EE324
HW8
Spring12
Problem 5.22 A long cylindrical conductor whose axis is coincident with the z-axis
has a radius a and carries a current characterized by a current density J = ẑJ0 /r,
where J0 is a constant and r is the radial distance from the cylinder’s axis. Obtain an
expression for the magnetic field H for
(a) 0 ≤ r ≤ a
(b) r > a
Solution: This problem is very similar to Example 5-5.
(a) For 0 ≤ r1 ≤ a, the total current flowing within the contour C1 is
¶
Z r1
ZZ
Z 2π Z r1 µ
ẑJ0
· (ẑr dr d φ ) = 2π
J0 dr = 2π r1 J0 .
I1 =
J · ds =
r
φ =0 r=0
r=0
Therefore, since I1 = 2π r1 H1 , H1 = J0 within the wire and H1 = φ̂φJ0 .
(b) For r ≥ a, the total current flowing within the contour is the total current flowing
within the wire:
¶
ZZ
Z 2π Z a µ
Z a
ẑJ0
· (ẑr dr d φ ) = 2π
I=
J · ds =
J0 dr = 2π aJ0 .
r
φ =0 r=0
r=0
Therefore, since I = 2π rH2 , H2 = J0 a/r within the wire and H2 = φ̂φJ0 (a/r).
Problem 5.27 In a given region of space, the vector magnetic potential is given by
A = x̂5 cos π y + ẑ(2 + sin π x) (Wb/m).
(a) Determine B.
(b) Use Eq. (5.66) to calculate the magnetic flux passing through a square loop
with 0.25-m-long edges if the loop is in the x–y plane, its center is at the origin,
and its edges are parallel to the x- and y-axes.
(c) Calculate Φ again using Eq. (5.67).
Solution:
× A = ẑ5π sin π y − ŷπ cos π x.
(a) From Eq. (5.53), B = ∇×
(b) From Eq. (5.66),
Φ=
ZZ
B · ds =
Z 0.125 m
Z 0.125 m
y=−0.125 m x=−0.125 m
(ẑ5π sin π y − ŷπ cos π x) · (ẑ dx dy)
¶¯0.125
¯
cos π y ´¯¯0.125
¯
=
−5π x
¯
¯
π
x=−0.125
y=−0.125
µ ³ ´
µ
¶¶
−5
π
−π
=
cos
− cos
= 0.
4
8
8
µ³
(c) From Eq. (5.67), Φ =
Z
A · dℓℓ, where C is the square loop in the x-y plane with
n
C
sides of length 0.25 m centered at the origin. Thus, the integral can be written as
Φ=
Z
A · dℓℓ = Sfront + Sback + Sleft + Sright ,
n
C
where Sfront , Sback , Sleft , and Sright are the sides of the loop.
Sfront =
=
Z 0.125
x=−0.125
Z 0.125
x=−0.125
=
Sback =
³
(x̂5 cos π y + ẑ(2 + sin π x))|y=−0.125 ·(x̂ dx)
5 cos π y|y=−0.125 dx
¶
µ
³π ´
5
5
−π
= cos
= cos
,
x=−0.125
4
8
4
8
´¯0.125
¯
(5x cos π y)|y=−0.125 ¯
Z 0.125
x=−0.125
Z 0.125
(x̂5 cos π y + ẑ(2 + sin π x))|y=0.125 ·(−x̂ dx)
5 cos π y|y=0.125 dx
=−
x=−0.125
³
´¯0.125
¯
= (−5x cos π y)|y=0.125 ¯
³π ´
5
= − cos
,
4
8
x=−0.125
Sleft =
Z 0.125
y=−0.125
Z 0.125
=−
Sright =
=
Z
(x̂5 cos π y + ẑ(2 + sin π x))|x=−0.125 ·(−ŷ dy)
y=−0.125
0.125
y=−0.125
Z 0.125
y=−0.125
0|x=−0.125 dy = 0,
(x̂5 cos π y + ẑ(2 + sin π x))|x=0.125 ·(ŷ dy)
0|x=0.125 dy = 0.
