Problem solutions:
1)
For
the electric field vector E:
¶ 2E / ¶ x2 = moεo ¶2 E / ¶ t2
If
we compare the above to the generic wave equation for propagation of transverse
waves, say on a string, we find:
¶
2y/ ¶
x2 = 1/ v2 ¶ 2y/ ¶ t2
Where
v is the wave velocity. For the Maxwell wave equations, however, we have v = c.
And hence we can equate:
1/
c2 = moεo
c
2 = 1/ moεo and: c = 1/ Ö( moεo )
c
= 1/ Ö( 4 π x 10-7 H/m) (8.85 x 10-12 F/m )
So
that c = 2.99792 x 108 m/s
2)a) For a particular
electromagnetic wave traveling in free space the power is given as P = 1300 Wm-2
. Find the magnitudes of the wave vector
components, E y and H z if:
P = E y H z / c
Hint:
The impedance of free space is given by:
Ömo / Ö εo
= E y / H z
b)Find
the magnitude of the magnetic flux density:
B z
Solution:
a)
P = E y H z / c so: Pc =
E y H z
= (1300 Wm-2 )
( 3.0 x 108 m/s) = 3.9 x 1011
Wms-1
But: Ömo / Ö εo = E
y / H z
So: E
y = Ömo / Ö εo H z
Ömo / Ö εo = Ö( 4 π x 10-7 H/m) / Ö (8.85 x 10-12 F/m )
= 377 W (Impedance of free space)
So: E y = 377 W (H z )
Subs. This into Pc = E y H z
To get: P c
= 377
W (H z) 2
Then:
H z = Ö P c /
Ö377
W = Ö (3.9 x 1011 Wms-1/377 W )
H z = 3.2 x 104 Am-1
And: E y = 377 W (H z ) =
377 W (3.2 x 104 Am-1 )
= 1.2
x 10 7 Vm-1
b) B z = mo H z
= (4 π
x 10-7 H/m) (3.2 x 104 Am-1 )
B z = 0.04
T (Tesla)
3)The
general wave equation for E is:
¶
2E / ¶
x2 = moεo ¶2 E / ¶ t2
Writing
out all component wave eqns.:
¶
2E x / ¶ x2 = moεo ¶2 E x / ¶ t2
¶
2E y / ¶ x2 = moεo ¶2 E y / ¶ t2
¶
2E z / ¶ x2 = moεo ¶2 E z / ¶ t2
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