Force on Charge - Electricity and Magnetism - Solved Exam, Exams of Electromagnetism and Electromagnetic Fields Theory

Download Force on Charge - Electricity and Magnetism - Solved Exam and more Exams Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! Solutions to PC2131 AY0506 Sem 2 Paper Part I 1. Based on the diagram with the image charge drawn Force on charge Q = 2 0 2 2 0 2 16)2(4 D Q D Q   (downwards) 2. Maxwell equations: 0   E  0 B  t B E     t E JB     000  yctxkEE ˆ)(cos0   zctxkBB ˆ)(cos0   For electromagnetic waves, 0 ,0  J   0)(cos0    ctxkE y E  0)(cos0    ctxkB z B  yctxkkcEyctxkE tt E zctxkkcBzctxkB tt B ˆ)(sinˆ)(cos ˆ)(sinˆ)(cos 00 00               yctxkkByctxkB x ctxkB zyx zyx B zctxkkEzctxkE x ctxkE zyx zyx E ˆ)(sinˆ)(cos )(cos00 ˆˆˆ ˆ)(sinˆ)(cos 0)(cos0 ˆˆˆ 00 0 00 0                           Since t B E     , )(sin)(sin 00 ctxkkcBctxkkE  0 0 B E c  Since t E t E JB        00000  , 00 2 0 0 00 0000 1 1 )(sin)(sin       c cc B E ctxkkcEctxkkB D D Q Q (image) National University of Singapore Physics Society 2009 3. When charges are stored, let the inner cylinder have line charge density  , the outer cylinder have line charge density  For region between the 2 coaxial cylinders Using Gauss’ law,   0 r encQAdE  r E l rlE r r 0 0 2 )2(       mmrmmrrdEV BA r r B A 1 ,2 where    2ln 22 0 1 2 0     rr r r dr r rdEV B A    Capacitance per unit length VVl Q l C      1 12 0 186.0 2ln )1085.8)(32.2(2 2ln 2       nFm V r  Capacitance per unit length 1186.0  nFm 4. Using Faraday’s law, t B E        AdBdt d ldE  For ,ar  BrrBdrrB T rE r    22 0 2 1 2 1 )2( 2 )2( 1 )2(     ˆ 4 1 rBE   For ,ar  BadrrB T rE a  2 0 2 1 )2( 1 )2(     ˆ 4 1 2 B r a E   The polarization is only in the dielectric itself only, there is no polarization outside the dielectric.    ˆ 4 )1( )1( 000 rBEEP r re    rBrBP rrb       4 )1( 4 )1( 00            0ˆ  nPb   (direction of zn ˆˆ  for top and bottom of disk, direction of ̂ˆ n for the side of disk, both directions are perpendicular to ̂ ) 1 mm 2 mm 32.2r a  B  s National University of Singapore Physics Society 2009 2(b) 0)(   dVBAdB    belowabove BB Since MBH  0 1  , )(   belowabovebelowabove MMHH Using encf IldH   fbelowabove fbelowabove KHH lKlHH   |||| |||| )( For parallel component of H  , parallel to K  , we draw another amperian loop which side l is parallel to K  , the current enclosed is 0. nKHH fbelowabove ˆ ||||   2(c) dt d dt dI M   When current flows through solenoid A, Total flux experienced by solenoid B, BAa A Btotal NrIa N BAN 20   (shown) 2 0 2 0 2 0 a rNN M dt dI a rNN dt dI M dt dI a rNN dt d ABA aABAa aABA       2(d) a rNN M ABA 2 0  3(a) For ,cosrV  rrR )( ,  cos)(      c d d d d d d d d d d d d r dr d rdr dR r dr d R c                   2)cossin2( sincos 1 sin sincos 1 sin sincos 1)(cos sin sincos 1)( sin sin 1 2)( 11 2 2 22           cosr is a solution to Laplace equation. || aboveH || belowH National University of Singapore Physics Society 2009 For 2 cos r V   , 2 1 )( r rR  ,  cos)(  2 22 )) 2 (( 1 2 22 3 222            r r rdr d r r r dr d r dr dR r dr d R c c d d d d       2 )( sin sin 1    2 cos r  is a solution to Laplace equation. (shown) 3(b) Using spherical coordinates, there is no  dependence. At ar  , 0V For large r, placing the E  field in the z direction, cos00 rEzEV  Here cannot assume that V is zero at infinity. Solution for Laplace equation:        0 1 0 )(cos)(cos),( n n n n n n n n PrBPrArV  Since the solution for large r is cos0rE , no larger power of r is permissible. For smaller powers, we need terms which can cancel out cos so that V can be 0 on the surface. 2 3 00 cos cos),( r a ErErV   By uniqueness theorem, since the above solution satisfy the boundary conditions and the Laplace equation, the potential outside a grounded spherical conductor in uniform electric field will have the above solution. 3(c) Using Poisson’s equation, 0 2    V 0 2 2    dx Vd vJ  By conservation of energy, eVmv 2 2 1 eV m J v J m eV v 2 2    dx dV V e mJ dx Vd dx dV V e mJ dx Vd 2 1 0 2 2 2 1 0 2 2 2 2       s V = 0 0VV  National University of Singapore Physics Society 2009 Integrating both sides with respect to x, CV e mJ dx dV cV e mJ dx dV             2 1 0 2 2 1 0 2 2 4 )2( 22 1   When 0V , 0 dx dV E , 0C 2 0 2 3 0 2 1 0 2 2 1 2 2 2 1 2 1 0 2 22 3 3 2 2 3 2 3 2 1 2 4 x e mJ V x e mJ dx dV V e mJ dx dV V dx Vd V V e mJ dx dV                    At 0 , VVsx  (shown) 2 9 4 22 3 3 2 2 3 02 0 2 0 2 3 0 V m e s J s e mJ V     National University of Singapore Physics Society 2009