Thus,
Φ=
Z
A · dℓℓ = Sfront + Sback + Sleft + Sright =
n
c
³π ´ 5
³π ´
5
− cos
+ 0 + 0 = 0.
cos
4
8
4
8
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
∇ A = x̂ ∇ Ax + ŷ ∇ Ay + ẑ ∇ Az = ẑ
−µ0 (x + y )
+
+
∂ x2 ∂ y2 ∂ z2
4
J0
= −ẑ µ0 (2 + 2) = −ẑ µ0 J0 .
4
2
2
2
µ
2
Hence, ∇2 A = −µ0 J is verified.
(b)
¶
µ
¶
µ
¶¸
· µ
∂ Ay ∂ Ax
1
∂ Az ∂ Ay
∂ Ax ∂ Az
1
×A =
H=
−
−
−
+ ŷ
+ ẑ
∇×
x̂
µ0
µ0
∂y
∂z
∂z
∂x
∂x
∂y
¶
µ
∂ Az
∂ Az
1
=
− ŷ
x̂
µ0
∂y
∂x
¶
¶¸
µ
µ
·
∂
∂
1
J0
J0
=
−µ0 (x2 + y2 ) − ŷ
−µ0 (x2 + y2 )
x̂
µ0
∂y
4
∂x
4
J0 y
J0 x
= −x̂
+ ŷ
(A/m).
2
2
(c)
Z
H · dl = I =
n
C
Z
S
J · ds,
φ̂φ Hφ · φ̂φ2π r = J0 · π r2 ,
r
H = φ̂φ Hφ = φ̂φ J0 .
2
z
r
J0
Figure P5.28: Current cylinder of Problem 5.28.
We need to convert the expression from cylindrical to Cartesian coordinates. From
Table 3-2,
y
x
φ̂φ = −x̂ sin φ + ŷ cos φ = −x̂ p
+ ŷ p
,
2
2
2
x +y
x + y2
p
r = x2 + y2 .
Hence
H=
Ã
y
x
−x̂ p
+ ŷ p
x2 + y2
x2 + y2
!
which is identical with the result of part (b).
·
yJ0
xJ0
J0 p 2
x + y2 = −x̂
+ ŷ
,
2
2
2
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) θ1 and θ2
(c) Evaluate H2 , θ1 , and θ2 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
(b)
q
2 + H2 ,
H1x
1y
q
2 + H2
H1x
1y
H1t
tan θ1 =
=
,
H1z
H1z
q
2 + H2
H1x
1y
µ2
H2t
tan θ1 .
=
tan θ2 =
=
µ
1
H2z
µ1
H1z
µ2
H1t =
(c)
1
H2 = x̂ 2 + ẑ · 4 = x̂ 2 + ẑ (A/m),
µ4 ¶
2
θ1 = tan−1
= 26.56◦ ,
4
µ ¶
2
θ2 = tan−1
= 63.44◦ .
1
Problem 5.33 Given that a current sheet with surface current density Js = x̂ 8 (A/m)
exists at y = 0, the interface between two magnetic media, and H1 = ẑ 11 (A/m) in
medium 1 (y > 0), determine H2 in medium 2 (y < 0).
Solution:
y
H1
n2
Js
x
H2
Figure P5.33: Adjacent magnetic media with Js on boundary.
Js = x̂ 8 A/m,
H1 = ẑ 11 A/m.
H1 is tangential to the boundary, and therefore H2 is also. With n̂2 = ŷ, from Eq.
(5.84), we have
n̂2 × (H1 − H2 ) = Js ,
× (ẑ 11 − H2 ) = x̂8,
ŷ×
× H2 = x̂ 8,
x̂ 11 − ŷ×
or
× H2 = x̂ 3,
ŷ×
which implies that H2 does not have an x-component. Also, since µ1 H1y = µ2 H2y
and H1 does not have a y-component, it follows that H2 does not have a y-component
either. Consequently, we conclude that
H2 = ẑ 3.
Problem 5.39 In terms of the dc current I, how much magnetic energy is stored in
the insulating medium of a 3-m-long, air-filled section of a coaxial transmission line,
given that the radius of the inner conductor is 5 cm and the inner radius of the outer
conductor is 10 cm?
Solution: From Eq. (5.99), the inductance per unit length of an air-filled coaxial
cable is given by
µ ¶
b
µ0
′
ln
(H/m).
