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6.641
�� Electromagnetic Fields, Forces, and Motion
Spring 2005
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Spring 2005
6.641 — Electromagnetic Fields, Forces, and Motion
Problem Set 3 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 3.1
A
(x,y,z)
z r
+
d
2
d
2
+q
r
r-
x
-q
Figure 1: Addition of potential contributions from 2 point charges. (Image by MIT OpenCourseWare.)
We can simply add the potential contributions of each point charge:
q
q
Φ=
−
4πε0 r+
4πε0 r−
�
d
r+ = x2 + y 2 + (z − )2
2
�
d
r− = x2 + y 2 + (z + )2
2
⎡
⎤
q ⎣
1
1
⎦
�
Φ=
−�
4π�0
d 2
d
2
2
2
2
2
x + y + (z − )
x + y + (z + )
2
2
B
p = qd. We must make some approximations. As r → ∞, �r+ , �r− , and �r become nearly parallel. Thus,
r+ ≈ r − a = r −
r+ ≈ r(1 −
d
cos θ
2
d
cos θ)
2r
Similarly,
r− ≈ r(1 +
d
cos θ)
2r
1
Problem Set 3
6.641, Spring 2005
z
r+
r
r-
θ
x
a= d cosθ
2
Figure 2: Differences in lengths between �r+ , �r− , and �r (Image by MIT OpenCourseWare.)
By part (a): Φ =
|
q
4πε0
d
cos θ| � 1
2r
�
1
r+
−
1
r−
�
. If |x| � 1, then
1
1+x
so
1
1
1
1
≈
≈
d
r+
r 1 − 2r cos θ
r
�
�
d
1+
cos θ
2r
1
1
1
1
≈
≈
d
r−
r 1 + 2r cos θ
r
�
⇒
1−
d
cos θ
2r
�
1
1
1d
d
−
≈
cos θ = 2
cos θ
r+
r−
rr
r
Φ≈
p cos θ
qd cos θ
=
4πε0 r 2
4πε0 r 2
C
∂Φ
1 ∂Φ
1 ∂Φ ˆ
−
→
E = −�Φ = −
îr −
îθ −
iφ
∂r
r ∂θ
r sin θ ∂φ
p cos θ ∂Φ
p sin θ
∂Φ
=−
;
=−
3
∂r
2πε0 r ∂θ
4πε0 r 2
∂Φ
=0
∂φ
p cos θ ˆ
1 p sin θ
ˆ
−
→
E =
ir +
iθ
2πε0 r 3
r 4πε0 r 2
�
�
p
−
→
2 cos θ îr + sin θ îθ
E =
3
4πε0 r
2
≈1−x
Problem Set 3
6.641, Spring 2005
D
Er
2 cos θ
dr
=
=
= 2 cot θ
rdθ
Eθ
sin θ
1
dr = 2 cot θdθ
r
�
�
1
dr = 2 cot θdθ
r
ln r = 2 ln(sin θ) + k
r = C sin2 θ;
when θ =
π
, r = C = r0
2
Thus,
C = r0 ,
r
= sin2 θ
r0
Figure 3.2b) Used with permission.
Zahn, Markus. In Electromagnetic Field Theory: A Problem Solving Approach, 1987.
3
Problem Set 3
6.641, Spring 2005
Plots of Equipotential and Field Lines
θ
0.75
0.5
0.25
-2
r0 = 1
r0 = .5
1
-1
r0 = 2
2
E Field Lines
-0.25
-0.5
-0.75
r0 = 0.25
Figure 4: Polar plot of r0 sin2 θ for 0 ≤ θ ≤ π and for r0 = 0.25, 0.5, 1, and 2 meters with
100 volt−1 -m−2 (Image by MIT OpenCourseWare.)
=
Φ = .0025 Equipotential
Lines
2
1
4πε0
p
Φ = .01
Φ = 0.04
Φ = .16
-1
-0.5
0.5
1
Φ=0
-1
-2
p cos θ
Figure 5: Polar plot of equipotential lines Φ = 4πε
2 for 0 ≤ θ ≤ π, Φ = 0, ±0.0025, ±0.01, ±0.04, ±0.16,
0r
−1
4πε0
−2
and ±0.64 volts with p = 100 volt -m (Image by MIT OpenCourseWare.)
4
Problem Set 3
6.641, Spring 2005
Figure 6: The superposition of the previous two plots of perpendicular equipotential and field lines (Image
by MIT OpenCourseWare.)
Problem 3.2
A
We can think of the bird as a perfectly conducting small sphere when it lands on the uninsulated wire, it
must become the same potential as the wire. This forces it to acquire a charge. When it flies away, the
charge stays with it because air is a poor conductor.
B, C
For B and C, use the method of images. We can use superposition to get total potential.
