Download Homework #6.EE135

Homework
#6.EE135
Problem 5.28 A uniform current density given by
J = ẑJ0
(A/m2 )
gives rise to a vector magnetic potential
A = −ẑ
µ0 J0 2
(x + y2 ) (Wb/m)
4
(a) Apply the vector Poisson’s equation to confirm the above statement.
(b) Use the expression for A to find H.
(c) Use the expression for J in conjunction with Ampère’s law to find H. Compare
your result with that obtained in part (b).
Solution:
(a)
$
"#
"2
"2
"2
J0 2
2
µ
+
+
+
y
)
−
(x
0
" x2 " y2 " z2
4
J0
= −ẑ µ0 (2 + 2) = −ẑ µ0 J0 .
4
!2 A = x̂ !2 Ax + ŷ !2 Ay + ẑ !2 Az = ẑ
!
Hence, !2 A = −µ0 J is verified.
(b)
# !
"
!
"
!
"$
" Ay " Ax
1
" Az " Ay
" Ax " Az
1
×A =
H=
−
−
−
!×
x̂
+ ŷ
+ ẑ
µ0
µ0
"y
"z
"z
"x
"x
"y
!
"
" Az
" Az
1
x̂
− ŷ
=
µ0
"y
"x
"
"$
#
!
!
"
J0
"
J0
1
x̂
−µ0 (x2 + y2 ) − ŷ
−µ0 (x2 + y2 )
=
µ0
"y
4
"x
4
J0 x
J0 y
+ ŷ
(A/m).
= −x̂
2
2
(c)
%
H · dl = I =
!
C
%
S
J · ds,
#̂# H# · #̂#2$ r = J0 · $ r2 ,
r
H = #̂# H# = #̂# J0 .
2
Homework
#6.EE135
Homework
#6.EE135
Problem 5.32
The x–y plane separates two magnetic media with magnetic
permeabilities µ1 and µ2 (Fig. P5.32). If there is no surface current at the interface
and the magnetic field in medium 1 is
H1 = x̂H1x + ŷH1y + ẑH1z
find:
(a) H2
(b) !1 and !2
(c) Evaluate H2 , !1 , and !2 for H1x = 2 (A/m), H1y = 0, H1z = 4 (A/m), µ1 = µ0 ,
and µ2 = 4µ0
z
θ1
H1
μ1
x-y plane
μ2
Figure P5.32: Adjacent magnetic media
(Problem 5.32).
Solution:
(a) From (5.80),
µ1 H1n = µ2 H2n ,
and in the absence of surface currents at the interface, (5.85) states
H1t = H2t .
In this case, H1z = H1n , and H1x and H1y are tangential fields. Hence,
µ1 H1z = µ2 H2z ,
H1x = H2x ,
H1y = H2y ,
and
H2 = x̂H1x + ŷH1y + ẑ
µ1
H1z .
µ2
Homework
#6.EE135
Homework
#6.EE135
Problem 5.40 The rectangular loop shown in Fig. P5.40 is coplanar with the long,
straight wire carrying the current I = 20 A. Determine the magnetic flux through the
loop.
z
20 A
30 cm
5 cm
20 cm
y
x
Figure P5.40: Loop and wire arrangement for
Problem 5.40.
Solution: The field due to the long wire is, from Eq. (5.30),
B = !ˆ
µ0 I
µ0 I
µ0 I
= −x̂
= −x̂
,
2" r
2" r
2" y
where in the plane of the loop, !ˆ becomes −x̂ and r becomes y.
The flux through the loop is along −x̂, and the magnitude of the flux is
#=
!
S
µ0 I 20 cm x̂
− · −x̂ (30 cm × dy)
2" 5 cm
y
! 0.2
µ0 I
dy
× 0.3
=
2"
0.05 y
"
#
0.2
0.3 µ0
× 20 × ln
=
= 1.66 × 10−6
2"
0.05
B · ds =
!
(Wb).
Homework
#6.EE135
Problem 6.2 The loop in Fig. P6.2 is in the x–y plane and B = ẑB0 sin ! t with B0
positive. What is the direction of I ("ˆ or −"ˆ ) at:
(a) t = 0
(b) ! t = # /4
(c) ! t = # /2
z
R
y
Vemf
I
x
Figure P6.2: Loop of Problem 6.2.
Solution: I = Vemf /R. Since the single-turn loop is not moving or changing shape
m = 0 V and V
tr
with time, Vemf
emf = Vemf . Therefore, from Eq. (6.8),
tr
/R =
I = Vemf
−1
R
!
