0,1 ampere/s. Tentukanlah (a) induktansidirikumparan, (b) GGL induksidirikumparan Jawab (a) (b) Jadiinduktansidirinya 250 henrydan GGL induktansidirisebesar 25 volt.

  35. Magnitude of magnetic inductions on a toroidis: B = 0i n = 0i N/l Whereas magnetic flux on toroid is:  = B A = A0i N/l. Li/N = A0i N/ L = Self-induced (henry) o= Magnetic permebility for vacum (4 10- 7weber/amp.m) • N= Number of Turn of wire = length of solenoid or coil (m), A = Cross section Areaof coil or solenoid (m2).

  36. ContohSoal 10 SebuahSolenoidadenganluaspenampang 5 cm2danpanjangnya50 cm dengan 500 buahlilitan. Berapakahlinduktansidirisolenoidatersebut. Jawab A = 5 cm2 = 5 10- 4 m2 ; N = 500 = 50 cm = 0,5 m ; o = 4 10- 7weber/amp.m

  37. Energyinthe inductor L i i b a When the inductor L flowed current I always change regarding to time, so defference potential between point a and b is: Thus power given in inductor is: During time interval dt, energygivenin inductor is:

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