�
�
1
q
1
Φ=
−
1
4πε0 [(x − U t)2 + (y − h)2 + z 2 ] 12
[(x − U t)2 + (y + h)2 + z 2 ] 2
where q is the charged bird modeled as a point charge.
D
By boundary condition found using Gauss’ Law
−
→a
−
→b
n̂ · (εa E − εb E ) = σs
at y = 0 ground plane boundary
−
→b
E =0
5
Problem Set 3
6.641, Spring 2005
y
+q
bird at (Ut, h, z)
Φ=0
on xz-plane
x
z
-q image at (Ut, -h, z)
Figure 7: Figure for 3.2 B, C. Method of Images for charged bird taken as a point charge (Image by MIT
OpenCourseWare.)
a
q
n
b
Figure 8: Figure for 3.2 D. Field lines from point charge above a perfectly conducting ground plane (Image
by MIT OpenCourseWare.)
−
→a
Because we can consider ground plane to be a perfect conductor, n̂ · E =
σs
ε0 .
σs
−
→
−
→
(îy ) · ( E (x, y = 0+ , z)) =
implies we only care about y components of E
ε0
Ey (x, y = 0+ , z) =
σs
ε0
�
∂
q
Ey = − Φ =
∂y
4πε0
(1)
(y − h)
3
[(x − U t)2 + (y − h)2 + z 2 ] 2
Evaluate at y = 0 and substitute into (1) above:
�
�
−2h
q
Ey (x, y = 0, z) =
4πε0 [(x − U t)2 + h2 + z 2 ] 23
So σs = ε0 Ey (x, y = 0, z)
σs =
−qh
3
2π [(x − U t)2 + h2 + z 2 ] 2
6
−
(y + h)
3
[(x − U t)2 + (y + h)2 + z 2 ] 2
�
Problem Set 3
6.641, Spring 2005
E
Q=
�
w
0
�
0
l
−qh
3
2π [(x − U t)2 + h2 + z 2 ] 2
dxdz
For w very small, σs does not change significantly from z = 0 to z = w, so integral in z becomes just
multiplication at z = 0.
Q=
�
0
l
−qhw
dx
2π [(x − U t)2 + h2 ]
Let x� = x − U t ⇒ dx� = dx So:
� l−Ut
−qhw
�
Q=
3 dx
�
2
2
2
−Ut 2π [((x ) + h ]
�
�
qw
l − Ut
Ut
�
Q=−
+�
2πh
(l − U t)2 + h2
(U t)2 + h2
q
(1)
(2)
l
Ut
(3)
Figure 9: Representative shape q = Q versus U t (Image by MIT OpenCourseWare.)
F
dQ
−qw
i=
=
dt
2πh
�
−U h2
�
U h2
3 +
3
[(l − U t)2 + h2 ] 2
[(U t)2 + h2 ] 2
�
�
qwR
U h2
−U h2
V = −iR =
+
3
2πh [(l − U t)2 + h2 ] 32
[(U t)2 + h2 ] 2
v
l /U
t
Figure 10: Voltage V versus time across small electrode resistance R (Image by MIT OpenCourseWare.)
7
Problem Set 3
6.641, Spring 2005
P = origin
II
y
a
I
III
x
IV
b
Figure 11: Magnetic field from rectangular line current (Image by MIT OpenCourseWare.)
Problem 3.3
A
1
−
→
H =
4π
1
−
→
H =
4π
→
J(−
r � ) × îr� r �
dv
→
→
|−
r −−
r � |2
�
a
y=
� 2
(I îy ) ×
�
b
2 îx
�
− yiˆy dy
b
x=
� 2
�
�
(I îx ) × −xîx − a2 îy dx
�� �
� �� �
� 12 +
a 2
a 2
2
2
+x
+x
2
2
�� �
� �� �
� 21 +
2
2
b
b
y=− a
x=− 2b
2
+ y2
+ y2
z=0
z=0
2
2
b
y= a
x=− 2
2
��
� �
��
�
�
� 2
−
→
−
→
normalization
for
î
�
|r −r |
r r
�
�
�
�
b
a
x=
y=
� 2 (−I îy ) × − b îx − y îy dy
� 2 (−I îx ) × −xîx + a îy dx
2
2
�� �2
�� �
� �� �2
� 12 +
� �� �
� 12
2
2
b
a
b
a
2
2
2
2
b
+
x
+
x
+
x
+
x
y=− a
x=−
2
2
2
2
2
2
x= 2b
z=0
⎡
y=− a
2
z=0
1 ⎢
⎣2
4π
�
x= 2b
�
⎤
y= a
2
a I(−îz )dx
b I(−îz )dy ⎥
� 32 + 2
� 23 ⎦
�� �2
a 2
2 �� a �2
y=− 2
b
2
2
+
x
+
x
2
2
� a2 ⎤
� 2b
⎡
�
�
�
�
I îz ⎢
ax
by
� ⎥
�
⎥
=−
+
⎣ � � �� �
1 �
1 �
�
�
�
⎦
� b �2 � b �2
4π
2 �
2 �
a 2
a 2
2
2
�
�
+
x
+
y
2
2
2
2
b
a
=
x=− 2b
−2
−2
⎡
⎤
4ab
I îz ⎢
−
→
⎥
�
�4ab
H =−
⎣ �� �
� 12 + �� �2 � � � 12 ⎦
�
�
4π
2
2
2
�
a2 − a2 + 2b
b2 2b + a2
� 2
�
2
−
→ −2I a + b îz
H =
1
πab(a2 + b2 ) 2
8
Problem Set 3
6.641, Spring 2005
1
−
→ −2I(a2 + b2 ) 2
H =
îz
πab
B
Figure 12: Line current in circular coil (Image by MIT OpenCourseWare.)
