Ampere’s Circuit Law
P3.14: A pair of infinite extent current sheets exists at z = -2.0 m and at z = +2.0 m.
The top sheet has a uniform current density K = 3.0 ay A/m and the bottom one has K
= -3.0 ay A/m. Find H at (a) (0,0,4m), (b) (0,0,0) and (c) (0,0,-4m).
1
We apply H  K  a N ,
2
1
1
(a) H   3a y   a z   -3a y   a z  0
2
2
(b)
1
1
H   3a y   a z   -3a y   a z
2
2
A
 3a x
m
(c) H = 0
Fig. P3.14
P3.15: An infinite extent current sheet with K = 6.0 ay A/m exists at z = 0. A
conductive loop of radius 1.0 m, in the y-z plane centered at z = 2.0 m, has zero
magnetic field intensity measured at its center. Determine the magnitude of the
current in the loop and show its direction with a sketch.
Htot = HS + HL
1
1
A
H S  K  a N   6a y   a z  3a x
2
2
m
For the loop, we use Eqn. (3.10):
I
H
az
2a
where here
I
I
HL 
 a x   a x
2a
2
(sign is chosen opposite HS).
So, I/2 = 3 and I = 6A.
Fig. P3.15
P3.16: Given the field H = 3y2 ax, find the current passing through a square in the x-y
plane that has one corner at the origin and the opposite corner at (2, 2, 0).
Referring to Figure P3.6, we evaluate the circulation of H around the square path.
 H dL  I
enc
b
c
d
a
a
b
c
d
    
b
 3 0
2
a x dxa x  0
a
c
 3y a
2
x
dya y  0
b
d
0
 3  2  a x dxa x  12 dx  24
2
c
2
a

0
d
So we have Ienc = 24 A. The negative
Sign indicates current is going in the
-az direction.
P3.17: Given a 3.0 mm radius solid wire centered on the z-axis with an evenly
distributed 2.0 amps of current in the +az direction, plot the magnetic field intensity H
versus radial distance from the z-axis over the range 0 ≤  ≤ 9 mm.
Figure P3.17 shows the situation along with the Amperian Paths. We have:
 H dL  I
enc
, where H  H a and dL   d a ;
2 H   I enc
This will be true for each Amperian path.

AP1: I enc
2
I
I
I 2
  J dS, J = 2 a z , I enc  2   d   d  2
a
a 0
a
0
I
 a for   a
2 a 2
I
AP2: Ienc = I, H 
a for   a
2
So: H 
Fig. P3.17a
%
%
Fig. P3.17b
MLP0317
generate plot for ACL problem
a=3e-3;
I=2;
N=30;
rmax=9e-3;
dr=rmax/N;
%radius of solid wire (m)
%current (A)
%number of data points to plot
%max radius for plot (m)
for i=1:round(a/dr)
r(i)=i*dr;
H(i)=(I/(2*pi*a^2))*r(i);
end
for i=round(a/dr)+1:N
r(i)=i*dr;
H(i)=I/(2*pi*r(i));
end
plot(r,H)
xlabel('rho(m)')
ylabel('H (A/m)')
grid on
P3.18: Given a 2.0 cm radius solid wire centered on the z-axis with a current density J
= 3 A/cm2 az (for  in cm) plot the magnetic field intensity H versus radial distance
from the z-axis over the range 0 ≤  ≤ 8 cm.
We’ll let a = 2 cm.
 H dL  I
enc
, where H  H a and dL   d a ;
2 H   I enc
AP1 ( < a): I enc   J dS   3 a   d  d a z  2 3
and H   2a for   a
AP2 ( > a): Ienc = 2a3, so H 
a3

a for   a
The MATLAB plotting routine is as follows:
%
MLP0318
%
generate plot for ACL problem
a=2;
%radius of solid wire (cm)
N=40;
%number of data points to plot
rmax=8; %max radius for plot
(cm)
dr=rmax/N;
for i=1:round(a/dr)
r(i)=i*dr;
H(i)=r(i)^2;
Fig. P3.18
end
for i=round(a/dr)+1:N
r(i)=i*dr;
H(i)=a^3/r(i);
end
plot(r,H)
xlabel('rho(cm)')
ylabel('H (A/cm)')
grid on
P3.19: An infinitesimally thin metallic cylindrical shell of radius 4.0 cm is centered
on the z-axis and carries an evenly distributed current of 10.0 mA in the +az direction.
(a) Determine the value of the surface current density on the conductive shell and (b)
plot H as a function of radial distance from the z-axis over the range 0 ≤ ≤ 12 cm.
(a) K s 
I
2 a

10mA
mA
mA
 39.8
; so K  40a z
2  0.04m 
m
m
(b) for  < a, H = 0. For  > a we have:
I
H
a
2
The MATLAB routine to generate the plot is as follows:
%
MLP0319
%
generate plot for ACL problem
a=4;
%radius of solid wire (cm)
N=120;
%number of data points to plot
I=10e-3; %current (A)
rmax=12; %max plot radius(cm)
dr=rmax/N;
for i=1:round(a/dr)
r(i)=i*dr;
H(i)=0;
end
for i=round(a/dr)+1:N
r(i)=i*dr;
H(i)=100*I/(2*pi*r(i));
end
plot(r,H)
xlabel('rho(cm)')
ylabel('H (A/m)')
grid on
Fig. P3.19b
P3.20: A cylindrical pipe with a 1.0 cm
wall thickness and an inner radius of 4.0
cm is centered on the z-axis and has an
evenly distributed 3.0 amps of current in
the +az direction. Plot the magnetic field
intensity H versus radial distance from
the z-axis over the range 0 ≤  ≤ 10 cm.
For each Amperian Path:
Fig. P3.19a
 H dL  I
enc
, where H  H a and dL   d a ;
2 H   I enc
Now, for  < a, Ienc = 0 so H = 0.
For a <  < b, I enc   J dS, where J 
I enc
I
a z and dS = d  d a z
 b  a2 
2

