Gauss Law - Electricity and Magnetism - Solved Exam, Exams of Electromagnetism and Electromagnetic Fields Theory

Download Gauss Law - Electricity and Magnetism - Solved Exam and more Exams Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! PC2131 Electricity and Magnetism I 07/08 Semester I Exam Solutions Part I 1. By Gauss’s Law,    encQadE  Outside, r r q E q r ˆ 4 ) E(4 2 00 2     Inside, r R q E R r qr  3 0 3 3 0 2 4 )( 1 ) E(4    Potential Energy of Configuration: R q drr r drr R rq dEW R R allspace 0 2 0 22 2 22 3 0 02 20 3 ) 1 ()()4)( 4 )( 2 (                 2. Electric Field of a Dipole )ˆsinˆcos2( 4 ),( 3 0     r r p rEp dip  (a) At (a,0,0),      ˆ 4 ,ˆ 4 , 2 , 3 0 3 0 a pq F a p Ear   (b) At (0,0,a), r a pq Fr a p r a p Ear ˆ 2 ,ˆ 2 ˆ)2( 4 ,0, 3 0 3 0 3 0     (c) Potential of a Electric Dipole 2 04 cos r p V    2 0 2 0 )0,0,(),0,0( 4 )0 4 ()( a pq a p qVVqqVW aa   3. (a) Electric Field: E r q DQadD fenc   24  Outside: 2 04 r q E   ; Inside: 2 0 2 )1(44 r q r q E e   Polarization: r r q EP e e e ˆ14 20        Bound Charge: )( 1 )(4 14 ˆ 14 33 2 rqr q r rq P e e e e e e b                 e e b R q nP       14 ˆ 2  (b) Total Bound Charge: 0)4( 14 )( 1 2 2 3       RR q drqdad e e e e bb        Total Bound Charge on Surface = e eq   1 4. By symmetry, magnetic forces on left and right wires cancel out each other, leaving two wires to be considered.   BldIF  R National University of Singapore Physics Society 2009 zIkaFFF z Ika xy a IkFz Ika xy a IkF lowerupper a alower a aupper ˆ ˆ 2 ˆˆ 2 ;ˆ 2 ˆˆ 2 2 22/ 2/ 22/ 2/          5. (a)   fencf IldHJH  , Outside,   ˆ 2 s I H   ; Inside,     ˆ 2 ˆ 2 2 2 2 s a I s a sI H   Outside,     ˆ 2 ,0 00 s I HBM   ; Inside,       ˆ 2 )1(ˆ 2 2 0 2 s a I s a I HB m    (b)     ˆ 2 2 s a I HM mm   z a I s a I MJ a I a a I nMK mm b mm b ˆ )ˆ 2 ( ˆ 2 ˆ 2 ˆ 22 2              (c) 0)()2( 2  IIaJaKI mmbbb  6. 44 25 4 2 0 3223 tkaa kat dyykatdxdytkyadB a    2 5kta dt d    Part 2 1. (a) 122 1 1 0 1 , ˆ 4 rrr r rld IB           212 adB  As 2 is proportional to 1B  , which is proportional to 1I , 1122 IM , 12M is a constant.      r ldldI ld r ldI ldAadAadB 21102 110 2121212 4 ) 4 ()(          r ldld M 21012 4    (b)        3 12 121 212 0 2 1 1 0 2212212 )( 4 ) ˆ 4 ( rr rrld ldII r rld IldIBldIF                                )( 1 4 )( 1 ) 1 ( 4 ) 1 ( 4 ) 1 ( 4 21 12 212 0 21 12 2 12 22112 0 12 21212 0 12 21212 0 ldld rr II ldld rrrr ldldII rr ldldII rr ldldII                     National University of Singapore Physics Society 2009