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182
CHAPTER 7
P.E. 7.1
z
α1
α2
2
27
−a x + a y
 −a − a y 
aφ = al × a ρ =  x
 × az =
2 
2

ρ = 5, cos α1 = 0, cos α 2 =
H3 =
  −a + a y 
10  2
− 0   x

 = −30.63a x + 30.63a y mA/m
4π (5)  27
2 

P.E. 7.2
2 
3 
1+

 az = 0.1458az A/m
4π (2) 
13 
12
ρ = 32 + 42 = 5, α 2 = 0, cos α1 = − ,
(b)
13
(a) H =
 3a − 4a z  4a x + 3a z
aφ = −a y ×  x
=
5
5


2  12   4a x + 3az 
1
H=
( 4ax + 3az )
1 + 
=
4π (5)  13 
5
 26π
= 48.97a x + 36.73a z mA/m
P.E. 7.3
(a) From Example 7.3,
Ia 2
H=
az
2(a 2 + z 2 )3/ 2
At (0,0,-1cm), z = 2cm,
50 × 10−3 × 25 ×10−4
H=
a z = 400.2a z mA/m
2(52 + 22 )3/ 2 × 10−6
(b) At (0,0,10cm), z = 9cm,
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H=
50 × 10−3 × 25 ×10−4
a z = 57.3a z mA/m
2(52 + 92 )3/ 2 × 10−6
P.E. 7.4
NI
2 × 103 × 50 × 10−3 (cosθ 2 − cosθ1 )a z
( cosθ 2 − cosθ1 ) a z =
2L
2 × 0.75
100
=
( cosθ 2 − cosθ1 ) az
1.5
0.75
(a) At (0,0,0), θ = 90o , cosθ 2 =
0.752 + 0.052
θ1
= 0.9978
100
H=
( 0.9978 − 0 ) az
1.5
H=
= 66.52 az A/m
(b) At (0,0,0.75), θ 2 = 90o ,cosθ1 = −0.9978
100
( 0 + 0.9978 ) a z
H=
1.5
= 66.52az A/m
−0.5
(c) At (0,0,0.5), cosθ1 =
= −0.995
0.52 + 0.052
0.25
cosθ1 =
= 0.9806
0.252 + 0.052
100
( 0.9806 + 0.995) a z
H=
1.5
= 131.7az A/m
P.E. 7.5
H=
(a)
(b)
θ2
θ1
θ2
θ1
θ2
1
K × an
2
1
H (0, 0, 0) = 50a z × (−a y ) = 25a x mA/m
2
1
H (1,5, −3) = 50a z × a y = −25a x mA/m
2
P.E. 7.6
 NI
, ρ − a < ρ < ρ + a, 9<ρ < 11

H =  2πρ

0,
otherwise

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(a) At (3,-4,0), ρ = 32 + 42 =5cm ‹ 9cm
H =0
(b) At (6,9,0), ρ = 62 + 92 = 117 ‹ 11
H =
103 × 100 ×10−3
= 147.1 A/m
2π 117 × 102
P.E. 7.7
(a) B = ∇ × A = (−4 xz − 0)ax + (0 + 4 yz )a y + ( y 2 − x 2 )az
B (−1, 2,5) = 20a x + 40a y + 3a z Wb/m2
(b) ψ =  B.dS =
4
1
 
( y 2 − x 2 )dxdy =
y =−1 x = 0
4

−1
1
y 2 dy − 5 x 2 dx
0
1
5
= (64 + 1) − = 20 Wb
3
3
Alternatively,
1
4
0
0
−1
1
ψ =  A.dl =  x 2 (−1)dx +  y 2 (1)dy +  x 2 (4)dx + 0
5 65
=− +
= 20 Wb
3 3
P.E. 7.8
z
h
R
y
k
dS
x
H =
kdS × R
,
4π R 3
dS = dxdy, k = k y a y ,
R = (− x, − y, h),
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185215
2
2π
−40
(b) k × IR==(Jhadx S+ =xa z )k y ,  ρ dφ d ρ , dS = ρ dφ d ρ a z
ρ =0 φ =0
k y (ha x + xaμzo)dxdy
H = 2
2π
3 2
2 2ρ
−40 4π ( x 2 + y 2 + −h40
2
)
d
d
=
=
ρ
ρ
φ
0 (2π )


μko ha0 ∞ ∞ 0
μo 2
k y az ∞ ∞
dxdy
xdxdy
= y x   −6
+
3
3


−80 × 2 × 10 2 + 2 + 2 2 4π
2
2
2 2
(
)
x
y
h
−∞ −∞ ( x + y + h )
= 4π −∞ −∞
=
−
400
A
4π × 10−7
The integrand in the last term is zero because it is an odd function of x.
Prob. 7.51
H = − ∇V
→ V = − H ⋅ dl = −mmf
m
m

k y h2π a x ∞ 2 2 2 − 3 2 d ( ρ 2 )
=
H=
( ρIa+ h )
3
  Example
H 4=
2
2 7.3,
2
π 0 2 2 3 2 a z 2
4π From
φ =0 ρ =0 ( ρ + h )
2 (z + a )


k h
k
−1
3
 ∞0 = yIaa2x
− Iz
= y ax 
2
2 − 2
1
2 m 2 =
dz =
−
z
+
a
+ c
 ( ρ 2 + hV
(
)
2
2
1

)


2
2 ( z2 + a 2 ) 2
1
Similarly, for
point
(0,0,-h),
H
=
−
k y ax
As z → ∞, Vm = 0 , 2i.e.
I
I
Hence,
0 = − + c → c =
1
2
k a ,
z >2 0

y
x
2
H =Hence,
1
z I< 0
 k a ,

z
 2 y Vxm =
1 −

2 
z2 + a 2 
P.E. 7.9
I
H =
aφ
2πρ
Prob. 7.1
But
H = − ∇Vm ( J = 0 )
(a) See text
I H = Hy + Hz 1 ∂Vm
I
(b) Let
aφ = −
aϕ → Vm = −
φ + C
ρ ∂φ
2πρ
2π
I
=
a
ρ = (π−3) 2 + 42 = 5
For H
I π
o z
φ =φ 60o =
⋅
+ C
At (10, 60 , 7 ) , 2πρ
, Vm = 0 → 0 = −
3
2π 3
(−3a +I 4a y ) (3a y − 4ax )
aφ =or−az ×C = x
=
56
5
I
I
20Vm = −
+
Hz =
(4ax + 32aπy )φ= 0.5093
6 ax + 0.382a y
2π (25)
π
At ( 4, 30o , − 2 ) , φ = 30o =
,
I
2
6
For H y =
aφ , ρ = (−3) + 52 = 34
2πρ
I π
I
12
I
Vm = −
⋅
+
=
=
12
12
(−3a + 5a )2π 36a + 5a x6
aφ = a y × Vm x = 1 zA = z
34
34
k y ha x
2π
∞
ρ dφ d ρ
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185
P.E. 7.10
∂
∂
k × R = (ha x + xa z )k y , ∂
∂y
∂z
(a ) B = ∇k y×(hAa x =
+ xa z )dxdy∂x
H =
2 3
xy 2 − xz3 − 6 xy + 2z 2 y 2
2 yz
4π ( x 2 + y 2 +2x
h 2 )y +
2
∞ + 4 x 2 y + 3 xz 2 )a + y + 6yz-4xy
∞ ∞
B k=y ha( x− ∞6xz
a y + y 2 − z 3 − 2 x 2 − z a z Wb/m 2
x k y az
dxdy
xdxdy
=
 −∞ ( x 2 + y 2 + h2 ) 3 2 + 4π −∞ −∞ ( x2 + y 2 + h2 ) 3 2
4
π
−∞
(b)
(
ψ =
)
(
)
x y + 3 xz ) dy dz
( −6 xz +in4the
 integrand
The
last term is zero because it is an odd function of x.
2
2
2
2
z=0 y=0
x =1
2
2
= k ha ( −
2π6 xz
∞ ) dy dz + 4  x y dy
∞ dz + 3  xz2 dy dz
k
h
2
π
a
3
0
0
0
ρ
φ
ρ
ρ
(
)
d
d
d
−
y
x
x
= y
H=
(2ρ 2 +2 h 2 ) 2
3



2 4π
4π 2 =0 ρ =20 ( ρ 2 + h 2 ) 22
2
= − 6 φdz
dy + 4 dz  y dy + 03 dy  z 2 dz

0
0
0
0
0
0
 ∞ ky
kyh 
−1

2
2
ax 
a
=
=
2
3
0
x
2y 
z 
 ( ρ 2 + h2 ) 12 
2
 ( 2) + 4(2)

= − 6(2)

 +3(2) 
 = -24+16+16
 2 0 1
 3 0
Similarly, for point (0,0,-h), H = − k y a x
2
ψ = 8 Wb
Hence,
∂A y
∂Ax 1
∂A
(c ) ∇ ⋅ A =
+
+z > 0 z = 4xy + 2xy − 6 xy = 0
,
k
a
∂x y x ∂y
∂z
H =2
∇ ⋅ B = − 6 z + 8 xy +1 3kz 3a+ ,6 z − 8zxy<+01 − 3z 3 − 1 = 0
2 y x
As a matter of mathematical necessity,
∇ • B = ∇ • (∇ × A) = 0
Prob. 7.1
7.43
∂
∂
(a) See∂ text
∂y
∂z = (cos x + sin y )a z
B1 = ∇ × A1 = ∂x
(b) Let H = Hy + Hz
0 (sin x + x sin y ) 0
I
For ∂H z = ∂ aφ∂ ρ = (−3) 2 + 42 = 5
2πρ
∂y ∂z = (cos x + sin y )a z
B2 = ∇ × A2 = ∂x
sin(−x3ax 0+ 4a y ) = (3a y − 4ax )
aφcos
= −yaz ×
5
5
B =B =B
1
2
20 the same B.
Hence, A 2 and A 2 give
Hz =
(4ax + 3a y ) = 0.5093ax + 0.382a y
2π (25)
∇ B = 0
showing that B is solenoidal.
I
For H y =
aφ , ρ = (−3)2 + 52 = 34
2πρ
aφ = a y ×
(−3a x + 5a z ) 3a z + 5a x
=
34
34
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Hy =
10
(5a x + 3a z ) = 0.234a x + 0.1404a z
2π (34)
H = Hy + Hz
= 0.7433ax + 0.382ay + 0.1404az A/m
Prob. 7.2
Idl × R
4π R 3
(a) At (1,0,0), R=(1,0,0) - (0,0,0) = (1,0,0)
4a × a
dH = x 3x = 0
4π (1)
(b) At (0, 1,0), R = ay
dH =
4a x × a y
dH =
4π (1)3
(c) At (0,0,1), R =az
dH =
= 0.3183a z A/m
4a x × a z
= −0.3183a y A/m
4π (1)3
(d) At (1,1,1), R=(1,1,1)
dH =
4a x × ( a x + a y + a z )
4π (3)3/ 2
= 61.26(−a y + a z ) mA/m
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Prob. 7.3
Let H = H1 + H 2
where H1 and H 2 are respectively due to the lines located at (0,0) and (0,5).
H1 =
I
2πρ
aφ ,
ρ = 5,
aφ = a × a ρ = a z × a x = a y
ay
10
ay =
2π (5)
π
I
ρ = 5 2, aφ = a × a ρ , a = −a z
H2 =
aφ ,
2πρ
5a x − 5a y a x − a y
=
aρ =
5 2
2
 a x − a y  -a x − a y
aφ = −a z × 
=
2 
2

10  -a x − a y  1
H2 =
( -a x − a y )