2
 2  a2 

I
 2  a2
I

 d   d  I 2 2 , H 
a
b a

  b2  a 2  a
2  b 2  a 2 
0
Fig. P3.20a
%
%
Fig. P3.20b
MLP0320
generate plot for ACL problem
a=4;
%inner radius of pipe (cm)
b=5;
%outer radius of pipe(cm)
N=120;
%number of data points to plot
I=3;
%current (A)
rmax=10;
%max radius for plot (cm)
dr=rmax/N;
aoverdr=a/dr
boverdr=b/dr
for i=1:round(a/dr)
r(i)=i*dr;
H(i)=0;
end
for i=round(a/dr)+1:round(b/dr)
r(i)=i*dr;
num(i)=I*(r(i)^2-a^2);
den(i)=2*pi*(b^2-a^2)*r(i);
H(i)=100*num(i)/den(i);
end
for i=round(b/dr)+1:N
r(i)=i*dr;
H(i)=100*I/(2*pi*r(i));
end
plot(r,H)
xlabel('rho(cm)')
ylabel('H (A/m)')
grid on
P3.21: An infinite length line carries current I in the +az direction on the z-axis, and
this is surrounded by an infinite length cylindrical shell (centered about the z-axis) of
radius a carrying the return current I in the –az direction as a surface current. Find
expressions for the magnetic field intensity everywhere. If the current is 1.0 A and
the radius a is 2.0 cm, plot the magnitude of H versus radial distance from the z-axis
from 0.1 cm to 4 cm.
 H dL  I
enc
Fig. P3.21a
; for 0<  a, H 
I
2
a and for   a, H  0.
The MATLAB routine used to
generate Figure P3.21b is as follows:
%
MLP0321
%
generate plot for ACL
problem
clc
clear
a=2;
%inner radius
of cylinder(cm)
N=80;
%number of data
points to plot
Fig. P3.21b
I=1;
%current (A)
rmax=4;
%max radius for plot (cm)
dr=rmax/N;
for i=1:40
r(i)=.1+(i-1)*dr;
H(i)=100*I/(2*pi*r(i));
end
for i=40:N
r(i)=i*dr;
H(i)=0;
end
plot(r,H)
xlabel('rho(cm)')
ylabel('H (A/m)')
grid on
P3.22: Consider a pair of collinear cylindrical shells centered on the z-axis. The inner
shell has radius a and carries a sheet current totaling I amps in the +az direction while
the outer shell of radius b carries the return current I in the –az direction. Find
expressions for the magnetic field intensity everywhere. If a = 2cm, b = 4cm and I =
4A, plot the magnitude of H versus radial distance from the z-axis from 0 to 8 cm.
 H dL  I
enc
; for 0<  a, H = 0;
for a<  b, H 
I
a ;
2
and for   b, H  0.
The MATLAB routine used to generate Figure P3.22b is as follows:
%
MLP0322
%
generate plot for ACL problem
a=2;
b=4;
N=160;
%inner radius of coax (cm)
%outer radius of coax(cm)
%number of data points to plot
Fig. P3.22a
Fig. P3.22b
I=4;
%current (A)
rmax=8;
%max radius for plot (cm)
dr=rmax/N;
aoverdr=a/dr
boverdr=b/dr
for i=1:round(a/dr)
r(i)=i*dr;
H(i)=0;
end
for i=round(a/dr)+1:round(b/dr)
r(i)=i*dr;
H(i)=100*I/(2*pi*r(i));
end
for i=round(b/dr)+1:N
r(i)=i*dr;
H(i)=0;
end
plot(r,H)
xlabel('rho(cm)')
ylabel('H (A/m)')
grid on
P3.23: Consider the toroid in Figure 3.55 that is tightly wrapped with N turns of
conductive wire. For an Amperian path with radius less than a, no current is enclosed
and therefore the field is zero. Likewise, for radius greater than c, the net current
enclosed is zero and again the field is zero. Use Ampere’s Circuital Law to find an
expression for the magnetic field at radius b, the center of the toroid.
 H dL  I
enc
;
Within the toroid, H = Ha, so
 H dL   H a   
b
da  2 bH .
Then, Ienc by the Amperian path is: Ienc = NI.
NI
H 
a .
2 b
4. Curl and the Point Form of Ampere’s Circuital Law
P3.24: Find  A for the following fields:
a. A = 3xy2/z ax
b. A = sin2 a – 2 z cos a
c. A = r2sin ar + r/cos a
(a)   3xy 2 z  a x 
Ax
A
3xy 2
6 xy
a y  x az 
ay 
az
2
z
y
z
z
(b)    sin 2 a    2 z cos a   
  2 cos a   0a 
3 z cos 
2
A
z
a 
A
z
a 
A 
1
 A     a z


  
 
1
 2sin  cos a z


  2 cos a  3 z cos   2sin  cos   a z
az 
(c)


r
   r 2 sin  a r 
a 
cos  


1  A 
1  1 Ar 
1    rA  Ar 
a

a



 a
r

r sin    
r  sin   
r  r
 
1
r   cos  
1   r2 
1  2

ar 
 r sin   a

 a 
r sin 

r r  cos  
r 

 2

 sin 
ar  
 r cos   a
2
sin  cos 
 cos 

P3.25: Find J at (3m, 60, 4m) for H = (z/sin) a – (2/cos) az A/m.
 A A 
1 Az
1 A
a      z  a 
a
 
  z
 z  
 H 
 1
  sin 
2 
z cos 
a  

a
 a 
2
cos 
 sin 2  z
 sin  cos  
Now find J by evaluating H at the given point:
A
J  10a   13a  0.89a z 2
m

P3.26: Suppose H = y2ax + x2ay A/m.
a. Calculate
 H dL around the path
A  B  C  D  A , where A(2m,0,0),
B(2m,4m,0), C(0,4m,0) and D(0,0,0).
b. Divide this
 H dL by the area S (2m*4m = 8m2).
c. Evaluate H at the center point.
d. Comment on your results for (b) & (c).
(a) Referring to the figure, we evaluate
B
4

  x2
C
0
A
0

  y2
D
0
B
2

  x2
A
2
C
4

  y2
D
0
 H dL  I
enc
B
C
D
A
A
B
C
D
    
dy  4  4   16C
x2
dx  32C
y 4
dy  0
x 0
dy  0
y 0
So we have  H dL  16C
Fig. P3.26
(b) dividing by S = 8m2, we have -2 C/m2
(c) Evaluating the curl of H:
 A A 
 H   y  x  a z   2 x  2 y  a z , and at the center point (x = 1 and y = 2) we
 x y 
have
C
  H center  2 2 a z
m
(d) In this particular case,   H 
 H dL
S , even though S is of appreciable size.
P3.27: For the coaxial cable example 3.8, we found:
for   a,
for a<  b,
for b<  c,
I
a ,
2 a 2
I
H
a ,
2
H
H=

I
2 c 2  b 2

 c2   2 

 a ,
  
and for c< , H =0.
a. Evaluate the curl in all 4 regions.
b. Calculate the current density in the conductive regions by dividing the current
by the area. Are these results the same as what you found in (a)?
1   I2 
I
(a)  H 
a  2 a z for   a

2  z
   2 a 
a
H 
1   I
   2

 a z  0 for a    b

2
2
1   I c    
I

 az 
 H 
a z for b<  c
2
2
2
2
   2  c  b  

c

b




  H = 0 for   c
I
(b) for   a, S   a 2 , J  I S a z  2 a z
a
I
for b<  c, S    c 2  b 2  , J 
az
  c2  b2 
 
Comment: H = J is confirmed.
P3.28: Suppose you have the field H = r cos  a A/m. Now consider the cone
specified by  = /4, with a height a as shown in Figure 3.56. The circular top of the
cone has a radius a.
a. Evaluate the right side of Stoke’s theorem through the dS = dSa surface.
b. Evaluate the left side of Stoke’s theorem by integrating around the loop.
1 
1   rH 

sin  H  ar 
a


r sin   
r r

ar derivative:
(a)  H 

 r sin  cos   r  cos2   sin 2   ;

cos 2   sin 2  

1 

ar
sin  H  ar 
r sin   
sin 
Fig. P3.28
a derivative:
2
1   r cos  
a  2 cos  a
r
r
cos 2   sin 2  

a r  2cos  a
So,   H 
sin 
Now we must integrate this over the a surface:
  cos 2   sin 2  




H
d
S

a

2
cos

a
r
  r sin  drd  a 


sin 


  2r sin  cos  drd 2  sin  cos     
4
(b)
 H dL   r r a
2a
2
 rdr  d  2 a
0
2
0
cos  a ad a  a 2 cos    
2
2
 d  2 a
2
4 0
Clearly in this case the circulation of H is the easiest approach.
5. Magnetic Flux Density
P3.29: An infinite length line of 3.0 A current in the +ay direction lies on the y-axis.
Find the magnetic flux density at P(7.0m,0,0) in (a) Teslas, (b) Wb/m2, and (c) Gauss.
H
I
2
a 
3A
mA
 a z   68 a z
2  7m 
m
H 
A   Wb 

9 Wb
B  o H   4 x107  68  a z 
a z  86 x109 Ta z
  86 x10
2
m 
m   HA 
m

10, 000G
B   86 x109 Ta z 
 860 x106 Ga z
T
P3.30: Suppose an infinite extent sheet of current with K = 12ax A/m lies on the x-y
plane at z = 0. Find B for any point above the sheet. Find the magnetic flux passing
through a 2m2 area in the x-z plane for z > 0.
1
H  K  aN ;
2
H m 
A

6 Wb
ay
12a x  a z   7.54 x10
2
2
m
m2


This is valid at any point above the sheet.
B
o
K  aN
 4 x10

7


Wb 

Now,    B dS  B S =  -7.54x10-6 2 a y  2m2  a y   15Wb
m


P3.31: An infinite length coaxial cable exists along the z-axis, with an inner shell of
radius a carrying current I in the +az direction and outer shell of radius b carrying the
return current. Find the magnetic flux passing through an area of length h along the zaxis bounded by radius between a and b.
H
I
2
a , B 
For a <  < b,    B dS 

o I
a ,
2
b
o I a
I
d  dza  o ln    a h

2 
2
o Ih  b 
ln   Wb
2
a
6. Magnetic Forces
P3.32: A 1.0 nC charge with velocity 100. m/sec in the y direction enters a region
where the electric field intensity is 100. V/m az and the magnetic flux density is 5.0
Wb/m2 ax. Determine the force vector acting on the charge.
m
Wb
Wb
F  q  E  u  B  ; u  B  100 a y  5 2  500
az
s
m ax
sm
V
Wb Vs  mN

F  109 C 100 a z  500
az 
 400nNa z
m
sm Wb  VC

P3.33: A 10. nC charge with velocity 100. m/sec in the z direction enters a region
where the electric field intensity is 800. V/m ax and the magnetic flux density 12.0
Wb/m2 ay. Determine the force vector acting on the charge.

