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6.641 Electromagnetic Fields, Forces, and Motion
Spring 2009
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6.641 — Electromagnetic Fields, Forces, and Motion
Spring 2009
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 that form an electric dipole. (Image by
MIT OpenCourseWare.)
We can simply add the potential contributions of each point charge:
Φ=
q
q
−
4πε0 r+
4πε0 r−
�
d
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 − 2 )
x + y + (z + 2 )
r+ =
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
1
Problem Set 3
6.641, Spring 2009
z
r+
r
r-
θ
x
a= d cosθ
2
Figure 2: Differences in lengths between �r+ , �r− , and �r (Image by MIT OpenCourseWare.)
Similarly,
r− ≈ r(1 +
By part (a): Φ =
|
d
cos θ)
2r
�
q
4πε0
1
r+
−
1
r−
�
. If |x| � 1, then
1
1+x
d
cos θ| � 1
2r
so
1
1
1
1
≈
≈
d
r+
r
r 1 − 2r
cos θ
�
1
1
1
1
≈
≈
d
r−
r
r 1 + 2r
cos θ
�
⇒
d
1+
cos θ
2r
d
1−
cos θ
2r
�
�
1
1
1d
d
−
≈
cos θ = 2 cos θ
r
r+
r−
rr
Φ≈
qd cos θ
p cos θ
=
, p = qd dipole moment
2
4πε0 r
4πε0 r2
C
∂Φ
1 ∂Φ ˆ
1 ∂Φ ˆ
→
−
E = −�Φ = − ˆir −
iθ −
iφ
∂r
r ∂θ
r sin θ ∂φ
∂Φ
p cos θ ∂Φ
p sin θ
=−
;
=−
∂r
2πε0 r3 ∂θ
4πε0 r2
∂Φ
=0
∂φ
p cos θ
1 p sin θ
→
−
E =
îr +
îθ
2πε0 r3
r 4πε0 r2
�
�
p
→
−
E =
2
cos
θ
î
+
sin
θ
î
r
θ
4πε0 r3
2
≈1−x
Problem Set 3
6.641, Spring 2009
D
dr
Er
2 cos θ
=
=
= 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: The equi-potential (dashed) and field lines (solid) for a point electric dipole calibrated for 4πε0 /p =
100. The equi-potential lines and the electric field lines are perpendicular to each other.
3
Problem Set 3
6.641, Spring 2009
Plots of Equipotential and Field Lines
θ
0.75
0.5
0.25
r0 = 1
r0 = .5
-2
E Field Lines
1
-1
r0 = 2
2
-0.25
-0.5
-0.75
r0 = 0.25
Figure 4: Polar plot of dipole electric field lines r0 sin2 θ for 0 ≤ θ ≤ π and for r0 = 0.25, 0.5, 1, and 2 meters
−1
0
-m−2 (Image by MIT OpenCourseWare.)
with 4πε
p = 100 volt
Φ = .0025 Equipotential
Lines
2
1
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 2009
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 the total potential for a charge q
at height h moving in the x direction at velocity U .
�
�
q
1
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
→b
−
n̂ · (εa E − εb E ) = σs at the y = 0 ground plane boundary where E = 0.
5
Problem Set 3
6.641, Spring 2009
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 flying over a ground
plane (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 the ground plane to be a perfect conductor, n̂ · E =
σs
ε0 .
σs
→
→
−
−
(îy ) · ( E (x, y = 0+ , z)) =
implies we only care about the y component 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 ] 32
6
−
�
(y + h)
3
[(x − U t)2 + (y + h)2 + z 2 ] 2
Problem Set 3
6.641, Spring 2009
So σs = ε0 Ey (x, y = 0, z)
σs =
−qh
3
2π [(x − U t)2 + h2 + z 2 ] 2
E
�
w
�
l
−qh
Q=
3
0
0
dxdz
2π [(x − U t)2 + h2 + z 2 ] 2
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−U t
−qhw
�
Q=
3 dx
−U t 2π [((x� )2 + h2 ] 2
⎡
⎤
⎢
⎥
⎥
qw ⎢
l − Ut
Ut
⎢
⎥
�
Q=−
+�
⎢
2
2
2
2
2πh ⎣ (l − U t) + h
(U t) + h ⎥
⎦
�
��
� �
��
�
(2)
(1)
q
(1)
(2)
l
Ut
(3)
Figure 9: Representative shape of total charge Q on the electrode versus U t. The dashed curves are the first
(2) and second (1) terms for Q and (3) is the sum (1) + (2). (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
7
Problem Set 3
6.641, Spring 2009
v
l /U
t
Figure 10: Voltage V versus time across small electrode resistance R (Image by MIT OpenCourseWare.)