=
2π 5 2 
2  2π
H1 =
H = H1 + H 2 =
Prob. 7.4
I
H=
aφ ,
2πρ
ay
π
+
1
(-a x − a y ) = −0.1592a x + 0.1592a y
2π
ρ = 5, I = 12
 3a + 4a y  4
3
aφ = a × a ρ = −a z ×  x
 = ax − a y
5
5

 5
12  4
3 
H=
 a x − a y  = 0.3056a x − 0.2292a y
2π (5)  5
5 
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188
Prob. 7.5
b
I
α2
a
α1
y
x
y
x
H=
I
4πρ
(cos α 2 − cos α1 )aφ
ρ = x 2 + y 2 , cos α1 =
a
a2 + ρ 2
, cos α 2 =
b
b2 + ρ 2
aφ = al × aρ = a z × a ρ = aφ . Hence,
H=

b
a
−

x2 + y 2 + a2
4π x 2 + y 2  x 2 + y 2 + b 2
I

aφ

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189
Prob. 7.6
y
1 A
α2
6A
P
ρ
B
O
H =
I
4πρ
α1
x
1
( cos α 2 − cos α1 ) aφ
1
2
2 =
2
2
 −a x + a y 
 −a x − a y 
1 -1 1 0
= 
×
=
= az



2 -1 -1 0

2
2
α 1 = 135o , α 2 = 45o , ρ =
aφ = al × a ρ
6
H =
4π
( cos 45
2
o
)
− cos135o a z =
3
π
az
2
H ( 0, 0, 0) = 0.954a z A/m
Prob. 7.7
(a) At (5,0,0), ρ = 5,
H=
aφ = a y ,
cos α 1 = 0,
cos α 2 =
2
10
(
)a y = 28.471a y mA/m
4π (5) 125
(b) At (5,5,0), ρ = 5 2,
aφ =
H=
cos α1 = 0,
cos α 2 =
−a x + a y
10
125
10
150
2
2
10  −a x + a y
(
)
4π (5 2) 150 
2

 = 13(−a x + a y ) mA/m

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190
(c) At (5,15,0), ρ = 250 = 5 10,
aφ =
cos α1 = 0,
cos α 2 =
5a y - 15a x
10
350
5 10
2
10  −15a x + 5a y 
(
)
 = −5.1a x + 1.7a y mA/m
4π (5 10) 350 
5 10

d) At (5,-15,0), by symmetry,
H=
H = 5.1a x + 1.7a y mA/m
Prob. 7.8
z
α1
x
A (2, 0, 0)
C (0, 0, 5)
y
α2
B (1, 1, 0)
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191
(a)
Consider the figure above.
AB
=
AC
=
(1, 1, 0)
( 0, 0, 5)
AB ⋅ AC
=
(
=
BA =
i.e.
)
al × a ρ
H2 =

10
0+

4π 27 
2
2
29 
2
=
( −1,
=
ρ =
×
( −1,
2 27
2
29
− 1, 5)
=
0
27
− 1, 5)
5
=
( 5, 5, 2)
=
27
( 5, 5, 2)
cos α 1 = −
→
2 29
−1 + 1
BC BA
=
( −1, − 1, 5) ,
( −1, 1, 0)
=
aφ =
2π 29
54
⋅
( 5, 5, 2)
27
A/m
27.37 a x + 27.37a y + 10.95 a z mA/m
H = H1 + H 2 + H 3 =
+
=
=
AB ⋅ AC
AB AC
=
BC ⋅ BA
BC BA
=
BC = ρ =
=
(b)
− ( 2, 0, 0 )
( 0, 0, 5) − (1, 1, 0)
(1, − 1, 0)
cos α 2
( −1, 1, 0)
( −2, 0, 5)
=
2, i.e AB and AC are not perpendicular.
cos 180o − α 1
BC
− ( 2, 0, 0)
( 0,
− 59.1, 0) +
( −30.63, 30.63, 0)
( 27.37,
27.37, 10.95)
− 3.26 a x − 1.1 a y + 10.95a z mA/m
Prob. 7.9
y
(a)
Let H = H x + H y = 2H x
Hx =
I
4πρ
( cos α 2 − cos α1 ) aφ
O
α2
5A
α1
x
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192
where aφ =
5
( cos 45
4π ( 2 )
Hx =
− a z , α1 = 180o , α 2 = 45o
− ax × a y =
o
− cos 180o
) ( −a )
z
= −0.6792 a z A/m
(b)
H = Hx + H y
5
where H x =
4π ( 2 )
(1 − 0)
aφ , aφ = − a x × −a y = a z
= 198.9a z mA/m
H y = 0 since α 1 = α 2 = 0
H = 0.1989 a z A/m
(c )
H = Hx + H y
where H x =
Hy =
5
(1 − 0) ( −ax × az )
5
(1 − 0)
4π ( 2 )
4π ( 2 )
(a
y
× az
)
= 198.9 a y mA/m
= 198.9 a x mA/m
H = 0.1989 a x + 0.1989 a y A/m.
Prob. 7.10
3
Let H = H1 + H 2 + H 3 + H 4
4
where H n is the contribution by side n.
(a)
H = 2H1 + H 2 + H 4 since H1 = H 3
I
1
10  6
1 
+

 az
4π ( 2)  40
2
H1 =
( cos α 2 − cos α1 ) aφ
4πρ
H2 =
10 
2 
2×

 az , H 4 =
4π ( 6 ) 
40 
=
2
10 
1 
2⋅

 az
4π ( 2 ) 
2
 5  3
1 
5
5 
= 
+
+
+
 a z = 1.964a z A/m


2
6π 10
2π 2 
 2π  10
At ( 4, 2, 0) , H = 2 ( H1 + H 4 )
H
(b)
H1 =
10
4π ( 2)
H
2 5  1
1 +  a =
π  4 z
=
8
az , H 4 =
20
10
4π ( 4 )
4
az
20
1.78a z A/m
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193
(c )
(d )
At
( 4, 8, 0) ,
H = H1 + 2H 2 + H 3
H1 =
10 
4 
2⋅
az , H 2 =

4π ( 8)  4 5 
H3 =
10  2 
4π ( 4)  2 
H
5
8π
=
( az )
10  8
1 
−

 az
4π ( 4 )  4 5
2
( −a z )
4
4 
 1
+
−


5
5
2
=
−0.1178a z A/m
At ( 0, 0, 2) ,
H1 =
10  8

− 0


4π ( 2)  68
H2 =
10
 4

 2a − 8a x 
− 0 a y ×  z




4π 68
84
68 
( ax × az )
=
−
10
ay
π 68
5 ( a x + 4a z )
=
17π 84
 2a x − 4a y 
10
 8

− 0  ax × 
 =
−
4π 20 
84
20 


−5a x
10 
4 
=
0 +
 ( −a y × a z ) =
π 20
4π 2 
20 
H3 =
H4
H
a y + 2a z
π 21
5
5 
10 
2 

 1
 20
= 
−
−
+
 az
 ax + 
 ay + 
 34π 21 π 21 
 34π 21 π 20 
 π 21 π 68 
= −0.3457 ax − 0.3165 a y + 0.1798 az A/m
Prob. 7.11
For the side of the loop along y-axis,
I
H1 =
( cos α 2 − cos α1 ) aφ
4πρ
where aφ =
− a x , ρ = 2 tan 30o =
5
3
cos 30o − cos 150o
4π 2
H = 3H1 = − 1.79a x A/m
H1 =
(
2
, α 2 = 30o , α 1 = 150o
3
) ( −a )
x
=
−
15
ax
8π
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Prob. 7.12
H = 4 H1 = 4
I
4πρ
(cos α 2 − cos α1 )aφ
ρ = a = 2cm, I = mA, α 2 = 45o , α1 = 90o + 45o = 135o
aφ = a × a ρ = a y × (−a x ) = a z
H=
1
2I
2 × 5 ×10−3
I 1
(
−−
)a z =
az =
a = 0.1125a z
πa 2
πa
π × 2 ×10−2 z
2
Prob. 7.13
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195
(a) Consider one side of the polygon as shown. The angle subtended by the Side At the
center of the circle
360° 2π
=
n
n
The filed due to this side is
I
H1 =
(cos α 2 − cos α1 )
4π ρ
where ρ = r ,
π
π
cos α 2 = cos(90 − ) = sin
n
n
π
cos α1 = − sin
H1 =
n
I
2 sin
4π r
π
π
nI
sin
2π r
n
π
3I
For n = 3, H =
sin
2π r
3
2
r cot 30o = 2 → r =
3
n
H = nH 1 =
(b)
H =
3× 5
2π 2
For n = 4, H =
⋅
3
3
2
45
8π
=
4I
π
sin
2π r
4
=
= 1.79 A/m.
4×5
⋅
2π ( 2 )
1
2
= 1.128 A/m.
(c)
As
n → ∞,
nI
π
sin
=
n →∞ 2π r
n
From Example 7.3, when h = 0,
H = lim
nI
π
⋅
2π r
n
=
I
2r
I
2r
which agrees.
H =
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Principles of Electromagnetics, 6e
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Prob. 7.14
4
β
1
Let H =
3
α2
2
H1 + H 2 + H 3 + H 4
I
10
az =
a z = 62.5 a z
4a
4 × 4 × 10−2
I
4
cos α 2 − cos 90o ) a z , α 2 = tan −1
= H4 =
−2 (
4π × 4 × 10
100
= 19.88 a z
H1 =
H2
H3 =
I
4π (1)
2 cos β a z , β = tan −1
100
4
= 2.29o
= 87.7o
10
2 cos 87.7 oa z = 0.06361 a z
4π
= ( 62.5 + 2 × 19.88 + 0.06361) a z
=
H
= 102.32 a z A/m.
Prob. 7.15
From Example 7.3, H due to circular loop is
H1 =
(a)
Iρ 2
az
2 ρ2 + z2
(
H ( 0, 0, 0) =
)
(
5 × 22
2 22 + 02
)
3
az +
2
(
5 × 22
2 22 + 42
)
3
2
az
= 1.36 a z A/m
(b)
H ( 0, 0, 2) = 2
(
5 × 22
2 22 + 22
)
3
2
az
= 0.884 a z A/m
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Principles of Electromagnetics, 6e
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Prob. 7.16
α2
H
=
nI
( cos θ 2 − cos θ1 )
2

cos θ 2 = -cos θ1 =
H
=
θ2
(
nI 
2
2 a + 
2
4
)
(
1
2
2
a + 
2
2
=
4)
1
2
0.5 × 150 × 2 × 10−2
2 × 10−3 × 42 + 102
= 69.63 A/m
(b)
α1
α 1 = 90o , tan θ 2 =
H
=
nI
cos θ 2 =
2
α2
a
4
=
= 0.2 → θ 2 = 11.31o
b
20
150 × 0.5
cos 11.31o = 36.77 A/m
2
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Prob. 7.17
e
y
• P (4, 3, 2)
x
H = Hl + H p
Let
Hl =
ρ =
aρ =
1
2πρ
aφ
( 4, 3, 2)
− (1, -2, 2) = (3, 5, 0),
3a x + 5a y
34
=
34
=
3a y − 5a x
34
20π  −5a x + 3a y 
x10-3 = ( − 1.47a y + 0.88a y ) mA/m


2π 
34

1
1
K × an =
100 × 10 −3 a z × ( -a x ) = − 0.05a y A/m
2
2
= H l + H p = −1.47a x − 49.12 a y mA/m
Hp =
H
ρ
, al = az
 3a x + 5a y 
aφ = a l × a ρ = a z × 

