Problem 5.19 Three long, parallel wires are arranged as shown in Fig. P5.19.
Determine the force per unit length acting on the wire carrying I3 .
I1 = 10 A
2m
I3 = 10 A
2m
2m
I2 = 10 A
Figure P5.19: Three parallel wires of Problem 5.19.
Solution: Since I1 and I2 are equal in magnitude and opposite in direction, and
z
I1 x
into the page (y)
R1
2m
B2
R2
x
I3
B1
2m
x
R1
R2 = R1
I2
out of the page (-y)
Figure P5.19: (a) B fields due to I1 and I2 at location of I3 .
I1 x
F'31
'
F32
θ
x
x
I3
I2
Figure P5.19: (b) Forces acting on I3 .
equidistant from I3 , our intuitive answer might be that the net force on I3 is zero.
As we will see, that’s not the correct answer. The field due to I1 (which is along ŷ) at
location of I3 is
µ0 I1
B1 = b̂1
2π R1
where b̂1 is the unit vector in the direction of B1 shown in the figure, which is
perpendicular to R̂1 . The force per unit length exerted on I3 is
F′31 =
µ0 I1 I3
µ II
× b̂1 ) = −R̂1 0 1 3 .
(ŷ×
2π R1
2π R1
Similarly, the force per unit length excited on I3 by the field due to I2 (which is
along −ŷ) is
µ0 I2 I3
F′32 = R̂2
.
2π R2
The two forces have opposite
components along
components
along ẑ.
√
√ x̂ and equal
√
−1
−1
◦
Hence, with R1 = R2 = 8 m and θ = sin (2/ 8) = sin (1/ 2) = 45 ,
¶
µ
µ0 I1 I3 µ0 I2 I3
′
′
′
+
sin θ
F3 = F31 + F32 = ẑ
2π R1
2π R2
¶
µ
4π × 10−7 × 10 × 20
1
√
= ẑ 2
× √ = ẑ 2 × 10−5 N/m.
2π × 8
2
Problem 5.21 Current I flows along the positive z-direction in the inner conductor
of a long coaxial cable and returns through the outer conductor. The inner conductor
has radius a, and the inner and outer radii of the outer conductor are b and c,
respectively.
(a) Determine the magnetic field in each of the following regions: 0 ≤ r ≤ a,
a ≤ r ≤ b, b ≤ r ≤ c, and r ≥ c.
(b) Plot the magnitude of H as a function of r over the range from r = 0 to
r = 10 cm, given that I = 10 A, a = 2 cm, b = 4 cm, and c = 5 cm.
Solution:
(a) Following the solution to Example 5-5, the magnetic field in the region r < a,
H = φ̂φ
rI
,
2π a2
and in the region a < r < b,
H = φ̂φ
I
.
2π r
The total area of the outer conductor is A = π (c2 − b2 ) and the fraction of the area
of the outer conductor enclosed by a circular contour centered at r = 0 in the region
b < r < c is
π (r2 − b2 ) r2 − b2
=
.
π (c2 − b2 ) c2 − b2
The total current enclosed by a contour of radius r is therefore
¶
µ
c2 − r2
r 2 − b2
=
I
,
Ienclosed = I 1 − 2
c − b2
c2 − b2
and the resulting magnetic field is
Ienclosed
I
H = φ̂φ
= φ̂φ
2π r
2π r
µ
c2 − r2
c2 − b2
¶
.
For r > c, the total enclosed current is zero: the total current flowing on the inner
conductor is equal to the total current flowing on the outer conductor, but they are
flowing in opposite directions. Therefore, H = 0.
(b) See Fig. P5.21.
Magnetic field magnitude H (A/cm)
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0.
1.
2.
3.
4.
5.
6.
7.
Radial distance r (cm)
Figure P5.21: Problem 5.21.
8.
9.
10.
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 5.33 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 5.20 A square loop placed as shown in Fig. P5.20 has 2-m sides and
carries a current I1 = 5 A. If a straight, long conductor carrying a current I2 = 10 A is
introduced and placed just above the midpoints of two of the loop’s sides, determine
the net force acting on the loop.
z
I2
4
1
a
a
3
y
I1
2
x
Figure P5.20: Long wire carrying current I2 , just above
a square loop carrying I1 (Problem 5.20).
