40.Electromagnetic waves

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```					                                        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



40.2

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Description: HC VERMA SOLUTIONS PHYSICS