34 
Hl =
ρ =
(
)
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Principles of Electromagnetics, 6e
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Prob. 7.18
(a) See text
(b)
I
a
b
For ρ < a,
 H ⋅ dl
For a < ρ < b,
= Ienc = 0 → H = 0
Hφ ⋅ 2πρ =
Hφ =
For ρ > b,
Iπ ( ρ 2 − a2 )
π ( b2 − a 2 )
 ρ 2 − a2 


2πρ  b2 − a2 
I
Hφ ⋅ 2πρ = I →
Hφ =
I
2πρ
Thus,
Hφ

 0,

 I
= 
 2πρ
 I

 2πρ
ρ <a
 ρ 2 − a2 
 2 2 ,
 b −a 
,
a < ρ <b
ρ >b
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Principles of Electromagnetics, 6e
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200
Prob. 7.19
x
y
-1
1
1
H =  K × an
2
1
1
= (20a x ) × (−a y ) + (−20a x ) × a y
2
2
= 10(−a z ) − 10(a z )
= −20a z A/m
Prob. 7.20
1
1
k × an = 10a x × a z = −5a y
2
2
HP =
HL =
I
2πρ
aφ =
I
I
(a x × −a z ) =
ay
2π (3)
6π
H P + H L = −5a y +
I
ay = 0
6π
⎯⎯
→ I = 30π = 94.25 A
Prob. 7.21
(a)
Applying Ampere's law,
πρ 2
→
π a2
Iρ
aφ
2π a 2
Hφ ⋅ 2πρ = I ⋅
i.e
H
=
Hφ = I ⋅
ρ2
2π a 2
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201
(b)
From Eq. (7.29),
 Iρ
 2π a 2 , ρ < a
Hφ = 
 I , ρ >a
 2πρ
At ( 0, 1 cm, 0 ) ,
3 × 1× 10−2
=
Hφ =
2π × 4 × 10−4
H = 11.94 aφ A/m
At
300
8π
( 0, 4 cm, 0 ) ,
3
2π × 4 × 10−2
11.94 aφ A/m
Hφ =
H =
=
300
8π
Prob. 7.22
For 0 < ρ < a

L
H dl = I enc =  J dS
H φ 2πρ = 
2π
φ =0
ρ

ρ
=0
Jo
ρ
ρ dφ d ρ
= J o 2πρ
Hφ = J o
For ρ > a
2π
a
 Hdl =  J dS = φ ρ
=0
Jo
=0
ρ
ρ dφ d ρ
H ρ 2πρ = J o 2π a
Hφ =
Joa
ρ
 J o , 0<ρ <a

Hence H φ =  J o a
 ρ , ρ >a

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Principles of Electromagnetics, 6e
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Prob. 7.23
2k
1 d
1 d
ρ2
( ρ H φ )a z =
( ko
)a z = o a z
a
a
ρ dρ
ρ dρ
(a) J = ∇ × H =
(b) For ρ>a,
 H ⋅ dl = I
2π
2 ko
2ko
ρ2 a
(2π )
ρ d ρ dφ =
=  J ⋅ dS =  
a
2 0
ρ =0 φ =0 a
a
enc
H φ 2πρ = 2π ko a
⎯⎯
→ Hφ =
a
H = ko   aφ ,
ρ
ρ >a
ko a
ρ
Prob. 7.24
∂
J = ∇ × H = ∂x
y2
∂
∂y
x2
∂
∂z = (2 x − 2 y )a z
0
At (1,-4,7), x =1, y = -4, z=7,
J = [ 2(1) − 2(−4) ] a z = 10a z A/m 2
Prob. 7.25
(a)
J = ∇× H =
1 ∂
1 ∂
( ρ Hφ )a z =
(103 ρ 3 )a z
ρ ∂ρ
ρ ∂ρ
= 3ρ ×103 a z A/m
2
(b)
Method 1:
2
2π
0
0
I =  J dS =  3ρ ρ dφ d ρ103 = 3 × 103  ρ 2 d ρ  dφ
S
= 3 × 103 (2π )
Method 2:
ρ 2
3
3 2
= 16π × 103 A = 50.265 kA
2π
I =  H dl =103  ρ 2 ρ dφ = 103 (8)(2π ) = 50.265 kA
L
0
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Prob. 7.26
Let H = H1 + H 2
where H1 and H 2 are due to the wires centered at x = 0 and x = 10cm respectively.
For H1 , ρ = 50 cm, aφ = al × a ρ = a z × a x = a y
(a)
H1 =
5
50
ay =
a
−2
π y
2π ( 5 × 10 )
For H 2 , ρ = 5 cm, aφ = − a z × −a x = a y , H 2 = H1
H = 2H1 =
100
π
ay
= 31.83 a y A/m
2a y − a x
 2a + a y 
For H1 , aφ = a z ×  x
 =
5 
5

 −a x + 2a y 
5
H1 =
 = − 3.183a x + 6.366a y
−2 
2π 5 5 ×10 
5

For H 2 , a ρ = − a z × a y = a x
(b)
H2 =
5
2π ( 5 )
a x = 15.915a x
H = H1 + H 2
= 12.3 a x + 6.366a y A/m
Prob. 7.27
(a) B =
μo I
aφ
2πρ
At (-3,4,5), ρ=5.
B=
4π × 10−7 × 2
aφ = 80aφ nW/m 2
2π (5)
μI
Ψ =  B • dS = o
2π
(b)

d ρ dz
ρ
6 4
4π ×10−7 × 2
=
ln ρ z
2 0
2π
= 16 ×10−7 ln 3 = 1.756 μ Wb
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Principles of Electromagnetics, 6e
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Prob. 7.28
(a) I =  J dS
=
2π
a
 
ρ2
J o (1 −
a2
φ =0 ρ =0
) ρ d ρ dφ
 ρ2 ρ4 
= 2π J o 
− 2
 2 4a 
1
= π a2 Jo
2
(b)
 H dl = I
enc
a
0
2π
a
0
0
= J o  dφ  ( ρ −
ρ3
a2
)d ρ
2π  2 a 2 
=
Jo  a − 
2
2 

=  J dS
For ρ < a,
H φ 2πρ =  J dS
 ρ2 ρ4 
= 2π J o 
− 2
 2 4a 
H ρ 2πρ = 2π J o
Hρ =
ρ2 
ρ2 
2
−


4 
a2 
Jo ρ 
ρ2 
2
−


4 
a2 
For ρ > a,
 Hdl = 
J o dS = I
1
H φ 2πρ = π a 2 J o
2
2
a Jo
Hφ =
4ρ
 Jo ρ 
ρ2 
2
−


, ρ < a
a2 
 4 
Hence Hφ = 
aJ o

, ρ >a

4ρ
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Principles of Electromagnetics, 6e
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205
Prob. 7.29
B =
μ0 I
aφ
2πρ
ψ =

=
Prob. 7.30
B ⋅ dS =
d+a
μ0 I
dρ dz
z = 0 2πρ
ρ 
=d
b
μ0 Ib d + a
In
2π
d
For a whole circular loop of radius a, Example 7.3 gives
H=
Ia 2 a z
2  a 2 + h 2 
3/2
→0
Let h ⎯⎯
I
az
2a
For a semicircular loop, H is halfed
I
H=
az
4a
μI
B = μo H = o a z
4a
H=
Prob. 7.31
∂Bx ∂By ∂Bz
+
+
=0
∂x
∂y
∂z
showing that B satisfies Maxwell’s equation.
(a) ∇ • B =
(b)
dS = dydza x
4
Ψ =  B • dS = 
1

z =1 y = 0
y 2 dydz =
4
y3 1
( z ) = 1 Wb
1
3 0
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Principles of Electromagnetics, 6e
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206
(c) ∇ × H = J
J = ∇×
⎯⎯
→
B
μo
∂
∂
∂
∇ × B = ∂x ∂y ∂z = −2 za x − 2 xa y − 2 ya z
y 2 z 2 x2
2
J = − ( za x + xa y + ya z ) A/m 2
μo
Prob. 7.32
h
( ρ − a)
6
where H1 and H 2 are due to the wires centered at x = 0 and x = 10cm respectively.
On the slant side of the ring, z =
ψ =
 B.dS
μo I
 2πρ
=
h
( ρ −a)
b
dρ dz
=
μo I
2π
=
μo Ih 
a + b
 b − a ln
 as required.
a 
2π b
a+b
ρ 
dz dρ
ρ
z=0
=a
=
μo Ih
2π b
 a
1−
dρ
=a 
 ρ 
a+b
ρ
If a = 30 cm, b = 10 cm, h = 5 cm, I = 10 A,
ψ =
4π × 10 −7 × 10 × 0.05 
4
 0.1 − 0.3 ln 
−2
3
2π 10 × 10
(
)
= 1.37 × 10 −8 Wb
Prob. 7.33
ψ =
 BdS
= μo 
0.2

50o
z=0 φ =0
106
ρ
sin 2φ ρ dφ dz
 cos 2φ 
ψ = 4π × 10 × 10 ( 0.2)  −


2 
−7
(
50o
6
=
0.04π 1 − cos 100
=
0.1475 Wb
o
)
0
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Principles of Electromagnetics, 6e
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207
Prob. 7.34
π /4 2
ψ =  B  dS = 

ρ
20
φ = 0 =1
S
= 20(1)
π /4

0
ρ
2
π /4
1
0
sin φ ρ d ρ dφ = 20 d ρ
2
 sin
2
φ dφ
π /4
1
1
(1 − cos 2φ )dφ = 10(φ − sin 2φ )
0
2
2
π
1
= 10( − ) = 2.854 Wb
4 2
Prob. 7.35
ψ =  B dS , dS = r 2sinθ dθ dφ ar
S
ψ = 
2π
2
cos θ r 2sinθ dθ dφ
= 2  dφ
r =1
r3
0
π /3
= 2(2π )  sin θ d (sin θ ) = 4π
0
π /3
 cos θ sin θ dθ
0
sin 2 θ π / 3
= 2π sin 2 (π / 3)
0
2
= 4.7123 Wb
Prob. 7.36
B = μo H =
μo J × R
dv
4π v R 3
Since current is the flow of charge, we can express this in terms of a charge moving with
velocity u. Jdv = dqu.
μo
4π
 qu × R 
 R 3 
In our case, u and R are perpendicular. Hence,
B=
μo qu 4π ×10−7 1.6 ×10−19 × 2.2 ×106
1.6 ×10−20
=
×
=
B=
4π R 2
4π
(5.3 ×10−11 ) 2
(5.3) 2 ×10 −22
= 12.53 Wb/m 2
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Principles of Electromagnetics, 6e
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208
Prob. 7.37
(a ) ∇A = − ya sin ax ≠ 0
∂
∂x
∇×A =
∂
∂y
y cos ax
∂
∂z
y + e-x
0
= a x + e − x a y − cos axa z ≠ 0
A is neither electrostatic nor magnetostatic field
1 ∂
1 ∂
ρ Bρ =
( 20) = 0
ρ ∂ρ
ρ ∂ρ
∇× B = 0
B can be E-field in a charge-free region.
(b)
∇⋅ B =
(c )
∇⋅ C =
(
)
1
∂ 2
(r sinθ ) = 0
r sin θ ∂φ
1
∂
1∂ 3
∇×C =
r 2 sin 2 θ ar (r sinθ )aθ ≠ 0
r sin θ ∂θ
r ∂r
C is possibly H field.
(
)
Prob. 7.38
(a) ∇⋅ D = 0
∂
∂x
∇× D =
y2 z
∂
∂y
∂
∂z
2(x + 1)yz -(x + 1)z 2
= 2(x + 1)ya x + . . . ≠ 0
D is possibly a magnetostatic field.
(b)
∇⋅ E =
∇× E =
∂  sin φ 
1 ∂
=0
( ( z + 1) cos φ ) + 
∂z  ρ 
ρ ∂ρ
1
ρ2
cos θ a ρ + . . . ≠ 0
E could be a magnetostatic field.
(c )
∇⋅ F =
1 ∂
1 ∂  sinθ 
( 2cosθ ) +