L =
2π
a
Over a length of 2 m, the inductance is
µ ¶
10
3 × 4π × 10−7
ln
= 416 × 10−9
L = 2L =
2π
5
′
(H).
From Eq. (5.104), Wm = LI 2 /2 = 208I 2 (nJ), where Wm is in nanojoules when I is in
amperes. Alternatively, we can use Eq. (5.106) to compute Wm :
1
Wm =
2
Z
V
µ0 H 2 d V .
From Eq. (5.97), H = B/µ0 = I/2π r, and
1
Wm =
2
Z 3m Z 2π Z b
z=0 φ =0 r=a
µ0
µ
I
2π r
¶2
r dr d φ dz = 208I 2
(nJ).
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 = φ̂φ
µ0 I
µ0 I
µ0 I
= −x̂
= −x̂
,
2π r
2π r
2π y
where in the plane of the loop, φ̂φ becomes −x̂ and r becomes y.
The flux through the loop is along −x̂, and the magnitude of the flux is
µ0 I 20 cm x̂
− · −x̂ (30 cm × dy)
Φ = B · ds =
2π 5 cm
y
S
Z 0.2
µ0 I
dy
=
× 0.3
2π
0.05 y
¶
µ
0.3 µ0
0.2
=
= 1.66 × 10−6
× 20 × ln
2π
0.05
Z
Z
(Wb).
Problem 5.41 Determine the mutual inductance between the circular loop and the
linear current shown in Fig. P5.41.
y
a
x
d
I1
Figure P5.41: Linear conductor with current I1 next to
a circular loop of radius a at distance d (Problem 5.41).
Solution: To calculate the magnetic flux through the loop due to the current in the
conductor, we consider a thin strip of thickness dy at location y, as shown. The
magnetic field is the same at all points across the strip because they are all equidistant
(at r = d + y) from the linear conductor. The magnetic flux through the strip is
dΦ12 = B(y) · ds = ẑ
=
µ0 I
· ẑ 2(a2 − y2 )1/2 dy
2π (d + y)
µ0 I(a2 − y2 )1/2
dy
π (d + y)
1
L12 =
dΦ12
I S
Z
µ0 a (a2 − y2 )1/2 dy
=
π y=−a
(d + y)
Z
Let z = d + y → dz = dy. Hence,
p
Z
a2 − (z − d)2
µ0 d+a
L12 =
dz
π z=d−a
z
p
Z
µ0 d+a (a2 − d 2 ) + 2dz − z2
dz
=
π d−a
z
Z √
µ0
R
dz
=
π
z
where R = a0 + b0 z + c0 z2 and
a0 = a2 − d 2
b0 = 2d
c0 = −1
∆ = 4a0 c0 − b20 = −4a2 < 0
From Gradshteyn and Ryzhik, Table of Integrals, Series, and Products (Academic
Press, 1980, p. 84), we have
Z √
Z
Z
√
b0
dz
dz
R
√
√ .
+
dz = R + a0
z
z
z R
R
For
√ ¯¯d+a
R¯
z=d−a
=
¯d+a
p
¯
a2 − d 2 + 2dz − z2 ¯
z=d−a
= 0 − 0 = 0.
dz
√ , several solutions exist depending on the sign of a0 and ∆.
z R
For this problem, ∆ < 0, also let a0 < 0 (i.e., d > a). Using the table of integrals,


d+a
Z
dz
2a0 + b0 z 
1
√ = a0  √
sin−1  q
a0
−a0
z R
z b2 − 4a c
For
Z
0
0 0
z=d−a
¶¸d+a
·
µ 2
2
p
−1 a − d + dz
2
2
= − d − a sin
az
z=d−a
p
= −π d 2 − a2 .
dz
√ , different solutions exist depending on the sign of c0 and ∆.
R
In this problem, ∆ < 0 and c0 < 0. From the table of integrals,
¸
·
Z
2c0 z + b0 d+a
b
b0
dz
−1
√ = 0 √
sin−1 √
z
2
−c0
−∆ z=d−a
R
µ
·
¶¸d+a
d −z
= −d sin−1
= π d.
a
z=d−a
For
Z
Thus
i
p
µ0 h
· π d − π d 2 − a2
πh
i
p
= µ0 d − d 2 − a2 .
L12 =