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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 4 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 4.1
A
y
x
z
Figure 1: Cartesian coordinate axes (Image by MIT OpenCourseWare.)
For region I, we use q and q �
�
�
q
q�
Φ=
1 +
1
4πε1 (x2 + (y − d)2 + z 2 ) 2
4πε1 (x2 + (y + d)2 + z 2 ) 2
EI = −�Φ
1
=
4πε1
�
q(xîx + (y − d)îy + z îz )
3
(x2 + (y − d)2 + z 2 ) 2
+
q � (xîx + (y + d)îy + z îz )
3
(x2 + (y + d)2 + z 2 ) 2
For region II, use q ��
Φ=
q ��
1
4πε2 (x2 + (y − d)2 + z 2 ) 2
EII = −�Φ
q �� (xˆix + (y − d)ˆiy + zˆiz )
=
3
4πε2 (x2 + (y − d)2 + z 2 ) 2
B
Tangential E components are equal:
1
�
Problem Set 4
6.641, Spring 2005
1
⎤
n̂ × [EI − EII ] = 0
⎥
ExI = ExII
⎥
⎥ at y = 0
EzI = EzII
⎥
⎦
Since there is no surface charge, i.e. σs = 0.
2
⎤
n̂ · (εI EI − εII EII ) = 0
⎦ at y = 0
εI EIy = εII EIIy
From 1,
1
4πεII
From 2,
1
ε1
4πεI
�
�
qx+q � x
3
(x2 +d2 +z 2 ) 2
−qd + q � d
(x2 + d2 + z 2 )
3
2
�
=
�
= εII
q �� x
3
4πεII (x2 +d2 +z 2 ) 2
. Therefore,
q �� (−d)
3
4πεII (x2 + d2 + z 2 ) 2
−q + q � = −q ��
Therefore,
q + q ��
q ��
εI ��
=
⇒q=
q − q�
εI
εII
εII
q �� = q − q �
Therefore,
εI
(q − q � ) − q �
εII
�
�
�
�
εI + εII
εII − εI
= −q �
q
εII
εII
�
�
εI + εII
q = −q �
εII − εI
q=
Or
εI ��
q − q�
εII
εI ��
=
q − q + q ��
εII
�
�
εI + εII
��
2q = q
εII
εI + εII ��
q=
q
2εII
q=
2
q+q �
εI
=
q ��
εII .
Problem Set 4
6.641, Spring 2005
C
¯
f¯ = qE
�
=q
q � (0îx + 2dîy + 0îz )
3
4πεI (02 + (2d)2 + 02 ) 2
�
q � q îy
2dq � q îy
=
4πεI · 8d3
4πεI · 4d2
� �
��
−εI
q −q εεII
2
I +εII
ˆiy = −q (εII − εI ) ˆiy
=
4πεI · 4d2
16πεI d2 (εI + εII )
2
q (εI − εII ) ˆ
=
iy
16πεI (εI + εII )d2
=
Problem 4.2
This is a charge relaxation problem, so we use, as shown in class, the equations (done in lecture 12)
∂ρf
−
→
�· Jf +
=0
∂t
−
→
We substitute in � · E =
ρf
ε
−
→
−
→
and J f = σ E to get
∂ρf
σ
−
→ ∂ρf
σ� · E +
=0⇒
+ ρf = 0
∂t
∂t
ε
So
t
ε
→
ρf = ρ(−
r , t = 0)e− τe ; τe =
σ
Thus
ρf (r, t) =
�
ρ0 r − τte
a e
0 < r < a0
r > a0
0
Notice: ρf (r, t) is 0 for r > a0 . Nonetheless, there is still conduction and displacement current for a0 < r < a1 .
� −
→ → �
By Gauss, S ε E · d−
a = V ρdV . Choosing S as a cylinder with radius r
�
−
→ →
ε E · d−
a = εEr r2πL
S
−
→ →
where L is the length of the cylinder. Note that E · d−
a = 0 on cylinder ends.
Now for RHS of Gauss
�
� L � r � 2π
� r
ρf dV =
ρf (r � , t)r � dφdr � dz = 2πL
ρf (r � , t)r � dr �
V
r < a0 :
�
0
ρf dV = 2πL
V
= 2πL
0
�
0
0
ρ0 (r � )2 − τt �
e e dr
a0
ρ0 r 3 − τt
e e
a0 3
3
Problem Set 4
6.641, Spring 2005
a 0 < r < a1 :
�
ρf dV = 2πL
v
�
a0
0
= 2πL
ρ0 (r � )2 − τt �
e e dr
a0
a30
ρ0 − τt
2πL
e e
=
ρ0 a20 e−t/τe
a0
3
3
r > a1 :
�
ρf dV = 2πL
V
= 2πL
�
0
a0
ρ0 (r � )2 �
dr
a0
ρ0 a30 −t/τe
2πL
e
=
ρ0 a20 e−t/τe
a0 3
3
So:
⎧ ρ0 r 2 − t
τ
⎪
⎨ 3a0 ε2 e te îr
−
→
ρ0 a0 − τ
e î
E =
r
3rε e
⎪
⎩ ρ0 a20
3rε0 îr
r < a0
a 0 < r < a1
r > a1
−
σsf = ε0 Er (r = a+
1 ) − εEr (r = a1 )
σsf =
t
ρ0 a20
(1 − e− τe )
3a1
Problem 4.3
A
−
→
−
→
−
→
There are no surface currents, so we have continuity of normal B and tangential H . Also, if µ → ∞, H = 0
−
→
inside, but B may still be nonzero. Equivalent image problem:
µ0
y
x
µ
∞
Figure 2: Magnetic field lines due to a line current above an infinitely magnetically permeable region (Image
by MIT OpenCourseWare.)