V
m
Wb
F  q  E  u  B   10 x109 C  800 a x  100 a z  12 2

m
s
m ay


  4  Na x

P3.34: A 10. nC charged particle has a velocity v = 3.0ax + 4.0ay + 5.0az m/sec as it
enters a magnetic field B = 1000. T ay (recall that a tesla T = Wb/m2). Calculate the
force vector on the charge.
m
Wb 

F  q  u  B   10 x109 C  3a x  4a y  5a z  1000 2 a y 
s
m


The cross-product:
az 
a x a y
3
4
5   5000a x  3000a z

 0 1000 0 
Evaluating we find: F = -50ax + 30az N
P3.35: What electric field is required so that the velocity of the charged particle in the
previous problem remains constant?
dv
 0 (constant velocity)
dt
F  q  E  u  B   0;
F  ma  m
E  u  B = -  3a x  4a y  5a z 
E  5a x  3a z
m
Wb
V
V
1000 2 a y  3000 a z  5000 a x
s
m
m
m
kV
m
P3.36: An electron (with rest mass Me= 9.11x10-31kg and charge q = -1.6 x 10-19 C)
has a velocity of 1.0 km/sec as it enters a 1.0 nT magnetic field. The field is oriented
normal to the velocity of the electron. Determine the magnitude of the acceleration
on the electron caused by its encounter with the magnetic field.
F  ma  q  u  B  ;
a
q
u  B  ;
m
quB  1.6 x10
a

m
19

C  1000 m


109 Wb 2
s
m  175 x103 m
31
s2
 9.11x10 kg 
P3.37: Suppose you have a surface current K = 20. ax A/m along the z = 0 plane.
About a meter or so above this plane, a 5.0 nC charged particle is moving along with
velocity v = -10.ax m/sec. Determine the force vector on this particle.
1
1
A
K  a N  20a x  a z  10a y
2
2
m
7
B  o H  10  4 x10  a y Wb 2  40 x107 a y Wb 2
m
m
m
Wb 

F  qu  b   5 x109 C   10a x  40 x107 a y 2   0.63 pNa z
s
m 

H
P3.38: A meter or so above the surface current of the previous problem there is an
infinite length line conducting 1.0 A of current in the –ax direction. Determine the
force per unit length acting on this line of current.
0
F12   I 2 dL2  B1  I 2  dxa x  40 x107 a y
L
Wb
m2
F12  I 2   L   40 x107 a z  ;
F12
N
 40 x107 a z  12.6
az
L
m
P3.39: Recall that the gravitational force on a mass m is F  mg, where, at the earth’s
surface, g = 9.8 m/s2 (-az). A line of 2.0 A current with 100. g mass per meter length
is horizontal with the earth’s surface and is directed from west to east. What
magnitude and direction of uniform magnetic flux density would be required to
levitate this line?
 Ns 2 kg 
m

Fg  mg; for 1 m Fg  100 g   9.8 2   a z  
  0.98 Na z
s 
kg
m
1000
g



1m
F
 IdL  B   F
g
 0.98 Na z
0
By inspection, B = Bo(-ax)
L
 Idya
y
 Bo  a x   ILBoa z
o
 Wb 
  2 A 1m   Bo 2   2 Bo  N 
 m 
The unit conversion to arrive at
Newtons is as follows:
Am Wb Vs W J Nm
N
m 2 Wb VA Ws J
So we have Bo = 0.490 Wb/m2, and
B = 0.490 Wb/m2 (-ax) (directed north)
Fig. P3.39
P3.40: Suppose you have a pair of parallel lines each with a mass per unit length of
0.10 kg/m. One line sits on the ground and conducts 200. A in the +ax direction, and
the other one, 1.0 cm above the first (and parallel), has sufficient current to levitate.
Determine the current and its direction for line 2.
F12 o I1I 2
4 x107 H m 200 A
Here we will use