 ≠ 0
2
r ∂r
rsinθ ∂θ  r 2 
1  ∂
2 sin θ 
−1
aθ ≠ 0
r
sin
θ
+
r  ∂r
r 2 
F can be neither electrostatic nor magnetostatic field.
∇×F =
(
)
Copyright © 2015 by Oxford University Press
POESM_Ch07.indd 212
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
213
209
Prob. 7.39
A=
μo Idl μo ILaz
=
4π r
4π r
This requires no integration since L << r.
∂A
1 ∂Az
B = ∇× A =
a ρ − z aφ
∂ρ
ρ ∂φ
But r = ρ 2 + z 2
μo ILa z
4π ( ρ 2 + z 2 )1/ 2
μ IL 1
∂Az μo IL ∂
( ρ 2 + z 2 )1/ 2 = o (− )( ρ 2 + z 2 ) −3/ 2 (2 ρ )
=
4π ∂ρ
4π
2
∂ρ
μo IL ρ aφ
μo IL ρ aφ
B=
=
4π ( ρ 2 + z 2 )3/ 2
4π r 3
A=
Prob. 7.40
y
2
a
1
I
3
0
a
dl
R
P
2a
x
4
Divide the loop into four segments as shown above. Due to segment 1,
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POESM_Ch07.indd 213
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
214
210
A1 = 
μo Idl
,
4π R
dl = dya y , R =
y2 + a2
)
(
a
a
μo I
μI
dy
= o a y ln( y + y 2 + a 2
A1 =
ay 
−a
4π
y 2 + a 2 4π
y =− a
=
 2 + 1  μo I
μo I
a y ln 
ln( 2 + 1) a y
 =
4π
 2 − 1  2π
By symmetry, the contributions due to sides 2 and 4 cancel. For side 3,
μ Idl
A3 =  o ,
dl = dy (−a y ), R = y 2 + (−3a) 2
4π R
A3 =
=
)
(
 10 + 1 
a μo I
μo I
(-a y ) ln( y + y 2 + 9a 2
(-a y ) ln 
=

−a 4π
4π
 10 − 1 
μo I  10 + 1 
ln 
 (-a y )
2π  3 
A = A1 + A2 + A3 + A4
=
=
μo I
ln( 2 + 1) a y −
2π
μo I  10 + 1 
ln 
 ay
2π  3 
μo I  3( 2 + 1) 
ln 
 ay
2π  10 + 1 
Prob. 7.41
∂
B = ∇ × A = ∂x
0
=−
π
2
sin
∂
∂y
0
πx
2
sin
∂
∂A
∂A
∂z
= z ax − z a y
∂y
∂x
Az ( x, y )
πy
2
ax −
π
2
cos
πx
2
cos
πy
2
ay
Prob. 7.42
∂
B = μ o H = ∇ × A = ∂x
0
∂
∂y
0
∂
∂A
∂A
∂z
= z a x − z a z = −2 μo kya x + 2 μo kxa y
∂y
∂x
Az ( x, y )
H = −2kya x + 2kxa y
Copyright © 2015 by Oxford University Press
POESM_Ch07.indd 214
10/14/2015 11:46:10 AM
(c )
Sadiku & Kulkarni ∇ ⋅ B
∇⋅ A =
x
∂x
+
y
∂y
+
=
z
∂z
4xy + 2xy − 6 xy = 0
= − 6 z + 8 xy + 3 z 3 + 6 z − 8 xy + 1 − 3z 3 − 1 =
As a matter of mathematical necessity,
∇ • B = ∇ • (∇ × A) = 0
Prob. 7.43
∂
B1 = ∇ × A1 = ∂x
0
∂
∂y
(sin x + x sin y )
∂
∂
∂y
B2 = ∇ × A2 = ∂x
cos y sin x
0
Principles of Electromagnetics, 6e
215
∂
∂z = (cos x + sin y )a z
0
∂
∂z = (cos x + sin y )a z
0
B1 = B2 = B
Hence, A 2 and A 2 give the same B.
∇ B = 0
showing that B is solenoidal.
212
Prob. 7.44
1 ∂Az
∂A
a ρ − z aφ
ρ ∂φ
∂ρ
B = ∇× A =
=
15
ρ
π
e − ρ cos φ a ρ + 15 e − ρ sin φ aφ
1
1


B  3, , -10  = 5 e −3
a ρ + 15 e −3
aφ
2
2
 4

B
107 15 −3  1

H =
e  a ρ + aφ 
=
μo
4π
2
3

H =
(14 a
ψ =
 B ⋅ dS
= 15 z
ρ
10
0
+ 42 aφ ) ⋅10 4 A/m
=
15
 ρ
π
( sin φ ) 0 2
e − ρ cos φ ρ dφ dz
e −5 = 150 e −5

ψ = 1.011 Wb
Prob. 7.45
 ∂
∂Aθ 

1  1 ∂Ar ∂
 ∂θ ( Aφ sin θ ) − ∂φ  ar + r  sin θ ∂φ − ∂r (rAφ )  aθ




1 ∂
∂A 
+  (rAθ ) − r  aφ
r  ∂r
∂θ 
1 10
1 ∂
2sin θ cos θ ar −
(10) sin θ aθ + 0aφ
=
r sin θ r
r ∂r
20
B = 2 cos θ ar
r
At (4, 60o , 30o ), r = 4, θ =60o
B = ∇× A =
H=
B
μo
=
1
r sin θ
1
 20

cos 60o ar  = 4.974 × 105 ar A/m
−7  2
4π × 10  4

Copyright © 2015 by Oxford University Press
POESM_Ch07.indd 215
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o
(14 a
H =
Sadiku & Kulkarni
ψ =
ρ
+ 42 aφ ) ⋅10 4 A/m
15
 B ⋅ dS =
= 15 z
10
0
 ρ
π
( sin φ ) 0 2
Principles of Electromagnetics, 6e
e − ρ cos φ ρ dφ dz
216
e −5 = 150 e −5

ψ = 1.011 Wb
Prob. 7.45
 ∂
∂Aθ 

1  1 ∂Ar ∂
 ∂θ ( Aφ sin θ ) − ∂φ  ar + r  sin θ ∂φ − ∂r (rAφ )  aθ




1 ∂
∂A 
+  (rAθ ) − r  aφ
r  ∂r
∂θ 
1 10
1 ∂
2sin θ cos θ ar −
(10) sin θ aθ + 0aφ
=
r sin θ r
r ∂r
20
B = 2 cos θ ar
r
At (4, 60o , 30o ), r = 4, θ =60o
B = ∇× A =
H=
B
μo
=
1
r sin θ
1
4π × 10−7
 20

o
5
 42 cos 60 ar  = 4.974 × 10 ar A/m
213
Prob. 7.46
Applying Ampere's law gives
H φ ⋅ 2πρ = J o ⋅ πρ 2
Jo
ρ
2
Hφ =
ρ
Bφ = μo H φ = μo
But B
=
−
∂AZ
∂ρ
∇× A
=
=
−
1
μ Jo ρ
2
or
a
Jo ρ
2
∂AZ
aϕ + . . .
∂ρ
⎯⎯
→
AZ =
− μo
Jo ρ 2
4
1
A = - μo J o ρ 2 a z
4
Prob. 7.47
∂
∂x
B = μo H = ∇ × A =
10sin π y
H=
∂
0
POESM_Ch07.indd 216
∂
∂z
= π sin π xa y − 10π cos π ya z
4 + cos π x
π 

sin π xa y − 10 cos π ya z 

μo 

π
∂x
J = ∇× H =
μo
J=
∂
∂y
0
∂
∂
π 

∂y
∂z
10π sin π ya x + π cos π xa z 
=

μo 

sin π x −10 cos π y
π2
(10sin π ya x + cos π xa z )
μo
Copyright © 2015 by Oxford University Press
10/14/2015 11:46:11 AM
But B
=
∂A
− Z
∂ρ
Sadiku & Kulkarni
∇× A
=
=
−
1
μ Jo ρ
2
or
∂ρ
aϕ + . . .
⎯⎯
→
AZ
=
− μo
Jo ρ 2
4
Principles of Electromagnetics, 6e
1
A = - μo J o217
ρ 2 az
4
Prob. 7.47
∂
∂x
B = μo H = ∇ × A =
10sin π y
H=
∂
∂z
= π sin π xa y − 10π cos π ya z
4 + cos π x
π 

sin π xa y − 10 cos π ya z 

μo 

∂
π
∂x
J = ∇× H =
μo
0
J=
∂
∂y
0
∂
∂
π 

∂y
∂z
10π sin π ya x + π cos π xa z 
=

μo 

sin π x −10 cos π y
π2
(10sin π ya x + cos π xa z )
μo
214
Prob. 7.48
1
∂
1 ∂
( Aφ sin θ )ar −
(rAφ )aθ
r sin θ ∂θ
r ∂r
A
1 Ao
=
(2sin θ cos θ )ar − o sin θ ( −r −2 )aθ
2
r sin θ r
r
A
= 3o (2 cos θ ar + sin θ aθ )
r
B = ∇× A =
Prob. 7.49
∂
∂
∂
∂z = (−2 yz − x 2 )a x + (2 xz − 2 xy )a z
(a) J = ∇ × H = ∂x ∂y
xy 2 x 2 z − y 2 z
At (2,-1,3), x=2, y=-1, z=3.
J = 2a x + 16a z A/m 2
(b) −
∂ρ v
= ∇ • J = 0 − 2x + 2x = 0
∂t
At (2,-1,3),
∂ρv
= 0 C/m3s
∂t
Prob. 7.50
(a) B = ∇ × A
POESM_Ch07.indd 217
∂Aρ 
 1 ∂Az ∂Aφ 
 ∂Aρ ∂Az 
1 ∂
a
a
az
=
−
+
−
+
(
A
)
−
ρ
ρ
φ
φ
 ∂z
∂z 
∂ρ 
∂φ 
ρ  ∂ρ
 ρ ∂φ

∂A
2
= − z aφ = 20 ρ aCopyright
© 2015 by Oxford University Press
φ μ Wb/m
∂ρ
B −20 ρ
10/14/2015 11:46:12 AM
(b) −
Sadiku & Kulkarni
∂ρ v
= ∇ • J = 0 − 2x + 2x = 0
∂t
At (2,-1,3),
Principles of Electromagnetics, 6e
∂ρv
= 0 C/m3s
∂t
218
Prob. 7.50
(a) B = ∇ × A
∂A 
 1 ∂Az ∂Aφ 
 ∂A ∂A 
1 ∂
a ρ +  ρ − z  aφ +  ( ρ Aφ ) − ρ  a z
=
−

∂z 
∂ρ 
∂φ 
ρ  ∂ρ
 ρ ∂φ
 ∂z
∂A
= − z aφ = 20 ρ aφ μ Wb/m 2
∂ρ
B −20 ρ
=
H=
aφ μ A/m
μo
μo
1 ∂
( ρ Aφ )a z
ρ ∂ρ
−40
1
=
(−40 ρ )a z =
a z μ A/m 2
μo ρ
μo
215
J = ∇× H =
I =  J dS =
(b)
2
−40
μo
2π
 