−
→
I = −I îφ
1
−
→
H =
4π
=
1
4π
�
π
0
�
�
2π
0
�
−I îφ × (−îr )adφ
a2
−îz Iadφ
a2
I
−
→
H = − îz
2a
C
I
a
P
Figure 13: Line current with semi-circular bump (Image by MIT OpenCourseWare.)
1
−
→
H =
4π
�
0
π
(−I îφ ) × (−ˆir )adφ
a2
9
Problem Set 3
6.641, Spring 2005
I
−
→
H = − îz
4a
D
Figure 14: Line current with rectangular bump (Image by MIT OpenCourseWare.)
As in part (c), contributions from segments I and V are zero (see Fig. 14). Segments II, III, and IV are
just like part (a), except integrals in y are from 0 to a and only one integral in x and ( a2 ) → a.
−
→
H =
�
� �2 � 12
−I a2 + 2b
πab
ˆiz
Problem 3.4
A
îr� r
�
π
�
2π
�
π
�
2π
(K0 îφ ) × ˆir� r R2 sin θdφdθ
→
→
|−
r −−
r � |2
0
0
−
→
→
−
→
r −−
r�
r
�
= −
=
−
= −î�r
�
→
→
|→
r −−
r |
|−
r �|
1
−
→
H =
4π
−
→ K0
H =
4π
K0
=−
4π
0
�
0
π
0
�
→
(−
r = 0)
2
�
(îφ ) × (−îr )R
�
sin θdφdθ
2
R�
�
2π
(sin θ)(cos θ cos φîx + cos θ sin φîy − sin θ îz )dφdθ
0
Any term� with an odd power of sin or cos in φ integrates to 0 in φ because integral is over one period.
−
→
−
→
π
0
H = îz 2πK
sin2 θdθ = K40 π îz = H
4π
0
10
Problem Set 3
6.641, Spring 2005
B
This requires us to integrate an infinite number of infinitesimal current shells of the type in (a) from r = R1
to R2 .
K0
−
→
H =
�
� �� �
(J0 dr) π ˆ
−
→ J0 π
iz dr ⇒ H =
(R2 − R1 )îz
4
4
R2
R1
Problem 3.5
Figure 15: Line current above perfectly conducting plane with image current (Image by MIT OpenCourseWare.)
A
−
→ µ0
A =
4π
�
�
∞
−∞
µ0
Idz � îz
�
−
4π
x2 + (y − d)2 + (z − z � )2
�
∞
−∞
�
Idz � îz
�
x2 + (y + d)2 + (z − z � )2
Let ξ ≡ z − z ⇒ dξ = dz . Both integrands are even functions in ξ.
�
� �
µ0 I ∞
1
1
−
→
�
A = îz
dξ
−�
2π 0
x2 + (y − d)2 + ξ 2
x2 + (y + d)2 + ξ 2
��ξ=∞
� ��
x2 + (y − d)2 + ξ 2
ξ
µ0 I
�
ln
+�
= îz
2π
x2 + (y − d)2
x2 + (y − d)2
ξ=0
�
��
��ξ=∞
x2 + (y + d)2 + ξ 2
µ0 I
ξ
�
− îz
ln
+�
2π
x2 + (y + d)2
x2 + (y + d)2
ξ=0
��
�
x2 + (y + d)2
−
→ µ0 I
ln �
îz
A =
2π
x2 + (y − d)2
11
Problem Set 3
6.641, Spring 2005
B
1
−
→
−
→
H =
�× A
µ0
�
�
�
�
I(y − d)
Ix
Ix
I(y + d)
−
îx −
−
îy
=
2π(x2 + (y + d)2 ) 2π(x2 + (y − d)2 )
2π(x2 + (y + d)2 ) 2π(x2 + (y − d)2 )
C
�
−Hx �y=0+ = Kz
−
→
K =−
Id
îz
+ d2 )
π(x2
D
Force comes from the image current
−
→
−
→
F = (Ilîz ) × (µ0 H (x = 0, y = d))
=
µ0 I 2 l
îy
4πd
F
µ0 I 2
=
îy
l
4πd
12