Solution: Since I2 is just barely above the loop, we can treat it as if it’s in the same
plane as the loop. For side 1, I1 and I2 are in the same direction, hence the force on
side 1 is attractive. That is,
F1 = ŷ
µ0 I1 I2 a
4π × 10−7 × 5 × 10 × 2
= ŷ
= ŷ 2 × 10−5 N.
2π (a/2)
2π × 1
I1 and I2 are in opposite directions for side 3. Hence, the force on side 3 is repulsive,
which means it is also along ŷ. That is, F3 = F1 .
The net forces on sides 2 and 4 are zero. Total net force on the loop is
F = 2F1 = ŷ 4 × 10−5 N.
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 5.26 With reference to Fig. 5-10:
(a) Derive an expression for the vector magnetic potential A at a point P located
at a distance r from the wire in the x–y plane.
(b) Derive B from A. Show that your result is identical with the expression given
by Eq. (5.29), which was derived by applying the Biot–Savart law.
Solution:
(a) From the text immediately following Eq. (5.65), that equation may take the
form
µ
A=
4π
Z
ℓ′
I ′ µ0
dl =
R′
4π
Z ℓ/2
I
p
ẑ dz′
2
′
2
z +r
´¯ℓ/2
³
p
µ0
¯
=
ẑI ln (z′ + z′ 2 + r2 ) ¯ ′
z =−ℓ/2
4π
q


2
2
(ℓ/2)
+
r
ℓ/2
+
µ0 I 

q
ln
= ẑ
4π
2
2
−ℓ/2 + (−ℓ/2) + r
!
Ã
√
µ0 I
ℓ + ℓ2 + 4r2
√
= ẑ
.
ln
4π
−ℓ + ℓ2 + 4r2
z′ =−ℓ/2
(b) From Eq. (5.53),
×A
B = ∇×
Ã
!!
Ã
√
µ0 I
ℓ + ℓ2 + 4r2
√
× ẑ
= ∇×
ln
4π
−ℓ + ℓ2 + 4r2
!
Ã
√
µ0 I ∂
ℓ + ℓ2 + 4r2
√
= −φ̂φ
ln
4π ∂ r
−ℓ + ℓ2 + 4r2
Ã
!
!
Ã
√
√
ℓ + ℓ2 + 4r2
µ0 I −ℓ + ℓ2 + 4r2 ∂
√
√
= −φ̂φ
4π
∂ r −ℓ + ℓ2 + 4r2
ℓ + ℓ2 + 4r2
!
Ã
√
µ0 I −ℓ + ℓ2 + 4r2
√
= −φ̂φ
4π
ℓ + ℓ2 + 4r2
!
Ã
√
√
√
√
(−ℓ + ℓ2 + 4r2 ) ∂∂r (ℓ + ℓ2 + 4r2 ) − (ℓ + ℓ2 + 4r2 ) ∂∂r (−ℓ + ℓ2 + 4r2 )
×
√
2
(−ℓ + ℓ2 + 4r2 )
!
Ã
√
√
4r
µ0 I (−ℓ + ℓ2 + 4r2 ) − (ℓ + ℓ2 + 4r2 )
√
√
√
= −φ̂φ
2
2
2
2
2
4π
(−ℓ + ℓ + 4r )(ℓ + ℓ + 4r )
ℓ + 4r2
µ0 I
= −φ̂φ
4π
µ
−2ℓ
4r2
¶
4r
µ Iℓ
√
√0
= φ̂φ
2
2
ℓ + 4r
2π r ℓ2 + 4r2
which is the same as Eq. (5.29).
(T).
Problem 5.29 A thin current element extending between z = −L/2 and z = L/2
carries a current I along +ẑ through a circular cross-section of radius a.
(a) Find A at a point P located very far from the origin (assume R is so much larger
than L that point P may be considered to be at approximately the same distance
from every point along the current element).
(b) Determine the corresponding H.
Solution:
z
L/2
P
R
θ
I
-L/2
Cross-section πa2
Figure P5.29: Current element of length L observed at distance R ≫ L.