ρ dφ d ρ , dS = ρ dφ d ρ a z
ρ =0 φ =0
2π
2
−40 ρ 2
=
ρ d ρ  dφ =
μo 0
μo 2
0
−40
=
2
0
(2π )
−80π × 2 × 10−6
= −400 A
4π × 10−7
Prob. 7.51
H =
− ∇Vm
→
Ia 2
From Example 7.3, H =
Vm =
−
Ia 2
2
 (z
−  H ⋅ dl = −mmf
Vm =
2 (z + a
2
+ a2 )
2
−3
2
2
)
3
2
az
dz =
− Iz
2 (z + a
2
2
)
1
2
+ c
As z → ∞, Vm = 0 , i.e.
0 = −
Hence,
Vm =
P.E. 7.9
H =
2πρ
I
2πρ
aφ =
−
I 
1 −
2 
→
c =
I
2


z +a 
z
2
2
aφ
H = − ∇Vm
But
POESM_Ch07.indd 218
I
I
+ c
2
( J = 0)
1 ∂Vm
I
aϕ → Vm = −
φ + C
Copyright
ρ ∂φ© 2015 by Oxford University Press
2π
At (10, 60o , 7 ) , φ = 60o =
π
, V
= 0→
0 = −
I
⋅
π
+ C
10/14/2015 11:46:12 AM
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
219
216
Prob. 7.52
∇'
1
R
=
=
=
1
R
=
=
=
(
)
1
2
2
2
2
= ( x − x') + ( y − y ') + z − z ' 


1
 ∂
∂
∂  
2
2
2 − 2

x
x'
y
y
'
z
z'
a
+
a
+
a
−
+
−
+
−
(
)
(
)
(
)
y
z
 ∂x x

∂y
∂z  
r − r'
R =
∇
1
 ∂
 
∂
∂
2
2
2 − 2

+
+
−
+
−
y
+
−
x
x'
y
'
z
z'
a
a
a
(
) (
) (
)

x
y
z 
∂y '
∂z '  
 ∂x '
3
2
2
2 − 2
 1
 −  ( −2 ) ( x − x' ) a x ( x − x' ) + ( y − y' ) + ( z − z')  + a y and a z terms
 2
R
R3
3
1
2
2
2 − 2
2 ( x − x') a x ( x − x') + ( y − y') + ( z − z')  + a y and az terms


2
− ( x − x') a z + ( y − y ') a y + ( z − z') a z 
R
= − 3
3
R
R
−
Copyright © 2015 by Oxford University Press
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
217220
CHAPTER 8
P.E. 8.1
∂u
= QE = 6a z N
∂t
∂u
∂
(b)
= 6a z = (u x , u y , u z ) 
∂t
∂t
∂u x
= 0 → ux = A
∂t
∂u y
= 0 → uy = B
∂t
∂u z
= 6 → u z = 6t + C
∂t
A=B=C=0
Since u ( t = 0 ) = 0 ,
(a) F = m
ux = 0 = uy, uz = 6t
∂x
ux =
=0→ x= A
∂t
∂y
uy =
=0→ y= B
∂t
∂z
uz =
= 6t → z = 3t 2 + C1
∂t
At t = 0, (x,y,z) = (0,0,0) → A1 = 0 = B1 = C1
Hence , (x,y,z) = (0,0,3t2),
u = 6ta z at any time. At P(0,0,12), z = 12 =3t2 → t =2s
t =2s
(c) u = 6ta z = 12a z m/s .
∂u
a=
= 6a z m 2
s
∂t
(d) K .E =
1
1
2
m u = (1)(144 ) = 72 J
2
2
P.E. 8.2
(a)
ma = eu × B = (eBouy, -eBoux, 0)
d 2 x eBo dy
dy
=
=ω
2
dt
m dt
dt
(1)
Copyright © 2015 by Oxford University Press
POESM_Ch08.indd 220
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
221
218
d2y
eBo dx
dx
=−
= −ω
2
dt
m dt
dt
(2)
d 2z
dz
= 0; 
= C1
2
dt
dt
(3)
From (1) and (2),
d 3x
d2y
dx
= ω 2 = −ω 2
3
dt
dt
dt
(D2 + w2 D)x = 0 → Dx = (0, ±jω)x
x = c2 + c3cosωt +c4sinωt
dy 1 d 2 x
=
= −c3ω cos ωt − c4ω sin ωt
dt ω dt 2
At t = 0, u = (α , 0, β ) . Hence,
c1 = β , c3 = 0, c4 =
α
ω
dx
dy
dz
= α cos ωt , = −α sin ωt , = β
dt
dt
dt
(b)
Solving these yields
a
α
x = sin ωt , y = cos ωt , z = β t
ω
ω
The starting point of the particle is (0,
(c)
x2 + y2 =
α
,0)
ω
α2
, z=βt
ω2
showing that the particles move along a helix of radius α
ω placed along the z-axis.
P.E. 8.3
(a)
From Example 8.3, QuB = QE regardless of the sign of the charge.
E = uB = 8 x 106 x 0.5 x 10-3 = 4 kV/m
(b)
Yes, since QuB = QE holds for any Q and m.
Copyright © 2015 by Oxford University Press
POESM_Ch08.indd 221
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
222
219
P.E. 8.4
By Newton’s 3rd law, F12 = F21 , the force on the infinitely long wire is:
μIIb 1
1
Fl = − F = o 1 2 ( −
)a
2π
ρo ρo + a ρ
=
4π × 10−7 × 50 × 3  1 1 
 −  a ρ = 5a ρ μ N
2π
 2 3
P.E. 8.5
m = ISan = 10 × 10−4 × 50
(2, 6, −3)
7
= 7.143 x 10-3 (2, 6, -3)
= (1.429a x + 4.286a y − 2.143a z ) × 10−2 A-m 2
P.E. 8.6
(a)
(b)
T = m×B =
= 0.03a x − 0.02a y − 0.02a z N-m
T = ISB sin θ →
| T |max =
P.E. 8.7
(a)
μr =
(b)
H=
(c)
10 × 10−4 × 50 2 6 −3
6 4 5
7 × 10
T
max
= ISB
50 × 10 -3
| 6a x + 4a y + 5a z |= 0.04387 Nm
10
μ
= 4.6, χ m = μ r − 1 = 3.6
μo
10 × 10−3 e − y
a z A / m = 1730e − y a z A/m
−7
μ 4π ×10 × 4.6
M = χ m H = 6228e − y a z A/m
B
=
P.E. 8.8
3a x + 4a y 6a x + 8a y
an =
=
5
10
(6 + 32)(6a x + 8a y )
B1n = ( B1 • an )an =
1000
Copyright © 2015 by Oxford University Press
POESM_Ch08.indd 222
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
223
220
= 0.228a x + 0.304a y = B 2 n
B1t = B1 − B1n = −0.128a x + 0.096a y + 0.2a z
B2t =
μ2
B = 10 B1t = −1.28a x + 0.96a y + 2a z
μ1 1t
B2 = B2 n + B2t = −1.052a x + 1.264a y + 2a z Wb/m2
P.E. 8.9
(a)
B1n = B2 n → μ1 H1n = z μ2 H 2 n
or μ1 H1 • an 21 = μ2 H 2 • an 21
(6 H 2 x − 10 − 12)
(60 + 2 − 36)
= 2μo
μo
7
7
35 = 6 H 2 x
H 2 x = 5.833 A/m
(b)
K = ( H1 − H 2 ) × an12 = an 21 × ( H1 − H 2 )
= an 21 ×  (10,1,12) − (35 , −5, 4) 
6


=
2 −3
1 6
25
7 6 6 8
K = 4.86a x − 8.64a y + 3.95a z A/m
(c)
Since B = μ H , B1 and H1 are parallel, i.e. they make the same angle with the
normal to the interface.
H •a
26
cos θ1 = 1 n 21 =
= 0.2373
H1
7 100 + 1 + 144
θ1 = 76.27 o
cos θ 2 =
H 2 • an 21
13
=
= 0.2144
H2
7 (5.833) 2 + 25 + 16
θ 2 = 77.62o
P.E. 8.10
(a)
L ' = μo μr n 2 S = 4π × 10−7 × 1000 × 16 × 106 × 4 × 10−4
= 8.042 H/m
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Principles of Electromagnetics, 6e
224
221
(b)
Wm ' = 1 L' I 2 = 1 (8.042)(0.5 2 ) = 1.005 J/m
2
2
P.E. 8.11 From Example 8.11,
Lin =
μo l
8π
Lext =
2 wm 1
= 2
I2
I
=
=
1
4π
2
l
2π
0
0
μI 2
 4π 2 ρ 2 ρ d ρ dφ dz
b
2 μo
 dz  dφ  (1 + ρ ) ρ dρ
a
b
1
2μo
1 
dρ
• 2π l   −
2
ρ (1 + ρ ) 
4π
a 
1+ b 
μ ol  b
ln − ln

π  a
1 + a 
μ l μ l b
1+ b
L = Lin + Lext = o + o ln − ln
8π
π  a
1 + a 
=
P.E. 8.12
(a)
L’in =
μo
4π × 10−7
=
= 0.05 μH/m
8π
8π
L’ext = L’ – L’in = 1.2 – 0.05 = 1.15 μH/m
(b)
L’ =
ln
μo
2π
d − a
1
 4 + ln a 


d − a 2π L '
2π ×1.2 ×10−6
=
− 0.25 =
− 0.25
a
μo
4π x10−7
= 6 − 0.25 = 5.75
d −a
= e 5.75 = 314.19
a
d − a = 314.19a = 314.19 ×
d = 407.9mm = 40.79cm
2.588 × 10 −3
= 406.6mm
2
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Principles of Electromagnetics, 6e
225
222
P.E. 8.13
This is similar to Example 8.13. In this case, however, h=0 so that
μ I a 2b
A1 = o 1 3 aφ
4b
μ I a2
μ πI a 2
φ12 = o 12 • 2πb = o 1
4b
φ12
m12 =
=
μoπ a
2b
I1
= 2.632 μH
P.E. 8.14
Lin =
2b
2
=
4π ×10−7 × π × 4
2×3
μo
μ 2πρo
4π × 10−7 × 10 × 10−2
l= o
=
8π
8π
4
= 31.42 nH
P.E. 8.15
(a) From Example 7.6,
Bave =
μo NI
l
φ = Bave • S =
or I =
=
μo NI
2πρ o
μo NI
• πa 2
2πρ o
2 ρ oφ 2 × 10 × 10−2 × 0.5 × 10−3
=
μa 2 N
4π × 10− 7 × 10− 4 × 103
= 795.77A
Alternatively, using circuit approach
l
2πρo
2πρo
R=
=
=
μ S μo S μoπ a 2
φℜ 2 ρ oφ
ℑ = NI =
=
, as obtained before.
N
μa 2 N
ℜ=
2 ρo
2 × 10 × 10−2
=
= 1.591× 109
μa 2 4π × 10− 7 × 10− 4
ℑ = ϕℜ = 0.5x10-3x1.591x109=7.9577x105
ℑ
I = = 795.77 A as obtained before.
N
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c
ψ
Sadiku & Kulkarni
F
(b)
226
223
Principles of Electromagnetics, 6e
Ra
If μ=500μo,
795.77
I=
= 1.592 A
F = NI500
= 500 x 0.2 = 100 A.t
P.E. 8.16
l
42 × 10−2
42 × 106
Rc = 2c =
=
3 −4
−4
B S (1.5) 2 ×−710 × 10
22500
16
ℑ = μ Sa =4π × 10 × 10 −7× 4 ×=10
= π895.25N
2 μo
2 × 4π × 10
8π241
la
10−3
108
=
=
Ra =
−7
−4
16π
P.E. 8.17 μo S 4π ×10 × 4 × 10
μ NI
We may approximate the longer
solenoid as infinite so that B1 = o 1 1 . The flux linking
8
l1
1.42 × 10
Ra + Rc =
the
second
is:π
Prob.
8.1 solenoid 16
μ NI
ψ 2 =2 N 2 B1S1 = o−31 1 1 • π r1216 N22
16πl ×100 16)π (0.4 ×10 −10 ) = 14.576 nN
F =ψ
mω= r =F9.11×
μ Wb
=10 1 × (2 8×10
=
R
1.42
ψa2 + Rμc o N11.42
N 2 × 102
M =
=
• πr1
l1
Prob. 8.2 I1
−6
(a) we assume
ψ air-core
16π × 10
Here
solenoids.
Ba = =
= 88.5 mWb/m 2
−4 6
10
−
2
× 4 ×10
−3
F = Q(u × B )S= 101.42
= 10−3 (−50a x − 250a y )
0 0 25
Prob.
8.45
P.E. 8.18
= −0.05a x −I 0.25a y N
H = R =a ρ 
2πρ μ S
2
(b) Constant 1velocity2implies
that
a = 0.−2
1
1I acceleration
−2
w
|
|
=
μ
H
=
μ
πρ
π
π
2
2
(5
6)10
11

