EE140 Spring 2021 Mid Term 2 Dr. Ray Kwok

1. A ring made of thin wire is placed in a magnetic field ⃗ = (3 − 4̂ − 5̂) cos(60) Tesla. The
radius of the ring is oscillating as: r = 2 + sin(0.5t) meters. Find the magnitude and direction
of the induced current. The resistivity of the wire is  m










Φ = ∫ ⃗ ∙ = −5cos (60) 2
= −
Φ

= 300 sin(60) 2 − 5 cos(60) 2


= 300 sin(60) 2 − 5 cos(60) (0.5)

For small t, the dA/dt term dominates (changing area).
Lenz’s Law: increasing area more flux ⊗, induced B is ⊙, induced current is CCW at t = 0+.

=


=
5 cos(60) (0.5) − 300 sin(60) 2
2
=
2.5 cos(60) (0.5) − 150 sin(60)


CCW







x
y
r
o
EE140 Spring 2021 Mid Term 2 Dr. Ray Kwok

2. A LONG coaxial cable of radius “a” is carrying a non-uniform current density = 2̂ (out of the page),
where  is a constant. The outer shield is a grounded conductor and a dielectric r is filled between the radii
a and b. The outer shield is grounded as shown.
(a) What is the magnetic (B) field inside the cable between radii a and b?
(b) Use the formula sJHHn

=− )(ˆ 212 to derive the surface current density at r = b.









































a
b
c
( )
( )







ˆ
4
4
22)2(
ˆ
4
4
22)2(
4
4
0
2
3
4
0
2
r
a
B
a
rdrrIrB
bra
r
B
r
rdrradJrB
IdB
ar
o
o
a
oinsideo
o
o
r
oo
insideo
=
===

=
===
=









)__(
4
ˆ0ˆ
4
ˆ
0ˆ
4
ˆ
4
ˆ
)(ˆ
44
33
212
pageofoutJ
b
a
z
b
a
r
br
J
aa
r
ar
JHHn
s
s
s



=





−=





−





−
=
==











−





−
=
=−







CCW
Into the page
EE140 Spring 2021 Mid Term 2 Dr. Ray Kwok

z

3. A large block of material with permeability 2o is found to be tilted at an angle of 37
o
from vertical, as shown
in figure. The magnetic field inside and outside of the block are given below. Find the induced surface current
density vector. Note: cos(37
o
) = 4/5, sin(37
o
) = 3/5.







































y
37
o

1=o
2=o
n
37
o

zyxB
zyxB
ˆ5ˆ10ˆ6
ˆˆ2ˆ3
2
1
++=
−+=


( )







 −−

+−
=







 ++
−
−+

+−
=
−=
+−
=+−=
o
s
oo
s
s
oo
zyzy
J
zyxzyxzy
J
HHnJ
zy
zyn


ˆˆ3
5
ˆ4ˆ3
2
ˆ5ˆ10ˆ6ˆˆ2ˆ3
5
ˆ4ˆ3
ˆ
5
ˆ4ˆ3
37cosˆ37sinˆˆ
2
7
212
2



xx
zyx
J
ooo
s
ˆ
2
9
12
2
21
ˆ
5
1
30
430
ˆˆˆ
5
1
2
7

=





−−
−
=−
−
=

EE140 Spring 2021 Mid Term 2 Dr. Ray Kwok

4. A car top antenna can be modeled as a current element above an
infinite ground plane. Assuming the current is constant, find the
magnetic field at a point P away from the antenna, as shown in the
figure.








































csc
sin
csc'
'
tan
sin'
4
ˆ
sin'
4
ˆ
4
2
2
22
Rr
r
R
dRdz
zz
R
r
dzI
B
r
Idz
r
rlId
Bd
L
L
o
oo
=
=
=
−
=
=
=

=

−


I
z
o
L
P(x,y,z) r
R

I
 
222
2222
'
2222
2
2
)()(4
ˆ
)'(
'
4
ˆcos
4
ˆ
csc
sincsc
4
ˆsin'
4
ˆ
yxR
LzR
Lz
LzR
Lz
R
I
B
zzR
zz
R
I
R
I
R
dRI
r
dzI
B
o
L
Lz
ooo
L
L
o
+=








−+
−
−
++
+
=








−+
−
−=−=

==
−=
−
























































































































































































































































































































































































































































































































































































































































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