ay 
I 2a z  4 x103 I 2a z
L
2 y
2
0.01m
2
F mg
N
2 Ns

  0.10 kg m   9.8 m s 
 0.98
L
L
kgm
m
So solving for I2:
0.98
I2 
 245 A in the -a x direction.
4 x103
P3.41: In Figure 3.57, a 2.0 A line of current is shown on the z-axis with the current in
the +az direction. A current loop exists on the x-y plane (z = 0) that has 4 wires
(labeled 1 through 4) and carries 1.0 mA as shown. Find the force on each arm and
the total force acting on the loop from the field of the 2.0 A line.
o I1
a
2
A  B : dL2  5da , a  a  0
F12   I 2 dL2  B1 ; B1 =
C  D : a  a  0
3
B  C : F12  I 2  d  a  
5
o I1
 I I 3
a  o 1 2 ln   a z
2
2
5
Fig. P3.41
4 x107 H m
3
 2 A 103 A ln   a z  204 pNa z
2
5
So for B to C: F12 = -0.20 nN az
Likewise, from D to A: F12 = +0.20 nN az
F12 
P3.42: MATLAB: Modify MATLAB 3.4 to find the differential force acting from
each individual differential segment on the loop. Plot this force against the phi
location of the segment.
%MLP0342
%modify ML0304 to find dF acting from the field
%
of each segment of current; plot vs phi
clear
clc
I=1;
%current in A
a=1;
%loop radius, in m
mu=pi*(4e-7);
%free space permeability
az=[0 0 1]; %unit vector in z direction
DL1=a*2*(pi/180)*[0 1 0]; %Assume 2 degree increments
%DL1 is the test element vector
%F is the angle phi in radians
%xi,yi is location of ith element on the loop
%Ai & ai = vector and unit vector from origin
%
to xi,yi
%DLi is the ith element vector
%Ri1 & ri1 = vector and unit vector from ith
%
point to test point
for i=1:179
phi(i)=i*2;
F=2*i*pi/180;
xi=a*cos(F);
yi=a*sin(F);
Ai=[xi yi 0];
ai=unitvector(Ai);
DLi=(pi*a/90)*cross(az,ai);
Ri1=[a-xi -yi 0];
ri1=unitvector(Ri1);
num=mu*I*cross(DLi,ri1);
den=4*pi*(magvector(Ri1)^2);
B=num/den;
dFvect=I*cross(DL1,B);
dF(i)=dFvect(1);
end
plot(phi,dF)
xlabel('angle in degrees')
ylabel('the differential force, N')
Fig. P3.42
P3.43: MATLAB: Consider a circular conducting loop of radius 4.0 cm in the y-z
plane centered at (0,6cm,0). The loop conducts 1.0 mA current clockwise as viewed
from the +x-axis. An infinite length line on the z-axis conducts 10. A current in the
+az direction. Find the net force on the loop.
The following MATLAB routine shows the force as a function of radial position
around the loop. Notice that while there is a net force in the -y direction, the forces in
the z-direction cancel.
%
%
%
MLP0343
find total force and torque on a loop of
current next to a line of current
%
%
%
%
%
%
%
%
%
%
%
%
%
%
variables
I1,I2
current in the line and loop (A)
yo
center of loop on y axis (m)
uo
free space permeability (H/m)
N
number of segements on loop
a
loop radius (m)
dalpha differential loop element
dL
length of differential section
DL
diff section vector
B1
I1's mag flux vector (Wb/m^2)
Rv
vector from center of loop
to the diff segment
ar
unit vector for Rv
y,z
the location of the diff segment
clc
clear
%
initialize variables
I1=10;
I2=1e-3;
yo=.06;
uo=pi*4e-7;
N=180;
a=0.04;
dalpha=360/N;
dL=a*dalpha*pi/180;
ax=[1 0 0];
%
perform calculations
for i=1:N
dalpha=360/N;
Fig. P3.43
alpha=(i-1)*dalpha;
phi(i)=alpha;
z=a*sin(alpha*pi/180);
y=yo+a*cos(alpha*pi/180);
B1=-(uo*I1/(2*pi*y))*[1 0 0];
Rv=[0 y-.06 z];
ar=unitvector(Rv);
aL=cross(ar,ax);
DL=dL*aL;
dF=cross(I2*DL,B1);
dFx(i)=dF(1);
dFy(i)=dF(2);
dFz(i)=dF(3);
end
plot(phi,dFy,phi,dFz,'--k')
legend('dFy','dFz')
Fnet=sum(dFy)
Running the program we get:
Fnet = -4.2932e-009
>>
So Fnet = -4.3 nN ay
P3.44: MATLAB: A square loop of 1.0 A current of side 4.0 cm is centered on the x-y
plane. Assume 1 mm diameter wire, and estimate the force vector on one arm
resulting from the field of the other 3 arms.
%
MLP0344 V2
%
%
Square loop of current is centered on x-y plane.
Viewed
%
from the +z axis, let current go clockwise. We want
to
%
find the force on the arem at x = +2 cm resulting
from the
%
current in arms at y = -2 cm, x = -2 cm and y = +2 cm.
%
Wentworth, 12/3/03
%
%
%
%
%
%
%
%
%
%
%
%
%
%
Variables
a
side length (m)
b
wire radius (m)
I
current in loop (A)
uo
free space permeability (H/m)
N
number of segments for each arm
xi,yi
location of test arm segment (at x
xj,yj
location of source arm segment (at
xk,yk
location of source arm segment (at
xL,yL
location of source arm segment (at
dLi
differential test segment vector
dLj, dLk, dLL
diff vectors on sources
Rji
vector from source point j to test
aji
unit vector of Rji
clc;clear;
a=0.04;
b=.0005;
I=1;
uo=pi*4e-7;
N=80;
for i=1:N
xi=(a/2)+b;
yi=-(a/2)+(i-0.5)*a/N;
ypos(i)=yi;
dLi=(a/N)*[0 -1 0];
for j=1:N
xj=-(a/2)+(j-0.5)*a/N;
yj=-a/2-b;
dLj=(a/N)*[-1 0 0];
Rji=[xi-xj yi-yj 0];
aji=unitvector(Rji);
=
y
x
y
+2 cm)
= -2 cm)
= -2 cm)
= +2 cm)
point i
num=I*CROSS(dLj,aji);
den=4*pi*(magvector(Rji))^2;
H=num/den;
dHj(j)=H(3);
end
for k=1:N
yk=(-a/2)+(k-0.5)*a/N;
xk=-(a/2)-b;
dLk=(a/N)*[0 1 0];
Rki=[xi-xk yi-yk 0];
aki=unitvector(Rki);
num=I*CROSS(dLk,aki);
den=4*pi*(magvector(Rki))^2;
H=num/den;
dHk(k)=H(3);
end
for L=1:N
xL=(-a/2)+(L-0.5)*a/N;
yL=(a/2)+b;
dLL=(a/N)*[1 0 0];
RLi=[xi-xL yi-yL 0];
aLi=unitvector(RLi);
num=I*CROSS(dLL,aLi);
den=4*pi*(magvector(RLi))^2;
H=num/den;
dHL(L)=H(3);
end
H=sum(dHj)+sum(dHk)+sum(dHL);
B=uo*H*[0 0 1];
F=I*CROSS(dLi,B);
dF(i)=F(1);
end
Ftot=sum(dF)
plot(ypos,dF)
Running the program:
Ftot =
7.4448e-007
So Ftot = 740 nN
Fig. P3.44
P3.45: A current sheet K = 100ax A/m exists at z = 2.0 cm. A 2.0 cm diameter loop
centered in the x-y plane at z = 0 conducts 1.0 mA current in the +a direction. Find
the torque on this loop.
  m  B;


m  ISa N  103 A    0.01m  a z  314 x109 Am 2a z
B   o H; H 
2
1
1
A
A
K  a N  100a x    a z   50a y ;
2
2
m
m
Wb
;
m2
  m  B = -20 pNma x
B  50oa y
P3.46: 10 turns of insulated wire in a 4.0 cm diameter coil are centered in the x-y
plane. Each strand of the coil conducts 2.0 A of current in the a direction. (a) What
is the magnetic dipole moment of this coil? Now suppose this coil is in a uniform
magnetic field B = 6.0ax + 3.0ay + 6.0az Wb/m2, (b) what is the torque on the coil?

(a) m  NISa x  10  2 A   0.02m 
2
  25.1mAm a
(b)   m  B  25.1mAa z   6a x  3a y  6a z 
2
z
Wb
 0.151a y  0.075a x Nm
m2
  75a x  151a y mNm
P3.47: A square conducting loop of side 2.0 cm is free to rotate about one side that is
fixed on the z-axis. There is 1.0 A current in the loop, flowing in the –az direction on
the fixed side. A uniform B-field exists such that when the loop is positioned at  =
90, no torque acts on the loop, and when the loop is positioned at  = 180 a
maximum torque of 8.0 N-m az occurs. Determine the magnetic flux density.
At  = 90°, m  ISa N  0.0004 Am 2a x . Also, since   m  B  0, B is in direction of
m, and therefore B = ±Boax.
At  = 180°, m  ISa N  0.0004 Am2a y , and   m  B  8 x106 a z Nm.
Therefore, B = -Boax and mBo = 8x10-6, so
8 x106
mWb
mWb
Bo 
 20 2 , and B  20 2 a x .
0.0004
m
m
7. Magnetic Materials
P3.48: A solid nickel wire of diameter 2.0 mm evenly conducts 1.0 amp of current.
Determine the magnitude of the magnetic flux density B as a function of radial
distance from the center of the wire. Plot to a radius of 2 mm.
J
I
1A
kA
a 
a z  31.8 2 a z
2
2 z
a
m
 1x103 m 
 H dL  I
enc
  J dS
I
I
 d  d  2  2
2 
a
a
  I
I
for   a H 
a ; B  r o 2 a
2 
2 a
2 a
I
I
for   a H 
a ; B  o a
2
2
2 H  
%
%
MLP0348
generate plot for ACL problem
a=2e-3;
I=1;
N=30;
rmax=4e-3;
dr=rmax/N;
uo=pi*4e-7;
ur=600;
%radius of solid wire (m)
%current (A)
%number of data points to plot
%max radius for plot (m)
for i=1:round(a/dr)
r(i)=i*dr;
B(i)=(ur*uo*I/(2*pi*a^2))*r(i);
end
for i=round(a/dr)+1:N
r(i)=i*dr;
B(i)=uo*I/(2*pi*r(i));
End
rmm=r*1000;
plot(rmm,B)
xlabel('rho(cm)')
ylabel('B (Wb/m^2)')
grid on
Fig. P3.48
8. Boundary Conditions
P3.49: A planar interface separates two magnetic media. The magnetic field in media
1 (with r1) makes an angle 1 with a normal to the interface. (a) Find an equation for
2, the angle the field in media 2 (that has r2) makes with a normal to the interface,
in terms of 1 and the relative permeabilities in the two media. (b) Suppose media 1
is nickel and media 2 is air, and that the magnetic field in the nickel makes an 89
angle with a normal to the surface. Find 2.
H1  H1N a N  H1T aT ; H 2T aT  H1T aT ; B1N a N  r1o H1N a N
B2 N a N  B1N a N  r1o H1N a N  r 2 o H 2 N a N ;  H 2 N 
tan  2 
 r1
H
 r 2 1N
H 2T
H1T