=
=
×
+
=
×10
m
2
2
F = ma = 0 =2Q( E + uo× B
2 ) 4π2 ρ
2
E = -u × B = S =
501.5
a x ×+10
250
a y −IV/m
1−2 (6
5)10−2 = 1.5 × 10−4 1
μ I 2 L ln(b / a)
W =  wm dv =  μ 2 2 ρ dφ d ρ dz =
2 4π ρ
4π ψ 
→ μ=
Prob. 8.3 1 F = NI = ψ−7R = ψ μ S −6 ⎯⎯
NIS
At P, x==4π
2, ×y4=×5,4πz ×=10
-3−3 (625 × 10 −2 )3ln(18 /12) = 304.1 pJ
12 × 10 (11π × 10 2 )
μa=x + (2) 2 (−3)a y + (2)
= 27.65 × 10−3 H/m
E = 2(2)(5)(−3)
−4 (5)a z = −60a x − 12a y + 20a z
500(2)1.5
10
×
Alternatively,
B =ψ
−3)−32 a y + 22 a z = 25a x + 9a y + 4a z
(5) 2 a12
x +×(10
B= =
= 80 Wb
F =Q
+ u××10B−)4 1 2 1 μ L b 2 μ I 2 L b
S ( E1.5
W = LI =
ln × I =
ln
1.4 3.2 −21
2 2π a
4π
a
u× B =
= 21.8a x − 30.6a y − 67.4a z
25 9
4
E + u × B = (−60, −12, 20) + (21.8, −30.6, −67.4) = ( −38.2, −42.6, −47.4)
F = Q( E + u × B ) = 4( E + u × B ) mN
= −152.8a x − 170.4a y − 189.6a z mN
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ℑ=
Sadiku & Kulkarni
B 2 a S (1.5) 2 × 10 × 10−4 22500
=
=
= 895.25N
2 μo
2 × 4π × 10−7
8π
Principles of Electromagnetics, 6e
227
Prob. 8.1
F = mω 2 r = 9.11×10−31 × (2 ×1016 ) 2 (0.4 ×10 −10 ) = 14.576 nN
Prob. 8.2
(a)
F = Q(u × B ) = 10−3
10 −2
6
0
25
0
= 10−3 (−50a x − 250a y )
= −0.05a x − 0.25a y N
(b) Constant velocity implies that acceleration a = 0.
F = ma = 0 = Q( E + u × B )
E = -u × B =
50a x + 250 a y V/m
Prob. 8.3
At P, x = 2, y = 5, z = -3
E = 2(2)(5)(−3)a x + (2) 2 (−3)a y + (2) 2 (5)a z = −60a x − 12a y + 20a z
B = (5) 2 a x + (−3) 2 a y + 22 a z = 25a x + 9a y + 4a z
F = Q( E + u × B)
u× B =
1.4 3.2 −1
= 21.8a x − 30.6a y − 67.4a z
25 9
4
E + u × B = (−60, −12, 20) + (21.8, −30.6, −67.4) = ( −38.2, −42.6, −47.4)
F = Q( E + u × B ) = 4( E + u × B ) mN
= −152.8a x − 170.4a y − 189.6a z mN
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Principles of Electromagnetics, 6e
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224
Prob. 8.4
F = qE = ma = m
du
dt
du qE 10 ×10−3
=
=
(30, 0, 0) ×103
dt
m
2
d
(u x , u y , u z ) = (150, 0, 0)
dt
Equating components gives
du x
= 150
⎯⎯
→ u x = 150t + c1
dt
du y
=0
⎯⎯
→ u y = c2
dt
du z
=0
⎯⎯
→ u z = c3
dt
At t = 0, u =(2,5,0) × 103 .
2000 = 0 + c1
⎯⎯
→ c1 = 2000
5000 = c2
0 = c3
Hence, u= (150t+2000,5000,0)
At t = 4s,
u = (2600,5000, 0) m/s
dx
= 150t + 2000
⎯⎯
→ x = 75t 2 + 2000t + c4
dt
dy
uy =
= 5000
⎯⎯
→ y = 5000t + c5
dt
dz
uz =
=0
⎯⎯
→ z = +c6
dt
At t=0, (x,y,z)=(0,0,0)
⎯⎯
→ c4 = 0 = c5 = c6
ux =
Hence,
( x, y, z ) = (75t 2 + 2000t ,5000t , 0)
At t = 4s, x=9,200, y=20,000, z=0.
i.e. ( x, y, z ) = (9200, 20000, 0)
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Principles of Electromagnetics, 6e
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225
Prob. 8.5
ma = Qu × B
10−3 a = −2 × 10−3
ux
0
uy
6
uz
0
d
(u x , u y , u z ) = (12u z ,0,−12u x )
dt
i.e.
du y
du x
= 12u z
dt
(1)
(2)
= 0 → u y = A1
dt
du z
= −12u x
dt
(3)
From (1) and (3),
ux = 12u z = −144u x
or
ux + 144u x = 0 → u x = c1 cos12t + c 2 sin 12t
From (1), uz= - c1sin12t + c2cos12t
At t=0,
Hence,
ux=5, uy=0, uz=0 → A1=0=c2, c1=5
u = (5cos12t , 0, −5sin12t )
u(t = 10s ) = (5cos120, 0, −5sin120) = 4.071a x − 2.903a z m/s
dx
= 5 cos12t → x = 5 sin 12t + B1
12
dt
dy
uy =
= 0 → y = B2
dt
dz
uz =
= −5 sin 12t → z = 5 cos12t + B3
12
dt
ux =
At t=0, (x, y, z) = (0, 1, 2) → B1=0, B2=1, B3=
19
12
19 
5
 5
( x , y , z) =  sin 12t ,1, cos 12t + 
 12
12 
12
(4)
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Principles of Electromagnetics, 6e
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226
At t=10s,
19 
5
 5
( x , y , z) =  sin 120 ,1, cos 120 +  = (0.2419, 1, 1.923)
 12
12 
12
By eliminating t from (4),
x 2 + ( z − 19 ) 2 = ( 5 ) 2 , y = 1 which is a circle in the y=1 plane with center at
12
12
(0,1,19/12). The particle gyrates.
Prob. 8.6
(a)
ma = −e(u × B )
−
u
m d
(u x , u y , u z ) = x
0
e dt
uy
0
uz


= u y Bo ax − Bo u x a y
Bo
du z
= 0 → uz = c = 0
dt
du x
Be
Be
= −u y o = −u y w , where w = o
m
dt
m
du y
= ux w
dt
Hence,
ux = − wu y = − w2u x
or ux + w2u x = 0 → u x = A cos wt + B sin wt
uy = −
u x
= A sin wt − B cos wt
w
At t=0, ux = uo, uy = 0 → A = uo, B=0
Hence,
u
dx
→ x = o sin wt + c1
dt
w
dy
u
u y = uo sin wt =
→ y = − o cos wt + c2
dt
w
u
At t=0, x = 0 = y → c1=0, c2= o . Hence,
w
uo
uo
x = sin wt , y = (1 − cos wt )
w
w
u x = uo cos wt =
2
u 2o
u
u 
(cos 2 wt + sin 2 wt ) =  o  = x 2 + ( y − o ) 2
2
w
w
w
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Principles of Electromagnetics, 6e
231
227
showing that the electron would move in a circle centered at (0,
uo
). But since the field
w
does not exist throughout the circular region, the electron passes through a semi-circle
and leaves the field horizontally.
(b)
d = twice the radius of the semi-circle
=
2u o 2u o m
=
w
Bo e
Prob. 8.7
F =  Idl × B =
0.2
 2dy(−a
y
) × (4a x − 8a z )
0
ax
ay
az
(−a y ) × (4a x − 8a z ) = 0
−1
0 = 8a x + 4a z
0
−8
4
F = 2(8a x + 4a z )(0.2) = 3.2a x + 1.6a z N
Prob. 8.8
qE
o
mg = qE
mg
mg 0.4 × 10−3 × 9.81
⎯⎯
→ q=
=
= 26.67 nC
1.5 ×105
E
Prob. 8.9
μ I I a ×a
F
= I1al × B2 = o 1 2 l φ
L
2πρ
−7
a × (−a y )4π ×10 (−100)(200)
F21 = z
= 4a x mN/m (repulsive)
2π
ℑ = IL × B → ℑ =
(a)
(b)
F12 = − F21 = −4a x mN/m (repulsive)
(c)
4
3
3
4
al × aφ = a z × (− a x + a y ) = − a x − a y , ρ = 5
5
5
5
5
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Principles of Electromagnetics, 6e
232
228
4π ×10−7 (−3 ×104 )  3
4 
F31 =
 − ax − a y 
2π (5)
5 
 5
= 0.72a x + 0.96a y mN/m (attractive)
(d)
F3 = F31 + F32
4π ×10−7 × 6 × 104 )
( az × a y ) = −4ax mN/m(attractive)
2π (3)
F3 = −3.28a x + 0.96a y mN/m
F32 =
(attractive due to L2 and repulsive due to L1)
Prob. 8.10
F=
μo I1 I 2 4π ×10−7 (10)10
=
= 100 μ N
2πρ
2π (20 ×10−2 )
Prob. 8.11
F =  Ldl × B = 3(2a z ) × cos φ
W = −  F • dl ,
F = 6 cos φ
2π
3