(a) Since R ≫ L, we can assume that P is approximately equidistant from all
segments of the current element. Hence, with R treated as constant, (5.65) gives
A=
µ0
4π
Z
V
′
J
µ0
dV ′ =
R′
4π R
Z
V
ẑ
′
I
µ0 I
π a2 dz = ẑ
(π a2 )
4π R
Z L/2
−L/2
dz = ẑ
µ0 IL
.
4π R
(b)
1
×A
∇×
µ0
¸
·
∂ Az
∂ Az
1
− ŷ
=
x̂
µ0
∂y
∂x
(
Ã
Ã
"
"
!#
!#)
1
∂ µ0 IL
∂ µ0 IL
1
1
p
p
x̂
=
− ŷ
µ0
∂ y 4π
∂ x 4π
x2 + y2 + z2
x2 + y2 + z2
¸
·
−x̂y + ŷx
IL
.
=
2
4π (x + y2 + z2 )3/2
H=
Problem 5.30 In the model of the hydrogen atom proposed by Bohr in 1913, the
electron moves around the nucleus at a speed of 2 × 106 m/s in a circular orbit of
radius 5 × 10−11 m. What is the magnitude of the magnetic moment generated by the
electron’s motion?
Solution: From Eq. (5.69), the magnitude of the orbital magnetic moment of an
electron is
¯
¯
|m0 | = ¯− 12 eur¯ = 12 × 1.6 × 10−19 × 2 × 106 × 5 × 10−11 = 8 × 10−24 (A·m2 ).
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 5.34 In Fig. P5.34, the plane defined by x − y = 1 separates medium 1 of
permeability µ1 from medium 2 of permeability µ2 . If no surface current exists on
the boundary and
B1 = x̂2 + ŷ3 (T)
find B2 and then evaluate your result for µ1 = 5µ2 .
Hint: Start by deriving the equation for the unit vector normal to the given plane.
y
Plane x − y = 1
x
(1, 0)
Medium 1
μ1
Medium 2
μ2
(0, −1)
Figure P5.34: Magnetic media separated by the plane
x − y = 1 (Problem 5.34).
Solution: We need to find n̂2 . To do so, we start by finding any two vectors in the
plane x − y = 1, and to do that, we need three non-collinear points in that plane. We
choose (0, −1, 0), (1, 0, 0), and (1, 0, 1).
Vector A1 is from (0, −1, 0) to (1, 0, 0):
A1 = x̂ 1 + ŷ 1.
Vector A2 is from (1, 0, 0) to (1, 0, 1):
A2 = ẑ 1.
Hence, if we take the cross product A2 × A1 , we end up in a direction normal to the
given plane, from medium 2 to medium 1,
n̂2 =
× (x̂ 1 + ŷ 1) ŷ 1 − x̂ 1
A2 × A1
ŷ
x̂
ẑ 1×
= √ −√ .
=
= √
|A2 × A1 |
|A2 × A1 |
1+1
2
2
In medium 1, normal component is
¶
µ
3
x̂
2
1
ŷ
√
√
· (x̂ 2 + ŷ 3) = √ − √ = √ ,
−
B1n = n̂2 · B1 =
2
2
2
2
2
B1n = n̂2 B1n =
µ
¶
x̂
ŷ x̂
1
ŷ
√ −√ ·√ = − .
2
2
2 2 2
Tangential component is
µ
ŷ x̂
B1t = B1 − B1n = (x̂ 2 + ŷ 3) −
−
2 2
¶
= x̂ 2.5 + ŷ 2.5.
Boundary conditions:
B1n = B2n , or
H1t = H2t , or
Hence,
B2t =
µ2
µ2
B1t =
(x̂ 2.5 + ŷ 2.5).
µ1
µ1
Finally,
B2 = B2n + B2t =
For µ1 = 5µ2 ,
ŷ x̂
− ,
2 2
B2t B1t
=
.
µ2
µ1
B2n =
µ
¶
ŷ x̂
µ2
−
+ (x̂ 2.5 + ŷ 2.5).
2 2
µ1
B2 = ŷ (T).