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Kelas : XII Semester : 1

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INDUKSI MAGNET. Mata Pelajaran : FISIKA. Kelas : XII Semester : 1. SK - KD. Standar Kompetensi 2 . Menerapkan konsep kelistrikan dan kemagnetan dalam berbagai penyelesaian masalah dan produk teknologi. Kompetensi Dasar - PowerPoint PPT Presentation
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Slide 1

Arah induksi magnet dapat ditentukan dengan menggunakan Kaidah Tangan Kanan Pertama sebagai berikut:

Arah ibu jari sebagai arah kuat arus listrik i Arah lipatan empat jari lain sebagai arah induksi magnet BMateri

Design By IT SUPPORT SMANTIEBahan AjarSMAN 3 [email protected] Soal: (1) Sebuah kawat lurus dialiri arus listrik 10 A. Tentukanlah induksi magnet di sekitar kawat itu yang berjarak 20 cm ! Penyelesaian : Diketahui : i = 10 A Ditanya: B ? a = 20 cm = 2 x 10 m = 4 x 10 Wb/Am Jawab : B = I 2 a = (4 x 10) (10) (2) (2 x 10) = 4 x 10 Wb/mMateriDesign By IT SUPPORT SMANTIEBahan AjarSMAN 3 [email protected] Soal: (2) i = 3 A iThe diagram on the side shows two long,Straight, and parallel wires. Current of 3 AAre flowing through the wires in opposite PDirections. Determine the magnitude of 0,4 mThe magnetic field induction at point P ! 0,5 mSolution : Electrical current I= 3 ADistance from P to wire 1, a = 0,5 0,4 = 0,1 mDistance from P to wire 2, a = 0,4 m i B B i MateriDesign By IT SUPPORT SMANTIEBahan AjarSMAN 3 [email protected] magnitudes 0f B and B are calculated :B = i B = i 2 a 2 a = (4 x 10) (3) = (4 x 10) (3) (2) (10) (2) (4 x 10) = 6 x 10 T = 1,5 x 10 TThe magnetic induction at P is the resultant of vectors B and B. Since B and B are vectors in the same direction , then their resultant is the algebraic sum of their magnitudes Bp = B + B = (6 x 10) + (1,5 x 10) = 7,5 x 10 THence, the magnitude of magnetic induction at point P is 7,5 x 10 TMateriDesign By IT SUPPORT SMANTIEBahan AjarSMAN 3 [email protected] PencapaianMendeskripsikan induksi magnetik di sekitar kawat berarus (lurus, melingkar, solenoida)

Design By IT SUPPORT SMANTIEBahan AjarSMAN 3 [email protected] Soal:Sebuah kumparan terdiri dari 10 liitan dengan jari jari 20 cm, dialiri arus sebesar 5 A. Tentukanlah induksi magnetik di pusat lingkaran ! a Penyelesaian : Diketahui : a = 20 cm P = 2 x 10 m i = 5 A Jawab : Bp = N i 2a = (10) (4 x 10) (5) 2 (2 x 10) = 5 x 10 T MateriDesign By IT SUPPORT SMANTIEBahan AjarSMAN 3 [email protected] HUKUM AMPERE 1.1. Induksi Magnet ik di Sekitar Kawat Lurus Beraru Besar Induksi magnetik di sekitar kawat lurus berarus dapat dirumuskan sebagai : i B P Bp = 0 i 2 a

Di mana: Bp = induksi magnet di titik P ( Wb/m) 0 = permeabilitas bahan = 4 . 10 Wb/Am) i = kuat arus (A) a = jarak titik sembarang terhadap kawat berarus (m)MateriDesign By IT SUPPORT SMANTIEBahan AjarSMAN 3 [email protected] & EditorPENYUSUNNama:AMI HASMI UTAMASARI, M.Pd.Institusi:SMA Negeri 3 Kota SukabumiE-mail:[email protected]:Design By IT SUPPORT SMANTIEBahan AjarSMAN 3 [email protected] : 1. Fisika 3 SMA Kelas 12, Purwoko, Pendi (Yudhistira) 2. LKS Fisika (SIMPATI SMA) XII Semester 1 ( Grahadi) 3. Physics for Senior High School 1 Semester , Marthen Kanginan (Erlangga)

ReferensiDesign By IT SUPPORT SMANTIEBahan AjarSMAN 3 [email protected]

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