Boundary conditions: Hx = Hz = 0 at y = 0
4
Problem Set 4
6.641, Spring 2005
B
Assume line current at origin:
�
I
→
−
→ −
H · d l = I ⇒ Hφ =
2πr
∂Az
Iµ0
−
→ −
→
�× A = B ⇒−
=
∂r
2πr
0
Suggesting: Az = − Iµ
2π ln(r) + constant. Assume line current at y = d:
Az = −
Iµ0 �
ln (y − d)2 + x2
2π
Now 2 line currents; one at y = d and one at y = −d.
Az = −
�
��
��
Iµ0 � �� 2
x + (y − d)2 + ln
x2 + (y + d)2
ln
2π
C
1
−
→ −
→
�× A = H
µ0
�
�
1
∂Az
∂Az
îx
− îy
=
µ0
∂y
∂x
�
�
I (y − d)îx − xîy
(y + d)îx − xîy
−
→
H =−
+ 2
2π x2 + (y − d)2
x + (y + d)2
D
Field line equation:
x
Hy
dy
x2 +(y−d)2 +
=
= − y−d
dx
Hx
x2 +(y−d)2 +
x
x2 +(y+d)2
y+d
x2 +(y+d)2
=
z
− ∂A
∂x
∂Az
∂y
�
��
�
∂Az
∂Az
dy = −
dx ⇒ Az = constant ⇒ x2 (y − d)2 x2 + (y + d)2 = constant
∂y
∂x
E
−
→
F
=
unit length
−
→
I
����
current at
y=d
×
−
→
B
����
−
→
B field
caused by image current
alone at x = 0, y = −d
��
�
��
−
→
F
−µ0 I
(y + d)îx − xîy �
�
= (I îz ) ×
�
unit length
2π
x2 + (y + d)2 �
x=0,y=d
��
�
�
2d
−µ0 I
îx
= (I îz ) ×
2π
(2d)2
−
→
µ0 I 2
F
=−
îy
unit length
4πd
5
Problem Set 4
6.641, Spring 2005
Problem 4.4
A
−
→
� · J = 0; by symmetry we just have x component of J
∂Jx
−
→
= 0 ⇒ J = J0 îx , J0 is constant
∂x
x
σx = σ0 e− s
Ex · σx = Jx ;
Jx
J0
J0 xs
=
e
x =
σx
σ0
σ0 e− s
� s
�
J0 s xs
J0 s xs ��s
J0 s
V0 =
Ex dx =
e dx =
e 0=
(e − 1)
σ0 0
σ0
σ0
0
Ex =
I0 = J0 · ld
R=
J0 s
σ0 (e
V0
=
I0
− 1)
J0 ld
=
s
(e − 1)
σ0 ld
B
� · (εE) = ρ ⇒ ρ = ε
ρ=
∂Ex
∂x
J0 ε x
es
σ0 s
At x = 0:
�
εJ0
σs = εEx �x=0 =
σ0
At x = s:
C
�
εJ0
σs = −εEx �x=s = −
e
σ0
qV = ld
�
s
ρdx =
0
ldJ0 ε
(e − 1)
σ0
Total surface charge
qS = (σs |x=s + σs |x=0 )ld = −
ldJ0 ε
(e − 1) = −qV
σ0
qS + qV = 0
6
Problem Set 4
6.641, Spring 2005
Problem 4.5
A
As no volume charge in the dielectric
� · D̄ = 0
as symmetry we just have r component
A
ε1 r
1 ∂(r 2 Dr )
= 0 ⇒ Dr = 2 , ε(r) =
r2
∂r
r
a
Dr
Aa
Aa
= 2
=
ε(r)
r ε1 r
ε1 r 3
�b
�
�
� b
� b
Aa
−Aa 1 ��
Aa 1
1
v=
Er dr =
dr =
=
− 2
3
2ε1 r 2 �a 2ε1 a2
b
a
a ε1 r
Dr = εEr ⇒ Er =
A=
2ε1 v
a
1
a2
1
−
1
b2
Er =
,
Ē = −�Φ ⇒ Er = −
Φ=
�
−Er dr = +
1
a2
∂Φ
=
∂r
v
−
1
b2
1
a2
2v
1
1 r3
− b2
1
a2
2v
1
1 r3
− b2
1
r2
B
σs |r=a = ε(r)Er |r=a =
2ε1 v 1
− b12 a3
1
a2
σs |r=b = −ε(r)Er |r=b =
−2ε1 v ab 1
−2ε1 v 1
1 b3 = 1
1 ab2
1
a2 − b2
a2 − b2
C
q = 4πa2 σs |r=a = −4πb2 σs |r=b = �
C=
q
8πε1
�
= �1
1
v
a2 − b2 a
8πε1 v
�
− b12 a
1
a2
7