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40.Electromagnetic waves

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40.Electromagnetic waves Powered By Docstoc
					                                        ELECTROMAGNETIC WAVES
                                             CHAPTER - 40
     0 dE     0 EA
1.          
       dt     dt 4 0r 2
          M1L3 T 4 A 2       A1T1 L2       1
     =                                 =A
         M1L3 A 2             L2
                                       T
     = (Current)                   (proved).
            Kq
2.   E=             , [from coulomb’s law]
               x2
                         KqA
     E = EA =
                     x2
                dE        d kqA                d
     Id    = 0       0          0 KqA  x 2
                  dt      dt x 2                dt
                     1                        dx qAv
           = 0         q  A  2  x 3          .
                  4 0                       dt 2x 3
             Q
3.   E=          (Electric field)
            0 A
                          Q A    Q
      = E.A. =                
                         0 A 2 0 2
                    dE     d  Q  1  dQ 
     i0 = 0            0                  
                     dt     dt  0 2  2  dt 
                                     
                                                             td
              1d                    1      1  t / RC E RE0 
                (EC e t / RC )  EC     e           e    
              2 dt                  2     RC           2R
             Q
4.   E=          (Electric field)
            0 A
                          Q A    Q
      = E.A. =                
                         0 A 2 0 2
                    dE     d  Q  1  dQ 
     i0 = 0            0                  
                     dt     dt  0 2  2  dt 
                                     
5.   B = 0H
                    B
      H=
                    0
     E0 B0 /(0 0 C)     1
                     
     H0    B0 /  0     0 C
                               1
           =                 = 376.6 = 377 .
                           12
         8.85  10  3  108
                 1            1               1
     Dimension          1  1 3 4 2
                                        1 2 3 2 = M1L2T–3A–2 = [R].
               0 C [LT ][M L T A ] M L T A
6.   E0 = 810 V/m, B0 = ?
     We know, B0 = 0 0 C E0
     Putting the values,
                  –7          –12     8
     B0 = 4  10  8.85  10  3  10  810
                        –10       –6
        = 27010.9  10 = 2.7  10 T = 2.7 T.


                                                            40.1
                                                                                                Electromagnetic Waves
                                     15    –1
7.   B = (200 T) Sin [(4  10 5 ) (t – x/C)]
     a) B0 = 200 T
                               –6        8      4
        E0 = C  B0 = 200  10  3  10 = 6  10
                                                 1 2 (200  10 6 )2        4  10 8   1
     b) Average energy density =                    B0                  =               = 0.0159 = 0.016.
                                                20      2  4   10 7
                                                                           8  10 7 20
                    14      2
8.   I = 2.5  10        W/m
                         1     2
     We know, I =          0 E0 C
                         2
        2      2I                            2I
      E0 =                     or E0 =
              0 C                          0 C

                2  2.5  1014                                9                  8
     E0 =                                  = 0.4339  10 = 4.33  10 N/c.
            8.85  10 12  3  108
     B0 = 0 0 C E0
                        –7           –12         8        8
         = 4  3.14  10  8.854  10  3  10  4.33  10 = 1.44 T.
                          1    2
9.   Intensity of wave = 0 E0 C
                          2
                    –12                  8              2
     0 = 8.85  10 ; E0 = ? ; C = 3  10 , I = 1380 W/m
                                   –12      2            8
     1380 = 1/2  8.85  10               E0  3  10
        2            2  1380               4
      E0 =                   = 103.95  10
             8.85  3  10 4
                         2            3
      E0 = 10.195  10 = 1.02  10
       E0 = B0C
                          1.02  103                     –5                –5
      B0 = E0/C =                        = 3.398  10        = 3.4  10        T.
                           3  108

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                                                                   40.2

				
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Description: HC VERMA SOLUTIONS PHYSICS