## Thursday, March 5, 2015

### Problem Solutions (Electrodynamics Pt. 1)

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