;
H2N
 r1
H1N
r 2
tan  2 
r 2 H1T r 2

tan 1
r1 H1N r1
 r 2

tan 1 
  r1

 1

 2  tan 1 
tan 89   5.5
 600

 2  tan 1 
Fig. P3.49
P3.50: MATLAB: Suppose the z = 0 plane separates two magnetic media, and that no
surface current exists at the interface. Construct a program that prompts the user for
r1 (for z < 0), r2 (for z > 0), and one of the fields, either H1 or H2. The program is to
calculate the unknown H. Verify the program using Example 3.11.
%
M-File: MLP0350
%
%
Given H1 at boundary between a pair of
%
materials with no surface current at boundary,
%
calculate H2.
%
clc
clear
%
enter variables
disp('enter vectors quantities in brackets,')
disp('for example: [1 2 3]')
ur1=input('relative permeability in material 1: ');
ur2=input('relative permeability in material 2: ');
a12=input('unit vector from mtrl 1 to mtrl 2: ');
F=input('material where field is known (1 or 2): ');
Ha=input('known magnetic field intensity vector: ');
if F==1
ura=ur1;
urb=ur2;
a=a12;
else
ura=ur2;
urb=ur1;
a=-a12;
end
%
perform calculations
Hna=dot(Ha,a)*a;
Hta=Ha-Hna;
Htb=Hta;
Bna=ura*Hna;
%ignores uo since it will factor out
Bnb=Bna;
Hnb=Bnb/urb;
display('The magnetic field in the other medium is: ')
Hb=Htb+Hnb
Now run the program (for Example 3.11):
enter vectors quantities in brackets,
for example: [1 2 3]
relative permeability in material 1: 6000
relative permeability in material 2: 3000
unit vector from mtrl 1 to mtrl 2: [0 0 1]
material where field is known (1 or 2): 1
known magnetic field intensity vector: [6 2 3]
ans =
The magnetic field in the other medium is:
Hb = 6 2 6
For a second test, run the program for problem P3.52(a).
enter vectors quantities in brackets,
for example: [1 2 3]
relative permeability in material 1: 4
relative permeability in material 2: 1
unit vector from mtrl 1 to mtrl 2: [0 0 -1]
material where field is known (1 or 2): 1
known magnetic field intensity vector: [3 0 4]
ans =
The magnetic field in the other medium is:
Hb = 3 0 16
P3.51: The plane y = 0 separates two magnetic media. Media 1 (y < 0) has r1 = 3.0
and media 2 (y > 0) has r2 = 9.0. A sheet current K = (1/o) ax A/m exists at the
interface, and B1 = 4.0ay + 6.0az Wb/m2. (a) Find B2. (b) What angles do B1 and B2
make with a normal to the surface?
(a) B N 1  4a y
(b) B N 2  B N 1  4a y
(c) BT 1  6a z
(d) HT 1 
(e) HT 2 
BT 1
 r1o
3
o

2
o
az
a z (see below)
(f) BT 2  r o HT 2  27a z
(g) B 2  4a y  27a z
Wb
m2
Fig. P3.51
Now for step (e):
a 21   H1  H 2   K ;
  HT 1  HT 2  a x 
1
o
 a y   H T 1a z  H T 2a z  
a x ; HT 2  HT 1 
 B
Angles: 1  tan 1  T 1
 BN 1
1
o
1
o
ax
; HT 2 

 BT 2
1
  56 ;  2  tan 

 BN 2
1
o

2
o

3
o

  82

P3.52: Above the x-y plane (z > 0), there exists a magnetic material with r1 = 4.0 and
a field H1 = 3.0ax + 4.0az A/m. Below the plane (z < 0) is free space. (a) Find H2,
assuming the boundary is free of surface current. What angle does H2 make with a
normal to the surface? (b) Find H2, assuming the boundary has a surface current K =
5.0 ax A/m.
(a)
(1) H N 1  4a z , (2) B N 1  16oa z , (3) B N 2  B N 1  16oa z ,(4) H N 2 
(5) HT 1  3a x , (6) HT 2  HT 1  3a x , (7) H 2  3a x  16a z
 HT 2
 HN2
 2  tan 1 
A
m

  10.6

(b)
Now step (6) becomes a21   H1  H2   K , shere a21 = az.
BN 2
o
 16a z
Let’s let H 2  Aa x  Ba y  16a z , then a 21   3a x  4a z  Aa x  Ba y  16a z   5a x
Solve for A and B:
a y az 
 ax
 0
0
1   Ba x   A  3 a y  5a x ; so A = 3 and B = 5

3  A  B 12
Finally, H 2  3a x  5a y  16a z
A
.
m
P3.53: The x-z plane separates magnetic material with r1 = 2.0 (for y < 0) from
magnetic material with r2 = 4.0 (for y > 0). In medium 1, there is a field H1 = 2.0ax
+ 4.0ay + 6.0az A/m. Find H2 assuming the boundary has a surface current K = 2.0ax
– 2.0az A/m.
(1)H1N  4a y ,(2)B1N  8oa y ,(3)B 2 N  B1N  8oa y ,(4)H 2 N 
8 o
a y  2a y
4 o
(5)H1T  2a x  6az ,(6)a21   H1  H2   K, let HT 2  H xa x  H zaz ,
so  a y    2  H x  a x   6  H z  a z   2a x  2a z ,
ax
ay
0
2  Hx
az
1
0   H z  6  a x   2  H x  a z  2a x  2a z , so H z  8, H x  4
0 6  Hz
(7)H 2T  4a x  8a z ,
 H 2  4a x  2a y  8a z
A
m
P3.54: An infinite length line of 2 A current in the +az direction exists on the z-axis.
This is surrounded by air for  ≤ 50 cm, at which point the magnetic media has r2 =
9.0 for  > 50 cm. If the field in media 2 at  = 1.0 m is H = 5.0a A/m, find the sheet
current density vector at  = 50. cm, if any.
Method 1:
From just the line of current we would have H1 
I1
2 1
a  1a .
Now, since HTOT  5a  H1  H 2 , then H 2  4a is the contribution from the sheet
current.
H2 
I2
2 1
K  8
a  4a , so I 2  8 A, then K 
A
az
m
I2
8
A
az 
 8 az
2 a
2  0.5
m
Method 2:
From I1 at boundary we have
2 a
H1 
 2a , but 5a at  = 1.0m corresponds to 10a at  =0.5m
2  0.5 
varies as 1/. So
a    2a  10a   K ,
K  8
A
az
m
 a    8a   8a z  K
since
H
Faraday’s law and EMF
P4.9: The magnetic flux density increases at the rate of 10 (Wb/m 2)/sec in the z
direction. A 10 cm x 10 cm square conducting loop, centered at the origin in the x-y
plane, has 10 ohms of distributed resistance. Determine the direction (with a sketch)
and magnitude of the induced current in the conducting loop.
dB
Wb
 10 2 a z
dt
ms
dB
Wb
Vemf   
dS    10 2 a z dxdya z
dt
ms
Wb Vs
Vemf  0.1
 0.1V
s Wb
0.1V
 10mA
I= I 
10
I=10mA clockwise (when viewed
From +z)
Fig. P4.9
P4.10: A bar magnet is dropped through a conductive ring. Indicate in a sketch the
direction of the induced current when the falling magnet is just above the plane of the
ring and when it is just below the plane of the ring, as shown in Figure 4.22.
Refer to Figure P4.10.
When the north pole first goes through the loop, flux is increasing and the current
induced to oppose this change in flux is as shown.
When the south pole is exiting the loop, flux is decreasing and the current induced
acts to oppose this change in flux.
Fig. P4.10
P4.11: Considering Figure 4.7, suppose the area of a single loop of the pair is 100
cm2, and the magnetic flux density is constant over the area of the loops but changes
with time as B  Bo e   t a z , where Bo = 4.0 mWb/m2 and  = 0.30 Np/sec. Determine
VR at 1, 10, and 100 seconds.
Vemf   N 
B
dB
dS;
   Bo e   t a z ;
t
dt
Vemf  2  Bo e   t S
2
 1m  0.30t
VR  2  Bo Se   t  2  0.30   4 x103 100cm 2  
 24 x106 e 0.30t
 e
 100cm 
at t = 1 sec, VR = 17.8 V
at t = 10 sec, VR = 1.20 V
at t = 100 sec, VR = 2.25x10-18 V
P4.12: Sometimes a transformer is used as an impedance converter, where impedance
is given by v/i. Find an expression for the impedance Z1 seen by the primary side of
the transformer in Figure 4.11 that has a load impedance Z2 terminating the
secondary.
We have i2 
N1
N
i1 and v2  2 v1
N1
N2
 N2 
v1 
2