aφ N
W = −  6 cos φ
0
= -1.8sin
3
ρo dφ = −6 ρo × 3sin φ 3
2π
0
3
aφ
J
2π
= -1.559 J
3
Prob. 8.12
6
(a)
μo I1 I 2
4π × 10−7
d ρ a ρ × aφ =
F1 = 
(2)(5) ln 6 a z
2
πρ
π
2
2
ρ =2
= 2 ln 3a z μ N = 2.197a z μ N
(b)
F2 =  I 2 dl2 × B1
μo I1 I 2
2π
μII
= o 1 2
2π
=
1
 ρ  d ρ aρ + dza  × aφ
1
z
 ρ d ρ a
z
− dzaρ 
But ρ = z+2, dz=dρ
2
4π ×10−7
1
(5)(2)   d ρ a z − dza ρ 
F2 =
2π
ρ =4 ρ
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Principles of Electromagnetics, 6e
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229
2 ln 2 (a z − a ρ ) μ N = 1.386aρ − 1.386a z a z μ N
4
μo I1 I 2 1
 d ρ a z − dza ρ 
F3 =
2π  ρ 
But z = -ρ + 6, dz = -dρ
4
4π × 10−7
1
(5)(2)   d ρ a z − dza ρ 
F3 =
2π
ρ =6 ρ
2 ln 4 (a z + a ρ ) μ N = −0.8109a ρ − 0.8109a z μ N
6
F = F1 + F2 + F3
= aρ (ln 4 + ln 4 − ln 9) + a z (ln 9 − ln 4 + ln 4 − ln 9)
= 0.575a ρ μ N
Prob. 8.13
A
From Prob. 8.7,
f =
C
μo I1 I 2
aρ
2πρ
60o
30
o
fBC
fAC
B
f = f AC + f BC
4π ×10−7 × 75 ×150
| f AC |=| f BC |=
= 1.125 × 10−3
2π × 2
o
f = 2 × 1.125cos 30 a x mN/m
= 1.949a x mN/m
Prob. 8.14
The field due to the current sheet is
B=
μ
2
K × an =
μo
2
10a x × (−a z ) = 5μ o a y
L
F = I 2  dl2 × B = 2.5 dxa x × (5μo a y ) = 2.5L × 5μo (a z )
0
F
= 12.5 × 4π × 10−7 (a z ) = 15.71a z μ N/m
L
Prob. 8.15
F =  Idl × B = IL × B = 5(2a z ) × 40a x 10−3 = 0.4a y N
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Principles of Electromagnetics, 6e
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230
Prob. 8.16
T = m × B = [ 0.4(0.6)(3)a x ] × (0.5a x + 0.8a y ) = 0.72(0.8)a z
= 0.576a z Nm
Prob. 8.17
F =  Idl × B
⎯⎯
→ F = IB = 520 × 0.4 × 10 −3 × 30 × 10 −3
F = 6.24 mN
Prob. 8.18
m = IS
I=
⎯⎯
→ I=
m
m
= 2
S πr
8 × 1022
= 6.275 × 108 = 627.5 MA
3 2
π (6370 ×10 )
Prob. 8.19
Let F = F1 + F2 + F3
0
F1 =  Idl × B =  2dxa x × 30a z mN
5
=-60a y x
0
= 300a y mN
5
5
F2 =  2dya y × 30a z mN
0
=60a x y
5
0
= 300a x mN
5
F3 =  2(dxa x + dza z ) × 30a z mN
0
=60(-a y ) x
5
= −300a y mN
0
F = F1 + F2 + F3 = 300a y +300a x -300a y mN=300a x mN
1
T = m × B = ISan × B = 2( )(5)(5)a y × 30a z 10−3 = 0.75a x N.m
2
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235
231
Prob. 8.20
For each turn, T = m × B, m = ISan
For N turns,
T = NISB = 50 × 4 × 12 × 10 −4 × 100 × 10 −3 = 24 mNm
Prob. 8.21
f ( x, y, z ) = x + 2 y − 5 z − 12 = 0
an =
∇f = a x + 2a y − 5a z
a x + 2a y − 5a z
∇f
=
| ∇f |
30
m = NISan = 2 × 60 × 8 ×10−4
Prob. 8.22
M = χmH = χm
B
μo μr
=
(a x + 2a y − 5a z )
30
M = χm H = χm
B
μo μ
4999
1.5
×
5000
4π × 10 −7
M =
= 17.53a x + 35.05a y − 87.64a z mAm
χm B
μo (1 + χ m )
Prob. 8.23
(a)
⎯⎯
→
=
1.193 × 106 A/m
N
(b)
M =
m
k =1
k
Δv
If we assume that all mk align with the applied B field,
M =
Nmk
Δv
→
mk =
Nmk
N
Δv
=
1.193 × 106
8.5 × 1028
m k = 1.404 × 10 −23 A ⋅ m 2
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Principles of Electromagnetics, 6e
236
232
Prob. 8.24
μr = χ m + 1 = 6.5 + 1 = 7.5
M = χm H
⎯⎯
→ H=
M
χm
=
24 y 2
az
6.5
At y = 2cm,
24 × 4 × 10−4
a z = 1.477a z mA/m
6.5
∂
∂
∂
∂x ∂y
∂z
48 y
=
ax
J = ∇× H =
2
6.5
24 y
0
0
6.5
At y=2cm,
H=
J=
48 × 2 ×10−2
a x = 0.1477a x A/m 2
6.5
Prob. 8.25
(a)
(b)
(c)
(d)
χ m = μr − 1 = 3.5
4y a z × 10 −3
= 707.3y a z A/m
μ
4π × 10 −7 × 4.5
M = χ m H = 2.476y a z kA/m
H =
B
=
Jb = ∇ × M =
∂
∂x
0
∂
∂
∂y
∂z
M z (y)
0
=
dM z
ax
dy
= 2.476ax kA/m 2
Prob. 8.26
When H = 250,
2H
2(250)
B=
=
= 1.4286 mWb/m 2
100 + H 100 + 250
But B=μo μr H
μr =
B
1.4286 × 10−3
=
= 4.54
μo H 4π ×10−7 × 250
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
237
233
Prob. 8.27
 H ⋅ dl
=
Ienc
πρ 2
Hφ ⋅ 2πρ =
⋅I
π a2
M = χm H =
Jb = ∇ × M =
→
Iρ
aφ
2π a 2
( μr − 1)
1 ∂
ρ ∂ρ
Iρ
2π a 2
Hφ =
( ρM ) a
φ
z
=
( μr − 1)
I
a
π a2 z
Prob. 8.28
(a)
From H1t – H2t = K and M = χmH, we obtain:
M 1t
χ m1
−
M 2t
χm2
=K
Also from B1n – B2n = 0 and B = μH = (μ/χm)M, we get:
μ1 M 1n μ 2 M 2 n
=
χ m1
χ m2
(b)
From B1cosθ1 = B1n = B2n = B2cosθ2
B sin θ1
B sin θ 2
= H1t = K + H 2t = K + 2
and 1
μ1
μ2
(1)
(2)
Dividing (2) by (1) gives
tan θ1
k
tan θ 2 tan θ 2 
kμ 2 

1 +
=
+
=
μ1
B2 cosθ 2
μ2
μ 2  B2 sin θ 2 
i.e.
tan θ 1 μ1 
kμ 2
1 +
=
tan θ 2 μ 2  B2 sin θ 2



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Principles of Electromagnetics, 6e
238
234
Prob. 8.29
B2 n = B1n = 1.8a z
H 2t = H1t
B2t =
⎯⎯
→
B2t
=
μ2
B1t
μ1
4 μo
μ2
B1t =
(6a x − 4.2a y ) = 9.6a x − 6.72a y
μ1
2.5μo
B2 = B2 n + B2t = 9.6a x − 6.72a y + 1.8a z mWb/m 2
H2 =
B2
μ2
=
10−3 (9.6, −6.72,1.8)
4 × 4π ×10−7
= 1,909.86 a x − 1,336.9a y + 358.1a z A/m
z
B2n
θ2
B2t
B2 n
1.8
=
= 0.1536
B2t
9.62 + 6.722
θ 2 = 8.73o
tan θ 2 =
Prob. 8.30
(a)
B1n = B 2n = 15aφ
B1t
H1t = H 2t →
B1t =
=
μ1
B2t
μ2
μ1
2
B2t =
10a ρ − 20a z
μ2
5
(
)
= 4a ρ − 8 a z
Hence,
B1 = 4aρ + 15aφ − 8a z mWb/m 2
(b)
w m1 =
1
B1 ⋅ H1 =
2
B12
2 μ1
w m1 = 60.68 J / m3
w m2 =
B22
2 μ2
=
(10
2
=
(4
2
)
+ 152 + 82 × 10 −6
2 × 2 × 4π × 10 −7
)
+ 152 + 202 × 10 −6
2 × 5 × 4π × 10
−7
= 57.7 J / m3
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
239
235
Prob. 8.31
B2 n = B1n = 40a x
B2 n = μ2 H 2 n
H 2t = H1t
B2t =
⎯⎯
→ H 2n =
⎯⎯
→
B2t
μ2
=
40a x
μ2
=
40a x
50 μo
B1t
μ1
μ2
B
μ1 1t
B2t
H 2t =
μ2
=
B1t
μ1
=
H 2 = H 2 n + H 2t =
(−30a x + 10a y )
μo
1 40
10−3
( , −30,10) ⋅10−3 =
(0.8, −30,10)
μo 50
4π ×10−7
H 2 = 0.6366a x − 23.87a y + 7.957a z kA/m
Prob. 8.32
H 2t = H1t = α a x + δ a z
B2 n = B1n
⎯⎯
→ μ2 H 2 n = μ1 H1n
μ1
μ
H1n = r1 β a y
μ2
μr 2
μ
H = α a x + r1 β a y + δ a z
μr 2
H 2n =
Prob. 8.33
B2 n = B1n = 0.6a y
H 2t = H1t
B1t =
⎯⎯
→
B2t
μ2
=
B1t
μ1
μ
μ1
B2t = o (1.4a x − 2a z ) = 0.1167a x − 0.1667a z
12μo
μ2
B1 = B1n + B1t = 0.1167a x + 0.6a y − 0.1667a z Wb/m 2
H1 =
B1
μ1
=
10−3 (0.1167, 0.6, −0.1667)
4π ×10−7
= (0.0929 a x + 0.4775a y − 0.1327a z ) ⋅106 A/m
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
240
236
Prob. 8.34
f ( x, y , z ) = x − y + 2 z
∇f = a x − a y + 2a z
an =
(a)
∇f
1
=
(a x − a y + 2a z )
| ∇f |
6
H1n = ( H1 an )an = (40 − 20 − 60)
(a x − a y + 2a z )
6
= −6.667a x + 6.667a y − 13.333a z A/m
(b)
H 2 = H 2 n + H 2t
⎯⎯
→
B 2 n = B1n
μ2 H 2 n = μ1 H1n
B2 = μ2 H 2 = μ2 H 2 n + μ 2 H 2t = μ1 H1n + μ 2 H 2t = μo (2 H1n + 5 H 2t )
But
= 4π × 10−7 [ (−13.333,13.333, −26.667) + (233.333, 66.666, −83.333]
= 4π × 10−7 (220,80, −110)
= 276.5a x + 100.5a y − 138.2a z μ Wb/m 2
Prob. 8.35
an = a ρ
B2 n = B1n = 22 μo a ρ
H 2t = H1t
B2t =
⎯⎯
→
B2t
μ2
=
B1t
μ1
μ2
μo
B1t =
(45μo aφ ) = 0.05625μo aφ
μ1
800 μo
B2 = μo (22a ρ + 0.05625aφ ) Wb/m 2
Prob. 8.36
r = a is the interface between the two media.
B2 n = B1n
⎯⎯
→
Bo1 (1 + 1.6) cos θ ar = Bo 2 cos θ ar
2.6 Bo1 = Bo 2
(1)
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
241
237
H 2t = H1t
⎯⎯
→
B2t
μ2
=
B1t
μ1
μ2 B1t = μ1 B2t
μ2 Bo1 (−0.2) sin θ aθ = μo Bo 2 (− sin θ )aθ
μB
(2)
μ2 = o o 2
0.2 Bo1
Substituting (1) into (2) gives
μ2 =
μo
0.2
(2.6) = 13μo
Prob. 8.37
(a)
H = 1 K × an = 1 (30 − 40)a x × (−a z ) = −5a y A/m
2
2
B = μo H = 4π × 10−7 (−5a y ) = −6.28a y μ Wb/m2
(b)
H = 1 (−30 − 40)a y = −35a y A/m
2
B = μo μr H = 4π ×10−7 (2.5)( −35a y ) = −110a y μ Wb/m 2
(c)
H = 1 (−30 + 40)a y = 5a y
2
B = μo H = 6.283a y μ Wb/m2
Prob. 8.38
H1n = −3a z ,
H1t = 10a x + 15a y
H 2t = H1t = 10a x + 15a y
H 2n =
μ1
1
H1n =
(−3a z ) = −0.015a z
μ2
200
H 2 = 10a x + 15a y − 0.015a z
B2 = μ2 H 2 = 200 × 4π × 10−7 (10,15, −0.015)
B2 = 2.51a x + 3.77a y − 0.0037a z mWb/m2
tan α =
B2 n
B2t
or α = tan −1
0.0037
2
2.51 + 3.77
2
= 0.047o
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
242
238
Prob. 8.39
(a)
The square cross-section of the toroid is shown below. Let (u,v) be the local
coordinates and ρ o =mean radius. Using Ampere’s law around a circle passing
through P, we get
v
u
(0, ρ o )
H (2π )( ρ o + v) = NI
⎯⎯
→
H=
The flux per turn is
Ψ=
a/2

a/2

u =− a / 2 v =− a / 2
L=
(b)
Bdudv =
NI
2π ( ρo + v)
μo NIa  ρo + a / 2 
ln 

2π
 ρo − a / 2 
N Ψ μo N 2 a  2 ρo + a 
=
ln 

2π
I
 2 ρo − a 
The circular cross-section of the toroid is shown below. Let (r,θ) be the local
coordinates. Consider a point P( r cos θ , ρ o + r sin θ ) and apply Ampere’s law
around a circle that passes through P.
H (2π )( ρ o + r sin θ ) = NI
H=
⎯⎯
→
r
NI
NI  r sin θ 
≈
1 −

