40.Electromagnetic waves

ELECTROMAGNETIC WAVES CHAPTER - 40 1.

0 dE 0 EA  dt dt 4 0r 2 =

M1L3 T 4 A 2

M1L3 A 2 = (Current) 2.

E=

Kq x2

3.

A1T1 L2 1 =A  2 T L (proved).

, [from coulomb’s law]

E = EA = Id



KqA

x2 dE d kqA d = 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

E=

Q (Electric field) 0 A

 = E.A. = i0 = 0

Q A Q  0 A 2 0 2

dE d  Q  1  dQ  0    dt dt  0 2  2  dt   td

 4.

1d 1 1  t / RC E RE0    (EC e t / RC )  EC  e  e 2 dt 2 RC 2R

E=

Q (Electric field) 0 A

 = E.A. = i0 = 0 5.

Q A Q  0 A 2 0 2

dE d  Q  1  dQ  0    dt dt  0 2  2  dt 

B = 0H  H=

B 0

E0 B0 /(0 0 C) 1   H0 B0 /  0 0 C 1

= 376.6 = 377 .

8.85  10  3  108 1 1 1 Dimension   1 2 3 2 = M1L2T–3A–2 = [R]. 1 1 3 4 2 0 C [LT ][M L T A ] M L T A =

6.

12

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 7.

15

–1

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 b) Average energy density =

8.

I = 2.5  10

14

E0 =

9.

2

W/m

We know, I =  E02 =

1 2 (200  10 6 )2 4  10 8 1 B0  = = 0.0159 = 0.016.  7 20 2  4   10 8  10 7 20

1 0 E02 C 2

2I 0 C

2I 0 C

or E0 =

2  2.5  1014

9

8.85  10 12  3  108

8

= 0.4339  10 = 4.33  10 N/c.

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 Intensity of wave = 0 E02 C 2 –12 8 2 0 = 8.85  10 ; E0 = ? ; C = 3  10 , I = 1380 W/m –12

1380 = 1/2  8.85  10

 E02  3  10

8

2  1380

4 = 103.95  10 8.85  3  10 4 2 3  E0 = 10.195  10 = 1.02  10 E0 = B0C

 E02 =

 B0 = E0/C =

1.02  103 3  108

= 3.398  10

–5

= 3.4  10

–5



40.2

T.