Problem Solutions (Electrodynamics Pt. 1)

Problem solutions:

1)     For the electric field vector E:

 ¶ ²E / ¶ x²  =   m_oε_o  ¶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    =  m_oε_o    

c ²   = 1/  m_oε_o    and: c =  1/ Ö( m_oε_o  )

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:

Öm_o  / Ö ε_o    = 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:  Öm_o  / Ö ε_o    = E _y  / H _z    

 

So:  E _y  =     Öm_o  / Ö ε_o     H _z    

Öm_o  / Ö ε_o    =  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  = m_o 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  =   m_oε_o  ¶² E / ¶ t²

Writing out all component wave eqns.:

¶ ²E _x  / ¶ x²  =   m_oε_o  ¶² E _x / ¶ t²

¶ ²E _y  / ¶ x²  =   m_oε_o  ¶² E _y / ¶ t²

¶ ²E _z  / ¶ x²  =   m_oε_o  ¶² E _z / ¶ t²