7/30/2019 40.Electromagnetic Waves

1/2

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

2/2

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.