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Unformatted text preview: KOLEJ KEJURUTERAAN SEMESTER II 2002/2003 PROGRAM : B.Eng. Hons (EE/EP) –Year 3 PROGRAM MATAPELAJARAN : ELECTROMAGNETICS SUBJECT KOD MATAPELAJARAN : EEEB354 SUBJECT CODE TARIKH : 19 hb Mac 2003 DATE MASA : . 2.30 pm –5.30 pm TIME Arahan kepada calon: 1. Jawab TUJUH (7) SOALAN sahaja . EMPAT (4) SOALAN DARI BAHAGIAN A DAN TIGA(3) SOALAN DARI BAHAGIAN B. Markah yang diberi untuk semua soalan adalah sama. 2. Jumlah keseluruhan soalan ini ialah: 10 (SEPULUH) -BAHAGIAN A : LIMA (5) SOALAN. BAHAGIAN B : LIMA (5) SOALAN 3. Sebarang buku rujukan tidak boleh dibawa masuk ke dalam dewan peperiksaan. 4. Semua lukisan, lakaran & pengiraan mesti dibuat di dalam buku jawapan peperiksaan. 5. Masa yang dibenarkan ialah 3 jam. JANGAN BUKA BUKU KERTAS SOALAN INI SEHINGGA ANDA DIARAH BERBUAT DEMIKIAN Instruction to candidates: 1. Answer SEVEN (7) questions only. FOUR (4) QUESTIONS FROM SECTION A AND THREE (3) QUESTIONS FROM SECTION B . Equal marks will be given to all questions. 2. Total number of questions : 10 (TEN) – SECTION A : FIVE (5) QUESTIONS SECTION B : FIVE (5) QUESTIONS 3. This is a closed book examination. Any reference books are not allowed into the examination hall. 4. All drawings, sketching & calculations are to be done in the examination answer booklet. 5. Maximum time allowed is three hours. DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE INSTRUCTED TO DO SO. THIS QUESTION PAPER CONSISTS OF TWELVE (12) PRINTED PAGES INCLUDING THIS PAGE . TWO (2) ADDITIONAL PAGES CONTAINING A LIST OF FORMULAE ARE ALSO PROVIDED SECTION A SOLUTION Page 1 of 12 Question 1 Part (a) By using Ampere’s Circuital Law, find the magnetic flux density B at a point located at a distance r from an infinite straight wire carrying a current I. (3 marks) By applying Ampere’s law, I r H = π 2 r I H π 2 = Assuming free space, B = μ H, r I B π μ 2 = or Tesla 2 φ a B r I π μ = Part (b) Two infinitely long thin wires separated by a distance 6 w, each carries a current I in opposite directions. Obtain an expression for the magnetic flux density B at a point P located at a distance of 2 w from one of the wires (between wires). (3 marks) Using the expression for B in part a) and assuming that the currents are flowing in the a z and –a z direction respectively, we can now calculate the value of B due to each wire separately. The value of B due at a distance of 2 w from the current flowing in the a z direction is φ φ 1 a a B w I w I π μ π μ 4 2 2 = = The value of B due at a distance of 4 w from the current flowing in the - a z direction is φ φ 2 a a B w I w I π μ π μ 8 4 2 = = Based on the principle of superposition, the resultant value of B due to the 2 currents in opposite directions is Tesla 8 3 8 2 8 4 2 1 w I w I I w I w I π μ π μ μ π μ π μ = + = + = + = φ φ φ a a a B B B Part (c) Currents in the inner and outer conductors of Figure 1 (c) are uniformly distributed. Use Ampere’s Law to find the magnetic field intensity H for the region b ≤ r ≤ c.region b ≤ r ≤ c....
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This homework help was uploaded on 04/13/2008 for the course ECE 221 taught by Professor N/a during the Fall '05 term at University of Wisconsin.
- Fall '05