A. Magnetic Field 1. Magnetic Field Around Electric Current In Oersted's experiment, a compass is placed directly over a horizontal wire (here viewed.

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

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

5 Biot – Savart Law to define magnetic field at around 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. P a i Jawab : i = 5 A a = 10 cm = 0,1 m  o = 4  10 - 7 weber/amp. meter B = 10 - 5 weber/m 2

6 Besarnya induksi magnetik di titik P adalah 10 - 5 weber/m 2. xxxx xxxx xxxx xxxx xxxx 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.

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

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

9 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 P Diketahui a = 3 cm = 0,03 m r = 5 cm = 0,05 m sin  = a/r = 3/5 = 1,44  10 - 5 weber/m 2

10 4. Magnetic Field of a Solenoid XXXXXXXXXXXXX C PA D F dx For example there is solenoid containd N turns of wire The number of turns per unit length is, The radius of solenoid is a Thus, magnetic field at point P in the axis of solenoid by wire element with length dx is

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

12 If point P at left end of the solenoid, that equation become (for  1 = 0 o and  2 = 90 o ) : Contoh Soal 3 Suatu solenoida panjangnya 2 meter dengan 800 lilitan. Bila solenoida itu dialiri arus sebesar 0,5 A. Tentukanlah induksi magnet pada ujung solenoida yang berjari-jari 2 cm. Jawab N = 800 lilitan dan l = 2 meter sehingga n = N/ l = 800/2 = 400 lilitan/m.  0 = 4  10 - 7 weber/amp.m dan I = 0,5 A, maka diperoleh besar medan magnet : B =  0 I n = (4  10 - 7 )(0,5)(400) = 8  10 - 5 weber/m 2

13 Magnetic Field of a Toroid Circular of a solenoid is a toroid The number of Magnetic Field of a Toroid is : B =  0 I n

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

15 Define the direction of Lorentz Force F  B i Contoh Soal 4 Sebuah partikel bermuatan 1  C bergerak dengan kelajuan 10 3 m/s dalam medan magnet homogen sebesar 10 - 5 weber/m 2. Arah gerak partikel tegak lurus terhadap arah medan magnet. Tentukan besarnya gaya Lorentz yang dialami partikel tersebut. Jawab q = 1  C = 10 - 6 C  = 90 o sehingga sin 90 o = 1 v = 10 3 m/sB = 10 - 5 weber/m 2 F = qvb sin  = (10 - 6 )( 10 3 )( 10 - 5 ) sin 90 o = 10 - 8 newton

16 + i2i2 i1i1 F2F2 F1F1 a (1)(2) B2B2 B1B1 At each wire length exert Lorentz force. At wire (1) At wire (2) Thus, the magnitude of F 1 = F 2 is

17 Contoh Soal 5 i2i2 i1i1 P (2) (1) Jawab Oleh kawat (1), dengan i 1 = 2 A dan r 1 = 10 cm = 0,1 m, diperoleh : Dua buah kawat penghantar berarus listrik sangat panjang, jarak antara keduanya 20 cm, bila i 1 = 2 A dan i 2 = 4 A, tentukan induksi magnet pada P yang berada tepat di antara kedua kawat. arahnya tegak lurus ke dalam bidang kertas Oleh kawat (2), dengan i 2 = 4 A dan r 2 = 10 cm = 0,1 m, diperoleh : arahnya tegak lurus ke luar bidang kertas Karena B 1 dan B 2 berlawanan arah, maka resultan keduannya adalah B 2 - B 1 = (8 10 - 6 - 4 10 - 6 ) = 4 10 - 6 weber/m 2 dengan arah sesuai dengan B 2.

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

19 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

20 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

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

22 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

23 b. Motional emf a’ d a b’ b c s 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 v A motional emf is one induced in a conductor moving through a constant magnetic field a b 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 v F + - Vectors at rod ab

24 For example, a conductor ab moving 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 i l.s  t = - B l.s  = - B l.(s/ t)  = - B l v  Is potential difference between a and b, can be regarded as an induced emf. Contoh Soal 6 xxx xxx xxx xxx xxx xxx xxx xxx P v Q Induksi magnet homogen B = 2 x 10 - 2 weber/m 2 tegak lurus masuk bidang kertas, kawat PQ panjangnya 0,5 meter digerakkan seperti pada Gambar dengan kecepatan 100 m/s. Tentukan besarnya GGL induksi yang timbul pada kawat PQ.