S
$B
· ds.
$t
If we take the surface normal to be +ẑ, then the right hand rule gives positive
flowing current to be in the +"ˆ direction.
I=
−AB0 !
−A $
B0 sin ! t =
cos ! t
R $t
R
(A),
where A is the area of the loop.
(a) A, ! and R are positive quantities. At t = 0, cos ! t = 1 so I < 0 and the
current is flowing in the −"ˆ direction (so as to produce an induced magnetic field
that opposes B).
√
(b) At ! t = # /4, cos ! t = 2/2 so I < 0 and the current is still flowing in the −"ˆ
direction.
(c) At ! t = # /2, cos ! t = 0 so I = 0. There is no current flowing in either direction.
Homework
#6.EE135
Problem 6.6 The square loop shown in Fig. P6.6 is coplanar with a long, straight
wire carrying a current
I(t) = 5 cos(2! × 104t) (A).
(a) Determine the emf induced across a small gap created in the loop.
(b) Determine the direction and magnitude of the current that would flow through
a 4-" resistor connected across the gap. The loop has an internal resistance of
1 ".
z
10 cm
I(t)
10 cm
5 cm
y
x
Figure P6.6: Loop coplanar with long wire (Problem 6.6).
Solution:
(a) The magnetic field due to the wire is
B = #̂#
µ0 I
µ0 I
= −x̂
,
2! r
2! y
where in the plane of the loop, #̂# = −x̂ and r = y. The flux passing through the loop
is
#
!
! 15 cm "
µ0 I
$ = B · ds =
· [−x̂ 10 (cm)] dy
−x̂
2! y
S
5 cm
−1
15
µ0 I × 10
ln
=
2!
5
4! × 10−7 × 5 cos(2! × 104t) × 10−1
× 1.1
=
2!
4
−7
= 1.1 × 10 cos(2! × 10 t) (Wb).
Homework
#6.EE135
Homework
#6.EE135
Problem 6.9 A rectangular conducting loop 5 cm × 10 cm with a small air gap in
one of its sides is spinning at 7200 revolutions per minute. If the field B is normal to
the loop axis and its magnitude is 6 × 10−6 T, what is the peak voltage induced across
the air gap?
Solution:
2" rad/cycle × 7200 cycles/min
= 240" rad/s,
60 s/min
2
A = 5 cm × 10 cm/(100 cm/m) = 5.0 × 10−3 m2 .
!=
From Eqs. (6.36) or (6.38), Vemf = A! B0 sin ! t; it can be seen that the peak voltage
is
peak
Vemf = A! B0 = 5.0 × 10−3 × 240" × 6 × 10−6 = 22.62 (µ V).
Homework
#6.EE135
Problem 6.18 An electromagnetic wave propagating in seawater has an electric
field with a time variation given by E = ẑE0 cos ! t. If the permittivity of water is
81"0 and its conductivity is 4 (S/m), find the ratio of the magnitudes of the conduction
current density to displacement current density at each of the following frequencies:
(a) 1 kHz
(b) 1 MHz
(c) 1 GHz
(d) 100 GHz
Solution: From Eq. (6.44), the displacement current density is given by
#
#
D=" E
#t
#t
and, from Eq. (4.67), the conduction current is J = $ E. Converting to phasors and
taking the ratio of the magnitudes,
!
! ! !
! !
!"
" !!
$
! J ! ! $E
.
!=
! !=!
" ! !"r "0
!"
Jd ! ! j!"r "0 E
Jd =
(a) At f = 1 kHz, ! = 2% × 103 rad/s, and
! !
!"
!
4
!J!
= 888 × 103 .
! !=
!"
Jd ! 2% × 103 × 81 × 8.854 × 10−12
The displacement current is negligible.
(b) At f = 1 MHz, ! = 2% × 106 rad/s, and
! !
!"
!
4
!J!
= 888.
! !=
!"
Jd ! 2% × 106 × 81 × 8.854 × 10−12
The displacement current is practically negligible.
(c) At f = 1 GHz, ! = 2% × 109 rad/s, and
! !
!"
!
4
!J!
= 0.888.
! !=
!"
Jd ! 2% × 109 × 81 × 8.854 × 10−12
Neither the displacement current nor the conduction current are negligible.
(d) At f = 100 GHz, ! = 2% × 1011 rad/s, and
! !
!"
!
4
!J!
= 8.88 × 10−3 .
! !=
!"
Jd ! 2% × 1011 × 81 × 8.854 × 10−12
The conduction current is practically negligible.