Problem 3.3
A
1
→
−
H =
4π
−
J(→
r � ) × îr� r �
dv
−
−
|→
r − →
r � |2
�
y= a
2
1
−
→
H =
4π
(I îy ) ×
�
�
b
2 îx
x= 2b
�
− yiˆy dy
�
�
�
(I îx ) × −xîx − a2 îy dx
�� �
� �� �
� 12 +
a 2
a 2
2
2
+
x
+
x
2
2
�� �
� �� �
� 12 +
2
2
b
b
x=− 2b
+ y2
+ y2
z=0
z=0
2
2
x=− 2b
y= a
2
�
��
� �
��
�
�
−
→
−
→
2
normalization for îr� r
|r −r |
�
�
�
�
y= a
x= 2b
� 2 (−I îy ) × − b îx − y îy dy
� (−I îx ) × −xîx + a îy dx
2
2
� � �2
� �� �2
� 12 +
�� �
� �� �
� 12
b
b
a 2
a 2
2
2
2
2
b
+
x
+
x
+
x
+
x
y=− a
x=−
2
2
2
2
2
y=− a
2
2
x= 2b
z=0
y=− a
2
z=0
⎡
x= 2b
�
⎤
y= a
2
a I(−îz )dx
b I(−îz )dy ⎥
3 + 2
�
�
�
� 32 ⎦
�
�
a
2 a 2
2
y=− 2 2 � b �2
2
2
+
x
+
x
2
2
� 2b
� a2 ⎤
⎡
�
�
�
� ⎥
I îz ⎢
by
ax
� ⎥
�
=−
+
⎣ � � �� �
1 �
1 �
�
�
�
⎦
�
�
�
�
�
�
4π
2
2
a 2
a 2
b 2
b 2
+ x2 �
+ y2 �
2
2
2
2
=
1 ⎢
⎣2
4π
�
x=− 2b
− 2b
−a
2
P = origin
II
y
a
I
III
x
IV
b
Figure 11: Magnetic field at centerpoint of rectangular line current (Image by MIT OpenCourseWare.)
8
Problem Set 3
6.641, Spring 2009
⎡
⎤
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
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
�
=
2π
�
�
−I îφ × (−îr )adφ
a2
0
2π
0
−îz Iadφ
a2
I
−
→
H = − ˆiz
2a
9
Problem Set 3
6.641, Spring 2009
I
a
P
Figure 13: Line current with semi-circular bump (Image by MIT OpenCourseWare.)
C
Contributions from left and right straight line segments are each zero because J(r� ) × ir� r = Iix × ir� r =
Iix × (±ix ) = 0
1
→
−
H =
4π
�
π
0
(−I îφ ) × (−îr )adφ
a2
(semi-circular bump)
I
−
→
H = − ˆiz
4a
D
b
III
y
P
a
IV
II
x
V
I
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 � 1
2
−I a2 + 2b
πab
ˆiz
10
Problem Set 3
6.641, Spring 2009
Problem 3.4
A
îr� r
�
π
�
2π
�
π
�
2π
�
π
(K0 îφ ) × îr� 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
�
−
(→
r = 0)
2
�
(iˆφ ) × (−ˆir )R
�
sin θdφdθ , iˆφ × iˆr = iˆθ = cos θ cos φiˆx + cos θ sin φiˆy − sin θiˆz
2
R�
�
2π
(sin θ)(cos θ cos φîx + cos θ sin φîy − sin θîz )dφdθ
0
0
Any term with an odd power of sin or cos in φ integrates to 0 in φ because integral is over one period.
− ˆ 2πK0 � π 2
→
→
−
H = iz 4π 0 sin θdθ = K40 π ˆiz = H
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 =
�
R2
R1
� �� �
(J0 dr) π
−
→ J0 π
îz dr ⇒ H =
(R2 − R1 )ˆiz
4
4
Problem 3.5
Figure 15: Line current above perfectly conducting plane at y = d with image current at y = −d (Image by
MIT OpenCourseWare.)
11
Problem Set 3
6.641, Spring 2009
A
µ0
−
→
A =
4π
�
∞
Idz � îz
−∞
µ0
�
−
2
2
�
2
4π
x + (y − d) + (z − z )
∞
�
Idz � îz
−∞
�
x2 + (y + d)2 + (z − z � )2
Let ξ ≡ z � − z ⇒ dξ = dz � . Both integrands are even functions in ξ.
→ ˆ µ0 I
−
A = iz
2π
∞
�
�
1
1
�
dξ
−�
x2 + (y − d)2 + ξ 2
x2 + (y + d)2 + ξ 2
�
�
��ξ=∞
�
�
µ0 I � �
= iˆz
ln ξ + x2 + (y − d)2 + ξ 2 − ln ξ + x2 + (y + d)2 + ξ 2
2π
ξ=0
0
�
��
�
x2 + (y + d)2 ˆ
→ µ0 I
−
A =
ln �
iz
2π
x2 + (y − d)2
B
�
�
1
1 ∂Az ˆ
∂Az ˆ
→
−
→
−
H =
�× A =
ix −
iy
µ0
µ0
∂y
∂x
�
�
�
�
I(y + d)
I(y − d)
Ix
Ix
−
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
ˆiz
+ d2 )
π(x2
D
Force comes from the image current
→
−
→
−
F = (Ilˆiz ) × (µ0 H (x = 0, y = d))
=
µ0 I 2 l
îy
4πd
F
µ0 I 2
=
îy
l
4πd
12