25 Jawab Dengan menggunakan persamaan  = - B l v, dapat dihitung:  = - (2 x 10 - 2 )(0,5)(100) = - 1 volt Jadi GGL induksi pada kawat PQ sebesar 1 volt dengan arah dari 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 A A B  = magnetic flux(weber), B = Dencity of magnetic force lines/magnetic field (weber/m 2 ), A = area of region to be covered by B (m 2 )

26 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 For the instantaneous induced is:  = - d  /dt 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.

27 Contoh Soal 7 Suatu kumparan dengan 500 lilitan diberikan medan magnet. Apabila terjadi perubahan fluks magnet sebesar 2 10 - 3 weber dalam waktu 1 detik, tentukan besarnya gaya gerak listrik induksi yang timbul pada ujung-ujung kumparan itu. Jawab  = -500 2.10 - 3 /1 = - 1 volt

28 3. Implementations of Magnetic Induction a. Transformer secondary (output) primary (input) V 1 = Potential at primary coils, V 2 = Potential at secondary coils, N 1 = Quantities of primary coils, N 2 = Quantities of secondary coils V 1 : V 2 = N 1 : N 2 Ideal transformer V 1 i 1 =V 2 i 2 i 1 = Current at primary coils, i 2 = Current at secondary coils

29 Actuality there is no ideal transformer, there is dissipation energy. Dissipation energy coused by (1) Joule warming, and (2) warming of eddy current. Magnitude of transformer efficiency is: Contoh Soal 8 Sebuah transformator step down mempunyai efisiensi 80%, jumlah lilitan primer 1000 lilitan, sedangkan sekundernya 500 lilitan, apabila daya yang diberikan pada primernya 2000 watt dengan kuat arus 4 ampere. Tentukan (a) daya pada sekundernya, dan (b) Kuat arus pada sekundernya.

30 Jawab a) 80% = P 2 /2000 x 100%, sehingga diperoleh P 2 = 1600 watt (b) Tegangan primer: P 1 = V 1 i 1 2000 = V 1 4, sehingga diperoleh V 1 = 500 volt Tegangan sekunder: V 1 : V 2 = N 1 : N 2 500 : V 2 = 1000 : 500, sehingga diperoleh V 2 = 250 volt Jadi kuat arus pada sekunder i 2 = P 2 /V 2 = 1600/250 = 6,4 ampere

31 b. Dynamo and Alternator Dynamo is device changing agent mechanical energy to electrical energy. Generally dynamo named generator. Alternator is dynamo produced alternating current. B A C D S K 

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

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

37 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)

38 Definition: Self-inductance is 1 henry, if at coils appear self-induced emf 1 volt with rates of current change 1 ampere per second. L = Self-induced of coil (henry), N = Quantities of coils,  = Magnetic flux on coils, i = Current on coils (ampere) Contoh Soal 9 Pada sebuah kumparan yang mempunyai 500 lilitan, terjadi perubahan cepat fluks magnetnya 0,05 weber/s dan perubahan cepat kuat arusnya 0,1 ampere/s. Tentukanlah (a) induktansi diri kumparan, (b) GGL induksi diri kumparan Jawab (a) (b)(b) Jadi induktansi dirinya 250 henry dan GGL induktansi diri sebesar 25 volt.

39 Magnitude of magnetic inductions on a toroid is: B =  0 i n =  0 i N/l Whereas magnetic flux on toroid is:  = B A = A  0 i N/l. Li/N = A  0 i N/ L = Self-induced (henry)  o = Magnetic permebility for vacum (4  10 - 7 weber/amp.m) N= Number of Turn of wire = length of solenoid or coil (m), A = Cross section Area of coil or solenoid (m 2 ).

40 Contoh Soal 10 Sebuah Solenoida dengan luas penampang 5 cm 2 dan panjangnya 50 cm dengan 500 buah lilitan. Berapakah linduktansi diri solenoida tersebut. Jawab A = 5 cm 2 = 5 10 - 4 m 2 ; N = 500 = 50 cm = 0,5 m ;  o = 4  10 - 7 weber/amp.m

41 Thus power given in inductor is: During time interval dt, energy given in inductor is: Energy in the 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: