Post on 14-Apr-2018
7/30/2019 40.Electromagnetic Waves
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40.1
ELECTROMAGNETIC WAVES
CHAPTER - 40
1. 0 E 02
0
d EA
dt dt 4 r
=
1 3 4 2 1 1 2
1 3 2 2
M L T A A T L
TM L A L
= A1
= (Current) (proved).
2. E =2
Kq
x, [from coulombs law]
E = EA = 2KqA
x
Id =2
0 0 02
d E d kqA dKqA x
dt dt dtx
= 30 30
1 dx qAvq A 2 x
4 dt 2 x
.
3. E =0
QA
(Electric field)
= E.A. =0 0
Q A Q
A 2 2
i0 =E
0 0
0
d d Q 1 dQ
dt dt 2 2 dt
0d
REt /RC t / RC1 d 1 1 E(ECe ) EC e e2 dt 2 RC 2R
4. E =0
Q
A(Electric field)
= E.A. =0 0
Q A QA 2 2
i0 =E
0 00
d d Q 1 dQ
dt dt 2 2 dt
5. B = 0H
H =0
B
0 0 0 0
0 0 0 0
E B /( C) 1
H B / C
=12 8
1
8.85 10 3 10
= 376.6 = 377 .
Dimension1 1 3 4 2 1 2 3 2
0
1 1 1
C [LT ][M L T A ] M L T A
= M
1L
2T
3A
2= [R].
6. E0 = 810 V/m, B0 = ?
We know, B0 = 0 0 C E0Putting the values,
B0 = 4 107 8.85 1012 3 108 810
= 27010.9 1010 = 2.7 106 T = 2.7 T.
7/30/2019 40.Electromagnetic Waves
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Electromagnetic Waves
40.2
7. B = (200 T) Sin [(4 1015 51) (t x/C)]a) B0 = 200 T
E0 = C B0 = 200 106 3 108 = 6 104
b) Average energy density =6 2
20 7
0
1 (200 10 )B
2 2 4 10
=
8
7
4 10 1
208 10
= 0.0159 = 0.016.
8. I = 2.5 1014
W/m2
We know, I = 20 01
E C2
20E =I
0
2
Cor E0 =
I
0
2
C
E0 =14
12 8
2 2.5 10
8.85 10 3 10
= 0.4339 109 = 4.33 108 N/c.
B0 = 0 0 C E0= 4 3.14 107 8.854 1012 3 108 4.33 108 = 1.44 T.
9. Intensity of wave = 20 01
E C2
0 = 8.85 1012
; E0 = ? ; C = 3 108
, I = 1380 W/m2
1380 = 1/2 8.85 1012 20E 3 108
20E = 42 1380
8.85 3 10
= 103.95 104
E0 = 10.195 102
= 1.02 103
E0 = B0C
B0 = E0/C =3
8
1.02 10
3 10
= 3.398 105 = 3.4 105 T.