Download Tutorial 1 + Answer

Tutorial 1
11th August 2008 + Answer
1. Consider the rotating conductor shown. The center of the 2a diameter bar is fixed at
the origin, and can rotate in x-y plane with B = Boaz. The outer ends of the bar make
conductive contact with a ring to make one end of the electrical contact to R; the other
contact is made to the center of the bar. Given Bo = 100 mWb/m2, a = 6 cm, and R =
50 , determine I if the bar rotates at 2π radians per second.
Figure indicates one of the paths for the circulation integral.
a
a
Vemf    a  Boa z  d  a     Bo   d  
0
I
Vemf
R
0

 Bo a
 Bo a 2
2
1  rev  rad 
2  1  Vs A
3 Wb 
 1
0.06m  
 2
100 x10

2 
2R
2  s 
rev 
m 
 50  Wb V
2
I  22.6  A
2. A conductive rod, of length 6.0 cm, has one end fixed on a grounded origin and is free
to rotate in the x-y plane. It rotates at ω = 120π radian per second in a magnetic field
B = 100 mT az. Find the voltage at the end of the bar.


   60
u
rev  2 rad 
rad

  120
s  rev 
s
d
a   a
dt
l
Vemf    a  Boa z  d  a  
 Bo l 2
2
0
1
rad  Wb 
2 Vs
Vemf  120
 0.1 2   0.06m 
2
s 
m 
Wb
 68mV
We can confirm the sign by observing that a positive charge placed in the
middle of the bar would move to the ungrounded end by the Lorentz force
equation.
3. A 1.0 m long coaxial cable of inner conductor diameter 2.0 mm and outer conductor
diameter 6.0 mm is filled with an ideal dielectric with r = 10.2. A voltage v(t) =
10.cos(6x106 t) mV is placed on the inner conductor and the outer conductor is
grounded. Neglecting fringing fields at the ends of the coax, find the displacement
current between the inner and outer conductors. Capacitance for coaxial cable is given
by:
2 lVo cos t
2 l
for coax (from chapter 2): C=
so Q 
ln  b a 
ln  b a 
C
Q
, so Q  Cv(t )
V
2 l Vo cos t Vo cos t
V cos t

and D  o
a .
2 l ln  b a   ln  b a 
 ln  b a  
D Vo sin t

a  Jd
t
 ln  b a  
so D 
2
l
Vo sin t 1
2 l Vo sin t
id   J d dS 
  d  dz 
ln  b a   0
ln  b a 
0
Now evaluate id with the given parameters:
2  6 x106 rad s  10.2   8.854 x10 12 F m  1m  0.010V sin t  C As
id 
ln  0.006 0.002 
FV C
id  97 sin  6 x106 t   A
4. A magnetic field propagating in free space is given by


H( z , t )  20.sin   108 t   z a x
A
m
.
Find frequency, phase constant, wavelength and the electric field, E(z,t).
1
x108  50MHz
2
8
c
3x10
2  rad
 
 6m,  

8
f 0.5x10
 3 m
D
H 
(no J c )
t
  2 f   x108 , f 
ax
H 

x
H o sin t   z 
 H o  cos t   z  a y   o
 dE  E =
Ho
ay

y
0
az

z
0
E
t
 cos t   z  dta
o
 Ho
V
sin t   z  a y
so E( z, t ) 
 o
m
y


z 
Ho
 o
H o sin t   z   a y
sin t   z  a y


Inserting the given values we find: E( z, t )  7.5sin   x108 t 
3

 kV
z ay
m

5. A 10.0 MHz magnetic field travels in a fluid for which the propagation velocity is
1.0x108 m/sec. Initially, we have H(0,0)=2.0 ax A/m. The amplitude drops to 1.0
A/m after the wave travels 5.0 meters in the y direction. Find the general expression
for this wave.
The general expression for the wave is:
A
H( y , t )  H o e  y cos  t   y    a x
m
rad
  2 f  2 10 x106   20 x106
s
u
2
rad
u p  1x108 m s ;   p  10m;  
 0.2
f

m
A
H(0,0)  2.0a x ;  H o  2.0,   0
m
H ( y  5)  1  H oe  y  2.0e  (5) ; solving we get   0.14
Finally,
H ( y, t )  2.0e 0.14 y cos  20 x106 t  0.2 y  a x
A
m
6. In free space,




E( z, t )  10.cos  106 t   z a x  20.cos  106 t   z a y
V
.
m
Find H(z,t) using phasors.
The phasor version of E is: Es  10e j z a x  20e j z a y
ax
  Es  
x
10e j  z
ay

y
20e j  z
az

z
0



20e j  z a x  10e j  z a y
z
z
 Es  j  20e j z a x  j 10e j z a y   jo Hs
j  20  j z
 j  10  j  z
20  j  z
10  j  z
Hs 
e ax 
e ay 
e ax 
e ay
 jo
 jo
o
o
up 

  x106
rad
; 

 10.47 x103
8

u p 3x10
m
Inserting the appropriate values we then find:
mA
H s  53e j  z a x  27e j  z a y
,
m
and H( z, t )  53cos t   z  a x  27 cos t   z  a y
mA
m