v1
v2  N 1   N 2 
Z1  , Z 2  

 Z1
i1
i2  N1   N1 
 N i1 
 2 
2
N 
Z1   1  Z 2
 N2 
P4.13: A 1.0 mm diameter copper wire is shaped into a square loop of side 4.0 cm. It
is placed in a plane normal to a magnetic field increasing with time as B = 1.0 t
Wb/m2 az, where t is in seconds. (a) Find the magnitude of the induced current and
indicate its direction in a sketch. (b) Calculate the magnetic flux density at the center
of the loop resulting from the induced current, and compare this with the original
magnetic flux density that generated the induced current at t = 1.0 sec.
We find the distributed resistance of the loop and work the problem assuming this
resistance is lumped in one spot as shown in the figure.
(a) The induced current is Vemf divided by the distributed resistance of the wire loop.
4  0.04m) 
1 l
1m
R

 3.5m
7
 A 5.8 x10   0.0005m 2
dB
Wb
dB
Wb
 1 2 a z ; Vemf   
dS  1 2
dt
ms
dt
ms
0.04
0.04
0
0
 dx  dy  1.6mV
I ind 
1.6mV
 0.46 A (note that this answer has no time dependence)
3.5m
Fig. P4.13
(b) The field at the center of the loop from a single arm of the loop is found from Eqn.
(3.7):
I
H
4

z
 2 2
  z   2
So B  4 o H  13
a

I
  a z  
2
  a
Wb
m2
 0.46 
1
1
A
 -a z  
 a z   2.59a z ;
2
m
2a
2  0.02 
az .
P4.16: Referring to Figure 4.23, suppose a conductive bar of length h = 2.0 cm
moves with velocity u = -1.0 m/s a towards an infinite length line of current I = 4.0
A. Find an expression for the voltage from one end of the bar to the other when 
reaches 10 cm and indicate which end is positive.
In Figure P4.16, an imaginary circuit has been chosen. For the chosen circulation
direction, we have the sign for Vemf as shown. Then,
h

o I 
 m  o I
Vemf    u  B  dL    -1a  
a  dza z   1 
 dz ,
2 
 s  2 0

Vemf  160nV . Therefore, the bottom of the bar is positive.
Fig. P4.16
P4.17: Suppose we have a conductive bar moving along a pair of conductive rails as
in Figure 4.12, only now the magnetic flux density is B = 4.0ax + 3.0az Wb/m2. If R =
10. , w = 20. cm, and uy = 3.0 m/s, calculate the current induced and indicate its
direction.
 Wb  m 
Vemf    B u y dxa z    3 2  3   0.2m   1.8V
 m  s 
1.8V
I
 0.18 A (clockwise when viewed from the +z axis)
10
P4.18: The radius r of a perfectly conducting metal loop in free space, situated in the
x-y plane, increases at the rate of (r)-1 m/sec. A break in the loop has a small 2.0
ohm resistor across it. Meanwhile, there exists a magnetic field B = 1.0 az T.
Determine the current induced in the loop, and show in a sketch the direction of flow.
Here we’ve assumed dS = -dSaz to get iind and Vemf as shown. Our approach will be to
find , then Vemf = -d/dt.
   B dS   1a z    d  d a z 
r
 2   d    r 2
0
d
dr
 1 
 2 r
 2 r    2V
dt
dt
r 
Vemf  2V
I
2V
 1A, clockwise as shown
2
Fig. P4.18
P4.19: Rederive Vemf for the rectangular loop of Figure 4.16 if the magnetic field is
now B = Boaz.
We see in Figure P4.19a that  u  B  dL  0 for the 1  2 and 3  4 line sections.
For the 2  3 section we have:
3

2
   u  B  dL; B  Boa z , dL  d  a  , u 
Fig. P4.19a
d l  d a

 a
dt
dt
Fig. P4.19b
 Bo a 2
  u  B  dL   Bo a  d   2
0
a
  u  B  dL   Bo   d  
0
 Bo a 2
, so for the 2  3 section, the contributions to
2
Vemf cancel. This will also be the case for the 4  1 section, and therefore Vemf = 0;
no current is induced.
P4.20: In Figure 4.16, replace the rectangular loop with a circular one of radius a and
rederive Vemf.
   B dS   Boa y dS = Boa y  a 2   sin  a x  cos  a y   Bo a 2 cos 
d flux
dt
  Bo a 2 sin  ;
Vemf  
d flux
dt
  Bo a 2 sin 
P4.21: 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 60 revolutions per second in a
magnetic field B = 100. mT az. Find the voltage at the end of the bar.



   60
Vemf
u
rev  2 rad 
rad

  120
s  rev 
s
  u  B  dL
d
a   a
dt
l
Vemf    a  Boa z  d  a  
0
 Bo l 2
2
1
rad  Wb 
2 Vs
Vemf  120
 0.1 2   0.06m 
2
s 
m 
Wb
 68mV
Fig. P4.21
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.
P4.22: Consider the rotating conductor shown in Figure 4.24. 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 = 6.0 cm, and R = 50. , determine I if the bar rotates at 1.0
revolution per second.
d
a   a , dL  d  a 
dt
Figure P4.22 indicates one of the paths for the circulation integral.
a
a
 Bo a 2
Vemf    a  Boa z  d  a     Bo   d  
2
0
0
Vemf 
I
  u  B  dL;
Vemf

u
 Bo a 2
R
I  22.6  A
1  rev  rad 
2  1  Vs A
3 Wb 
 1
0.06m  
 2
100 x10

2 
2R
2  s 
rev 
m 
 50  Wb V
Fig. P4.22
P4.23: A Faraday Disk Generator is similar to the rotating conductor of P4.22, only
now the rotating element is a disk instead of a bar. Derive an expression of the Vemf
produced by a Faraday Disk Generator, and using the parameters given in problem
4.22, find I.
Worked exactly as P4.22.
P4.24: Consider a sliding rail problem where the conductive rails expand as they
progress in the y direction as shown in Figure 4.25. If w = 10. cm and the distance
between the rails increases at the rate of 1.0 cm in the x direction per 1.0 cm in the y
direction, and uy = 2.0 m/sec, find the Vemf across a 100.  resistor at the instant when
y = 10. cm if the field is Bo = 100. mT.
First we modify the figure so that the top rail is horizontal and all the spreading occurs
via the bottom rail. As before, our approach will be to find  and then d /dt. We
have:
   B dS   Boa z dxdya z
Now, notice that x and y are not independent and are in fact related: x=y+w
So we have
  Bo
y yw
y
1

dxdy  Bo   y  w dy  Bo  y 2  wy 
2

y 0 x 0
0
 
d
dy
Wb 

 m
  Bo  y  w    Bo  y  w  u y   0.100 2   0.1m  0.1m   2 
dt
dt
m 

 s
 40mV
Vemf  
Vemf
Alternate Method:
Vemf    u  B  dL    u y a y  Boa z  dxa x
1
 y
2
Vemf  u y Bo