2π ( ρ o + r sin θ ) 2πρo 
ρo 
θ
(0, ρ o )
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
243
239
a 2π

Flux per turn Ψ =
r =0 θ
L=
μ NI  r sin θ 
μ NI a 2
(2π )
1 −
rdrdθ =
2πρ o 
ρo 
2πρ o 2
N Ψ μ N 2a2
=
2 ρo
I
Or from Example 8.10,
L = L' l =
μo N 2lS
l2
=
μo N 2πa 2 μo N 2a 2
=
2πρ o
2 ρo
Prob. 8.40
1
2
a = 2 cm
ρo = (3 + 5) = 4cm
L=
μo N 2 a  2 ρ o + a 
ln 

2π
 2 ρo − a 
N2 =
2π L
2π (45 × 10−6 )
=
= 22, 023.17
 2 ρo + a 
8+ 2
−7
−2
μo a ln 
 4π ×10 (2 ×10 ) ln  8 − 2 
a
ρ
−
2
 o

N = 148.4 or 148
Prob. 8.41
μ
L= o
8π
⎯⎯
→
L 4π × 10−7
=
= 50 nH/m

8π
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
244
240
Prob. 8.42
Lin =
μo 
,
8π
μo 
ln(b / a )
2π
μo  μo 
⎯⎯
→
=
ln(b / a)
8π
π
Lext =
If Lin = 2L ext
1
b
= e1/ 8 = 1.1331
8
a
b = 1.1331a = 7.365 mm
ln(b / a ) =
Prob. 8.43
L' =
μ
2π
b  4π ×10−7
1
ln
[0.25 + ln(6 / 2.5)] = 225 nH
+
=
 4
a 
2π
Prob. 8.44
For N = 1,
M 12 =
=
Nψ 12
I
ψ 12
I1
=
b
μo I
μ Ib a + ρo
dzd ρ = o ln
ρo
2πρ
2π
=
z =0
N μo b a + ρ o
ln
=
ρo
2π
ψ 12 =  B1 • dS =
M 12 =
ρo + a
 
ρ ρ
o
μo b a + ρ o
ln
ρo
2π
4π × 10−7
(1) ln 2 = 0.1386 μ H
2π
Copyright © 2015 by Oxford University Press
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ψ 2 = N 2 B1S1 =
Sadiku & Kulkarni
M =
ψ2
=
I1
μo N1 I1
μo N1N 2
l1
l1
2
• π r1  N 2
Principles of Electromagnetics, 6e
2
r1
•π
245
Here we assume air-core solenoids.
Prob. 8.45
H=
I
2πρ
wm =
aρ
1
1
I2
μ | H |2 = μ 2 2
2
2 4π ρ
1
I2
1
μ I 2 L ln(b / a)
W =  wm dv =  μ 2 2 ρ dφ d ρ dz =
2 4π ρ
4π
1
=
× 4 × 4π × 10−7 (625 × 10−6 )3ln(18 /12) = 304.1 pJ
4π
Alternatively,
1 2 1 μ L b 2 242
μI 2L b
W = LI =
ln × I =
ln
2
2 2π a
4π
a
Prob. 8.46
μr = χ m + 1 = 20
1
1
μH ⋅ H
B1 ⋅ H1 =
2
2
1
μ 25x 4 y 2 z 2 + 100x 2 y 4 z 2 + 225x 2 y 2z 4
=
2
wm =
(
Wm =
)
 w dv
m
1
2
2
1
2
2
1 
μ  25 x 4 dx  y 2 dy  z 2 dz + 100 x 2 dx y 4 dy  z 2 dz
0
0
0
−1
−1
2  0
1
2
2
+ 225 x 2 dx  y 2 dy  zdz 

0
0
−1
1
2
2
1
2
2
25μ  x 5
y3
z3
x3
y5 z 3
=
+ 4

2  5 0 3 0 3 −1
3 0 5 0 3 −1

1
2
2
x3
y3 z 5 
+ 9

3 0 3 0 5 −1 

=
=
25μ  1 8 9
 ⋅ ⋅
2 5 3 3
=
25
3600
× 4π × 10 −7 × 20 ×
2
45
+
4 32 9
⋅
⋅
3 3 3
+
9 8 33 
⋅ ⋅

3 3 5
Wm = 25.13 mJ
Prob. 8.47
W=
POESM_Ch08.indd 245
1
1
2 University
−6
Copyright © 2015
Oxford
μ H 2 dv =  4.5
× 4π × 10−7by
+ 5002 ]10Press
dxdydz
[200

2v
2
10/14/2015 12:24:30 PM
25
3600
× 4π × 10 −7 × 20 ×
2
45
=
Sadiku & Kulkarni
Wm = 25.13 mJ
Principles of Electromagnetics, 6e
246
Prob. 8.47
1
1
μ H 2 dv =  4.5 × 4π × 10−7 [2002 + 5002 ]10−6 dxdydz

2v
2
6
= 2π (4.5)10 −7 (29 × 10 4 )10 −6 (2)(2)(2)10 −243
= 6.56 pJ
W=
Prob. 8.48
NI = Hl =
Bl
l
μ
ρo
1.5 × 0.6π
N=
=
μo μr I 4π ×10−7 × 600 ×12
= 313 turns
Bl
N
Prob. 8.49
F = NI = 400 x 0.5 = 200 A.t
Ra =
100
MAt/Wb,
4π
Fa =
R 1 = R2 =
6
MAt/Wb,
4π
R3 =
1.8
MAt/Wb
4π
Ra F
= 190.8 A.t
Ra + R3 + R1 // R2
Ha =
Fa
190.8
=
= 19080 A/m
l a 1 × 10 − 2
Prob. 8.50
Total F = NI = 2000 x 10 = 20,000 A.t
lc
(24 + 20 − 0.6) × 10 −2
=
Rc =
= 0.115 x 107 A.t/m
μ o μ r S 4π × 10 −7 × 1500 × 2 × 10 − 4
la
Ra =
=
0.6 × 10−2
= 2.387 x 107 A.t/m
−7
−4
4π × 10 (1) × 2 × 10
μo μ r S
R = Ra + Rc = 2.502 x 107 A.t/m
ψ=
ℑ
20,000
=ψa =ψc =
= 8 x 10-4 Wb/m2
R
2.502 × 107
Ra
2.387 × 20,000
ℑ=
= 19,081 A.t
2.502
R a + Rc
Rc
0.115 × 20,000
ℑc =
ℑ=
= 919 A.t
2.502
R a + Rc
ℑa =
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
247
244
Prob. 8.51
Rc
ψ
Ra
F
F = NI = 500 x 0.2 = 100 A.t
Rc =
lc
42 × 10−2
42 × 106
=
=
μ S 4π × 10−7 × 103 × 4 ×10−4
16π
Ra =
la
10−3
108
=
=
μo S 4π ×10−7 × 4 ×10−4 16π
Ra + Rc =
ψ=
Ba =
1.42 × 108
16π
F
16π ×100 16π
μ Wb
=
=
Ra + Rc 1.42 × 108 1.42
ψ
S
=
16π × 10−6
= 88.5 mWb/m 2
1.42 × 4 ×10−4
P.E. 8.18
R=

μS
1
 = 2πρo = 2π × (5 + 6)10−2 = 11π ×10−2
2
−2
S = 1.5 ×10 (6 − 5)10−2 = 1.5 × 10−4

ψ
F = NI = ψ R = ψ
⎯⎯
→ μ=
μS
NIS
μ=
B=
ψ
S
=
12 × 10−3 (11π × 10−2 )
= 27.65 × 10−3 H/m
−4
500(2)1.5 ×10
12 ×10−3
= 80 Wb
1.5 × 10−4
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
248
245
Prob. 8.52
F=
B2S
ψ2
4 × 10−6
=
=
= 53.05 kN
2 μo 2μo S 2 × 4π × 10−7 × 0.3 ×10−4
Prob. 8.53
(a)
F = NI = 200 x 10-3 x 750 = 150 A.t.
l
10−3
Ra = a =
= 3.183 × 107
μo S 25 × 10− 6 μo
2π × 0.1
lt
=
Rt =
= 6.7 x 107
μo μ r S μo × 300 × 25 × 10− 6
ψ=
ℑ
150
= 7
= 15.23 × 10 −7
Ra + Rt 10 (3.183 + 20 / 3)
F=
B2S
ψ2
2.32 × 10−12
=
=
2 μo 2μo S 2 × 4π ×10−7 × 25 × 10−6
= 37 mN
(b)
Rt
ℑ
Ra
ℑ
150
=
Ra 3.183 × 107
If μt → ∞, Rt = 0, ψ =
F2 = I 2 dl2 • B1 = I 2 dl2
ψ
ψ1
S
=
2 × 10−3 × 5 × 10−3 × 150
3.183 × 107 × 25 × 10− 6
F2 = 1.885 μN
Prob. 8.54
ψ2
ψ2
ψ1
ℑ
Ra
ψ 1 = 2ψ 2 ,ψ 1 =
Ra
ℑ
3 R
2 a
ψ1
=
ℑ
Ra

Ra/2
Ra
ℑ
2ℑ
→ψ2 =
3Ra
3Ra
2
 ψ 2  ψ
3ψ 1
ℑ2
ℑ = 2 2  + 1 =
=
2
 2μo S  2 μo S 4 μo S 3Ra μo S
Copyright © 2015 by Oxford University Press
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Principles of Electromagnetics, 6e
249
246
=
μo S ℑ2
3la
2
4π × 10−7 × 200 ×10−4 × 9 × 106
=
3 ×10−6
= 24π × 10 3 = mg → m =
24π × 10 3
= 7694 kg
9.8
Prob. 8.55
ℑ
ℑ = NI
Rs
Ra
Rs

Rs/2
Ra
Since μ → ∞ for the core (see Figure) , Rc = 0.
a
Rs  ψ ( 2 + x)

ℑ = NI = ψ  Ra +  =
2 
μo S

ψ (2 x + a)
=
2μo S
ℑ=
=
N 2 I 2 4μo2 S 2
B2S
1
1
=ψ 2
=
•
2 μo
2 μo S 2 μo S
(a + 2 x) 2
2 N 2 I 2 μo S
(a + 2 x)2
F = − Fa x since the force is attractive, i.e.
F=
−2 N 2 I 2 μo Sa x
(a + 2 x) 2
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POESM_Ch08.indd 249
10/14/2015 12:24:32 PM