Tutorial 1
Problem 1: Within a certain region,   1011 F/m and   105 H/m. If
Bx  2  104 cos105 t sin103 y T:
(a) use   H  
E
to find E ;
t
(b) find the total magnetic flux passing through the surface x  0 , 0  y  40 m,
0  z  2 m, at t  1 s ;
(c) find the value of the closed line integral of E around the perimeter of the given
surface.
Solution:
(a) H x 
Bx

H  

2  104 cos105 t sin103 y
 20cos105 t sin103 y
5
10
H x
H x
z
y  0.02cos105 t cos103 y z
y
z
A
A
m
m2
E   H 0.02cos105 t cos103 yz


 2  109 cos105 t cos103 yz
t

1011
V
ms
E
dt  2  104 sin105 t cos103 yz V
m
t
(b) We assume that the integral surface direction (direction of dS )is  x , same as
the flux density.
E
   B  dS  
40
0

2
0
2
40
0
0
Bx dzdy   dz  2  104 cos105 t sin103 ydy
S
40
cos103 y
 2  2  10 cos10 t  ( 
)  3.1996  104 cos105 t
3
10
0
4
5
(t  1 s)  3.1996  104 cos(105  106 )  3.18 104
Wb
Wb
(c) The line integral is taken along the counter clockwise direction (right hand rule
based on the surface direction).
2
E dl
0
Ez ( y
0
40)dz
2
Ez ( y
0)dz
L
2
( 2 104 sin105 t cos(0.04)
2 104 sin105 t )dz
0
2 2 104 sin105 t ( cos(0.04) 1)
2 2 104 sin(105 10 6 ) (1 cos(0.04))
3.19
V
Or
E dl
emf
L
d
dt
3.1996 10
4
105 sin(105 t )
3.1996 10 sin(105 10 6 )
3.19
V
Note that in the case of electrostatics, line integral about the closed loop is zero.
However, in the case of time-vary field, the system is not conservative due to the
presence of external magnetic source.
z
E
B
D
C
A
B
y
x
Figure 1
Problem 2 With reference to the sliding bar shown in Figure 2, let d=7cm, B=0.3 z
T, and v=0.1 y e20 y m/s. Let y=0 at t=0. Find
(a) v(t  0) ;
(b) y(t  0.1) ;
(c) v(t  0.1) ;
(d) V12 at t=0.1
Solution:
(a) v(t  0)  0.1e200  0.1
m
s
(b)
v  0.1e
20 y
y
t
dy
e 20 y

 e 20 y dy  0.1dt   e 20 y dy   0.1dt 
0
0
dt
20
 e20 y  1  2t  y  
1
ln(1  2t )
20
y(t  0.1)  0.05  ln(1  2  0.1)  0.0112m
(c) v 
dy
1 d
1

ln(1  2t ) 
dt
20 dt
10(1  2t )
y
 0.1t 0
t
0
v(t  0.1) 
1
 0.125 m
s
10(1  2  0.1)
(d) V12  emf  
d
dy
  B0 d   B0vd  0.3  0.125  0.07  0.002625V
dt
dt
Figure 2
Problem 3 A rectangular loop of wire containing a high-resistance voltmeter has
corners initially at (a/2,b/2,0), (-a/2,b/2,0), (-a/2,-b/2,0), and (a/2,-b/2,0). The loop
begins to rotate about the x axis at constant angular velocity  , with the first-named
corner moving in the z direction at t=0. Assume a uniform magnetic flux density
B  B0z . Determine the induced emf in the rotating loop and specify the direction of
the current.
Solution:
Assume the direction of the loop wire’s surface is z at t=0, then the angle between
surface direction vector and the flux density should be   t , then
   B  dS  B0 S cos( )  B0ab cos(t )  emf  
S
d
 B0ab sin(t ) V
dt
Decide the direction of current with Lenz’s law: if an induced current flows, its
direction is always such that it will oppose the change which produced it. In Figure 4:
1) When changes from 0 to / 2 , original flux through the loop decreases, this
change of flux will generate a current, and the current will generate a flux to
oppose the decreasing, so the induced flux must have the +z direction component.
From the right hand rule, the current should be A->B->C->D.
2) When changes from / 2 to , original flux through the loop increases, this
change of flux will generate a current, and the current will generate a flux to
oppose the increasing, so the induced flux must have the -z direction component.
From the right hand rule, the current should be A->B->C->D.
3) When changes from to 3 / 2 , original flux through the loop decreases, this
change of flux will generate a current, and the current will generate a flux to
oppose the decreasing, so the induced flux must have the +z component. From the
right hand rule, the current should be A->D->C->B.
4) When changes from 3 / 2 to , original flux through the loop increases, this
change of flux will generate a current, and the current will generate a flux to
oppose the increasing, so the induced flux must have the -z component. From the
right hand rule, the current should be A->D->C->B.
B'
B
z
A'

C
D
a
2
x
B
y
b
2
A
C'
D'
Figure 3
A( B )
A( B )
z
I
B'
z
B

I


B'
D (C)
A( B )
Figure 4
z
B
I
B'
y
y
D (C)
D (C)
D (C)
z
B
B
I
y

B'
y
A( B )