1
w y
2
Fig. P4.24
dx  u y Bo  w  y 
Uniform Plane Wave and Power Transmission
1. A wave with  = 6.0 cm in air is incident on a nonmagnetic, lossless liquid media.
In the liquid, the wavelength is measured as 1.0 cm. What is the wave’s frequency (a)
in air? (b) in the liquid? (c) What is the liquid’s relative permittivity?
3x108 m s
(a) f 
 
 5GHz
 
0.06m
(b) the frequency doesn’t change with the media (the wavelength does) so f = 5 GHz
(c)
1
m
c

u p   f   5 x109   0.01m   5 x107 
s
s
r

up
c
2
 3x108 
 r  
 36
8 
 0.5 x10 
_____________________________________________________________________
2. Given  = 1.0x10-5 S/m , r = 2.0, r = 50., and f = 10. MHz, find , , , and .
  jr o   j r o     j

jr o
  j r  o
jr o  j 2 10 x106   50   4 x107   j 3948
  j r  o  1x105  j 2 10 x106   2  8.854 x1012   1x105  j1.11x103
Inserting these into the expressions for  and ,
  9.4 x103  j 2.1 1 m ,   9.4 x103 Np m ,   2.1 rad m ,   1880e j 257 
_____________________________________________________________________
3. Suppose in free space, H(x,t) = 100.cos(2x107t – x + /4) az mA/m. Find E(x,t).
H s  0.100e j  x e j a z , a P  a x ,
   4 
E s   a P  H s  120 a x  0.100e j  x e j a z  12 e  j  x e j a y
E  12 cos t   x    a y
Since free space is stated,
2 2


 2 30 rad m
 c f
and then
2
 V

E  12 cos  2 x107 t 
x  ay
30
4 m

_____________________________________________________________________
4. A 100 MHz wave in free space propagates in the y direction with an amplitude of 1
V/m. If the electric field vector for this wave has only an az component, find the
instantaneous expression for the electric and magnetic fields.
From


up
the

given
information
we
have
  2 f  200 x106
2 rad
,
3 m
rad
s
and
2  V

y  az .
or E( y, t )  1cos  200 x106 t 
3  m

Now to find H.
1
1
1  j y
Es  1e j y a z , H s  a P  Es 
a y 1e j y a z 
e ax

120
120
So
1
2  A

H  y, t  
cos  200 x106 t 
y  ax
120
3  m

or
2  mA

H  y, t   2.7 cos  200 x106 t 
y  ax
.
3 
m

_____________________________________________________________________
5. In a lossless, nonmagnetic material with r = 16, H = 100 cos(t – 10y) az mA/m.
Determine the propagation velocity, the angular frequency, and the instantaneous
expression for the electric field intensity.
up 

c
3x108
m


 0.75 x108

s
r
16
  u p    0.75 x108  10   7.5 x108
rad
s
mA
H ( y, t )  100 cos  7.5 x108 t  10 y  a z
m
 j y
H s  0.100e a z ,
Es  a P  H s 
120
r
a y  0.100e j y a z  3 e j y a x
E( y, t )  9.4 cos  7.5 x108 t  10 y  a x
V
m
_____________________________________________________________________
6. In a media with properties  = 0.00964 S/m , r = 1.0, r = 100., and f = 100. MHz,
a 1.0 mA/m amplitude magnetic field travels in the +x direction with its field vector in
the z direction. Find the instantaneous form of the related electric field intensity.
 mA   x
 x  j  x
H  1
 e cos t   x  a z ; H s  H o e e a z
 m 
Es  aP  Hs  a x  Hoe x e j xa z   Hoe x e j xa y
j 2 100 x106  100   4 x107 
j


 2664e j 30 
6
12
  j
0.00964  j 2 100 x10  8.854 x10 

j   j   14.8  j 25.7
1
m
Finally,
E( x, t )  2.66e 15 x cos  200 x106 t  26 x  30  a y
V
m
_____________________________________________________________________
7. In seawater, a propagating electric field is given by E(z,t) = 20.e-z cos(xt –
z + 0.5) ay V/m. Assuming ’’=0, find (a)  and , and (b) the instantaneous form of
H.
For seawater we have r = 72,  = 5, and r = 1.
So: jo  j 7.896, j r o  j 0.004

j
 1.257e j 44.98 
  j
 
j   j   4.441  j 4.445 1 m
    4.4
1
m
V
V
 20e  z e   z e j 28.6 a y
m
m
1
1

20
A
H s  a P  Es  a z  20e  z e   z e j 28.6 a y 
e  z e   z e j 28.6 a x



m
A
H( z, t )  15.9e 4.4 z cos  2 x106 t  4.4 z  28.6  45  a x
m
or with appropriate significant digits:
A
H( z, t )  16e 4.4 z cos  2 x106 t  4.4 z  16  a x
m
_____________________________________________________________________
8. For Nickel ( = 1.45 x 107, r = 600), make a table of , , , up, and  for 1Hz,
1kHz, 1MHz, and 1 GHz.
Es  20e  z e   z e j 0.5radians a y
For Ni we have = 1.45x107S/m, r = 600
     f    f  Hz  600  4 x107 1.45x107   34.35x103 f ( Hz )
= 1/
 j 45
e   18.08 x106 f ( Hz )e j 45 

c
m
up 
 12 x106
s
r r
Table
f(Hz)=
1
103
185
5860
(Np/m)
185
5860
(rad/m)
j45º
18e 
570ej45º

5.4mm
170m

up(m/s)
12x106
12x106
 2
106
185x103
185x103
18ej45ºm
5.3m
12x106
109
5.9x106
5.9x106
0.57ej45º
170nm
12x106
9. A semi-infinite slab exists for z > 0 with  = 300 S/m, r = 10.2, and r = 1.0. At
the surface (z = 0),
E(0,t) = 1.0 cos( x 106t) ax V/m.
Find the instantaneous expressions for E and H anywhere in the slab.
The general expression for E is: E( z, t )  1.0e  z cos  x106 t   z  a x
j  j  x106  4 x107   j 3.948
V
m
j  j  x106  10.2   8.854 x1012   j 284 x10 6
Here,  (i.e. it is a good conductor), so
1
   f   24.3  
m
 j 45
  2 e  0.115e j 45 

So now we have
V
E( z, t )  1.0e 24 z cos  x106 t  24 z  a x
m
To find B we’ll work in phasors.
Es  1e z e j z a x , H s 
1

a P  Es 
1
a z 1e z e j z a x 

1

e z e j z a y
1
A
e 24 z cos  x106 t  24 z  45  a y
0.115
m
A
H( z, t )  8.7e 24 z cos  x106 t  24 z  45  a y
m
_____________________________________________________________________
10. A 600 MHz uniform plane wave incident in the z direction on a thick slab of
Teflon (r = 2.1, r = 1.0) imparts a 1.0 V/m amplitude y-polarized electric field
intensity at the surface. Assuming  = 0 for Teflon, find in the Teflon (a) E(z,t), (b)
H(z,t) and (c) Pav.
H( z, t ) 


E(0, t )  1cos 2  600 x106  t   z a y
E( z , t )  1e  z cos t   z  a y
Teflon:  = 0 so  = 0,
and     

c
r 
V
m
V
m
2  600 x106 
8
3x10
(a) E( z , t )  1cos 1.2 x109 t  18.2 z  a y
2.1  18.2
V
m
1
2.1
V
(b) H s  a P  Es 
a z 1e j z a y ,

120
m
mA
H ( z , t )  3.8cos 1.2 x109 t  18.2 z  a x
m
rad
m
1 1 2.1
mW
(c) Pavg 
a z  1.9 2 a z
2 120
m
_____________________________________________________________________
11. A 200 MHz uniform plane wave incident on a thick copper slab imparts a 1.0
mV/m amplitude at the surface. How much power passes through a square meter at
the surface? How much power passes through a square meter area 10. m beneath the
surface?
2
mV
1 Eo 2
f  200MHz , Eo  1
, Pavg 
m
2 
Cu:   2
 j 45
Np
e ,    f   214 x103
, so   5.22e j 45 m

m
3
1 10 
W
Pavg 
 96 2 ; P  Pavg S  96W
3
2 5.22 x10
m
Now at 10 m beneath the surface, we have
2
E ( z  10 m)  Eo e  (10  m )  103 e  (214 x10 )(10  m )  118 x106
3
V
m
1 118 x10 
W
Pavg 
 1.3 2 ; P  1.3W
3
2 5.22 x10
m
_____________________________________________________________________
12. Given E(z,t) = 10.cos(t-z)ax - 20.cos(t-z-45)ay V/m, find the polarization
and handedness.
6 2
The field can be rewritten as E(z,t) = 10.cos(t-z)ax + 20.cos(t-z-45-180°)ay
or E(z,t) = 10.cos(t-z)ax + 20.cos(t-z+135°)ay
_____________________________________________________________________
13. Given
E( z, t )  Exo cos t   z  ax  E yo cos t   z    ay ,
we say that Ey leads Ex for 0 <  < 180, and that Ey lags Ex when –180 <  < 0.
Determine the handedness for each of these two cases.
For 0 <  < 180°, we have LHP
For 180° <  < 360°, we have RHP
_____________________________________________________________________
14. Suppose a UPW in air carrying an average power density of 100 mW/m2 is
normally incident on a nonmagnetic material with r = 11. What is the time-averaged
power density of the reflected and transmitted waves?
 1
 1


120
11
  0.537
1  o  120; 2 
; 


11
 1
 1
11 

  1   0.463
r
i
Pavg
  Pavg
 28.8
2
mW
m2
2
1 Exo
mW
i
P 
  2 Pavg
11  71.2 2
2 2
m
_____________________________________________________________________
15. A UPW in a lossless nonmagnetic r = 16 media (for z < 0) is given by
E(z,t) = 10.cos(t-1z)ax + 20.cos(t-1z+/3)ay V/m.
This is incident on a lossless media characterized by r = 12, r = 6.0 (for z > 0). Find
the instantaneous expressions for the reflected and transmitted electric field
intensities.
t
avg
Eis  10e j1z a x  20e j1z e j 3a y
Ers  10e j1za x  20e j1z e j 3a y
120
12
 30; 2  120
 120 2
6
16
 
  2 1  0.700;   1    1.70
2  1
1 
Ers  7e j1z a x  14e j1z e j 3a y
 V

E(rz ,t )  7 cos t  1 z  a x  14 cos  t  1 z   a y
3
m

 j 2 z
 j 2 z j 3
 j 2 z
t
t
Es  10 e
a x  20 e
e a y , or Es  17e
a x  34e j2 z e j 3a y , so
 V

Et( z ,t )  17 cos t   2 z  a x  34 cos  t   2 z   a y
.
3
m

_____________________________________________________________________
16. The wave Ei = 10.cos(2x 108t - 1z) ax V/m is incident from air onto a copper
conductor. Find Er, Et and the time-averaged power density transmitted at the
surface.
For copper we have
2  2
 2 j 45
e 
2
where  2   f 2 2   108  4 x107  5.8 x107   151x103
Np
 2
m
so 2  3.7e j 45 m
22
2
 2  19.6 x106 e j 45
1   2 1
r
8
So E = -10.cos(2x 10 t + 1z) ax V/m
V
V
Ets  196  e  2 z e  j2 z e j 45 a x
, and Et  196e  2 z cos t   2 z  45  a x
m
m
We find   1, and  =
6
1 196 x10 V m 
W

cos  45  a z  3.7 2 a z .
3
2  3.7 x10  
m
2
t
avg
P
17. A wave specified by Ei = 100.cos(x107t-1z)ax V/m is incident from air (at z < 0)
to a nonmagnetic media (z > 0,  = 0.050 S/m, r = 9.0). Find Er, Et and SWR. Also
find the average power densities for the incident, reflected and transmitted waves.
rad
2 
rad
so f  5 x106 Hz, 1 
  0.105
s
1 c
m
In this problem we find in medium 2 (z > 0) that  = 0.0025 and  = 0.05. These
values are too close to allow for simplifying assumptions. Using (5.13) and (5.31),
we calculate:
Np
rad
 2  0.969
,  2  1.019
, 2  28.1e j 43.6  .
m
m
Then,
1 
 
  2 1  0.898e j174 , SWR 
 18.6,   1    0.141e j 40.8
2  1
1 
1  120,    x107
V
m
V
V
Ers  100e  j 1z a x  89.8e  j 1z e j174 a x ,
m
m
Eis  100e  j 1z a x
so Er ( z , t )  89.8cos  x107 t  0.105 z  174  a x
Ets  100 e j2 z a x
V
.
m
V
V
 14.1e j 2 z e j 40.8 a x ,
m
m
so Et ( z , t )  14.1cos  x107 t  1.02 z  40.8  a x
V
.
m
14.1  j 43.6  j 2 z j 40.8
A
A
e
e
e
a y  0.502e  j 2 z e  j 2.8 a y
28.1
m
m
1
W
P t  14.1 0.502  cos  40.8  2.8  a z  2.6 2 a z
2
m
H ts 
100   13.3 W a
P 
z
2 120 
m2
2
  89.8 
W
Pr 
 10.7 2  -a z 
2 120 
m
2
i
(check: 13.3 W/m2 = 10.7 W/m2 + 2.6 W/m2)
_____________________________________________________________________
18. A 100 MHz TM polarized wave with amplitude 1.0 V/m is obliquely incident
from air (z < 0) onto a slab of lossless, nonmagnetic material with r = 25 (z > 0). The
angle of incidence is 40. Calculate (a) the angle of transmission, (b) the reflection
and transmission coefficients, and (c) the incident, reflected and transmitted fields.
(a) The material parameters in this problem are the same as for P5.48. So, once again
we have t = 7.4°. Also, 1 = 2.09 rad/m and 2 = 10.45 rad/m.
(b)
 cos t  1 cos i
TM  2
 0.589
2 cos t  1 cos i
 TM 
22 cos i
 0.318
2 cos t  1 cos i
(c)
Incident:
Eis  1e j1.34 x e j1.60 z  cos 40 a x  sin 40 a z 
Ei ( z , t )   0.766a x  0.643a z  cos t  1.34 x  1.60 z 
H is 
V
m
1  j1.34 x  j1.60 z A
e
e
ay
120
m
H i ( z, t )  2.65cos t  1.34 x  1.60 z  a y
mA
m
Reflected:
Ers  0.589e j1.34 x e j1.60 z  cos 40 a x  sin 40 a z 
Er ( z , t )   0.452a x  0.379a z  cos t  1.34 x  1.60 z 
H rs 
0.589  j1.34 x  j1.60 z A
e
e
ay
120
m
H r ( z , t )  1.56 cos t  1.34 x  1.60 z  a y
V
m
mA
m
transmitted:
Ets  0.318e j1.35 x e j10.4 z  cos 7.4 a x  sin 7.4 a z 
Et ( z , t )   0.315a x  0.041a z  cos t  1.35 x  10.4 z 
V
m
mA
m
_____________________________________________________________________
H t ( z, t )  4.22 cos t  1.35 x  10.4 z  a y