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6.641
�� Electromagnetic Fields, Forces, and Motion
Spring 2005
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6.641 — Electromagnetic Fields, Forces, and Motion
Spring 2005
Problem Set 1 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 1.1
A
−
→
→ → −
−
→
F = q( E + −
v × B ) ← Lorentz Force Law
−
→
In the steady state F = 0, so
−
→
→
−
−
→
→
−
→
→
q E = −q −
v × B ⇒ E = −−
v ×B
�
pos. charge carriers
vyˆiy
→
−
v =
−vyˆiy neg. charge carriers
−
→
B = B0ˆiz
so
−
→
E =
�
−vy B0ˆix
vy B0ˆix
pos. charge carriers
neg. charge carriers
B
vH = Φ(x = d) − Φ(x = 0) = −
�
0
vH =
�
d
Ex dx =
�
0
Ex dx
d
pos. charges
vy B0 d
−vy B0 d neg. charges
C
As seen in part (b), positive and negative charge carriers give opposite polarity voltages, so answer is “yes.”
Problem 1.2
By problem:
�
ρb
ρ=
ρa
r<b
b<r<a
→
−
Also, no σs at r = b, but nonzero σs at r = a such that E = 0 for r > a.
1
Problem Set 1
6.641, Spring 2005
v
E
-
+
+
-
+
-
Force caused
by E
+
+
+
B
+
+
+
+
+
-
-
v
E
Force
caused
by B
­
Force caused
Force caused by E
by B
(which is caused by
charge on electrodes)
Figure 1: Figure for 1C. Opposite polarity voltages between holes and electrons (Image by MIT OpenCourseWare.)
A
By Gauss’s Law
�
�
→
− −
ε0 E · d→
a =
SR
SR = sphere with radius r
ρdV
VR
→
−
−
→
As shown in class, symmetry ensures E has only radial component: E = Er îr .
LHS of (1):
�
� 2π � π
− →
→
ε0 E · d−
a =
�0 (Er îr ) ·
r 2 sin θdθdφîr
�
��
�
SR
0
0
−
d→
a in spherical coordinates
2
�4πr
�� �
=
Er ε0
surface area of sphere of radius r
RHS of (1):
For r < b:
�
�
ρdV =
0
VR
=
�
r
4 3
πr
�3 �� �
2π
0
�
π
0
ρb
ρb r 2 sin θdθdφdr
�
��
�
dV : diff vol. element
vol of sphere
For r > b and r < a: (b < r < a):
�
� b � 2π � π
ρdV =
ρb r 2 sin θdθdφdr
VR
+
�
0
r
b
�
0
2π
0
�
0
π
ρa r 2 sin θdθdφdr
0
4πρa (r 3 − b3 )
4πρb b3
+
=
3
3
2
(1)
Problem Set 1
6.641, Spring 2005
Equating LHS and RHS:
� 3
4πr
2
3
ρb
4πr Er ε0 = 4πρ
3
bb
3
Er =
�
rρb
3ε0
b3 (ρb −ρa )
3ε0 r 2
r<b
+
4πρa (r 3 −b3 )
3
b<r<a
r<b
+
ρa r
3ε0
b<r<a
B
Again: n
ˆ · (ε0 E a − ε0 E b ) = σs
→
−
E (r = a+ ) = 0
� 3
�
→
−
b (ρb − ρa ) ρa a
−
E (r = a ) = +
+
îr
3ε0
3ε0 a2
→
−
σs = îr · (−ε0 E (r = a− ))
by part (a)
so
�
σs = −
b3 (ρb − ρa ) ρa a
+
3
3a2
�
C
4 3
Qσ (r = a) = σs 4πa2
πb ρb
3
4
b<r<a
Qa = π(a3 − b3 )ρa
3
QT = Qb + Qa + Qσ = 0
Qb =
r<b
Problem 1.3
a
We are told current going in +z direction inside cylinder r < b. Current going through cylinder
�
→ −
−
J · d→
a
= Itotal =
S
=
�
b
0
�
0
2π
(J0ˆiz ) · rdφdriˆz )
� �� � � �� �
→
−
J
−
a
d→
= J0 πb2
→
−
Total current in sheet
|K | =
length of sheet (i.e. circumference of circle of radius a)
→
−
→
−
Amps
Thus, K ’s units are Amps
m , whereas J ’s units are m2
→
−
J0 πb2
J0 b2
|K | =
=
2πa
2a
−
→
J0 b2
K =−
îz
2a
3
Problem Set 1
6.641, Spring 2005
z
a
b
3
Figure 2: Figure for Problem 1.3 Part A. A cylinder with volume current going in the +z direction for r < b.
(Image by MIT OpenCourseWare)
B
I
− −
→
H · d→
s =
C
Z
− −
→
J · d→
a +
S
d
dt
|
Z
→ →
−
ε0 E · d−
a
S
{z
}
(2)
→
−
no E field, so this term is 0
Choose C as a circle and S as the minimum surface that circle bounds. Now, solve LHS of Ampere’s Law
z
a
b
c
s
Figure 3: Choice of contour C and surface S (Image by MIT OpenCourseWare).
(2)
I
− −
→
H · d→
s =
C
Z
0
2π
Hφˆiφ · rdφîφ = 2πrHφ
| {z } | {z }
→
−
H
−
d→
s
H
→ →
−
We assumed Hz = Hr = 0. This follows from the symmetry of the problem. Hr = 0 because S µ0 H ·d−
a = 0.
In particular, choose S as shown in Figure 3. Hz is more difficult to see. It is discussed in Haus & Melcher.
The basic idea is to use the contour, C, to show that if Hz =
6 0 it would have to be nonzero even at ∞,
which is not possible without sources at ∞. Now for RHS of Ampere:
r<b:
4
Problem Set 1
6.641, Spring 2005
z
b
c
Figure 4: Illustration of the contour C (Image by MIT OpenCourseWare).
b
s
Figure 5: Diagram showing surface S (Image by MIT OpenCourseWare).
�
− −
→
J · d→
a =
�
2π
0
S
�
r
�
b
0
→
−
J
= J0 r 2 π
a>r>b:
�
�
→ −
−
J · d→
a =
2π
0
S
�
�
� � � �
J0 · îz · r dr dφîz
��
�
� �� � �
−
d→
a
�
�
(J0 îz ) · (r dr dφîz +
0
2
= J0 b π
�
�0
2π
�
b
�
J0 r
2 îφ
J0 b2
2r îφ
�
r<b
a>r>b
C
From text, Ampere’s continuity condition:
−
→
−
→
−
→
n̂ × (H a − H b ) = K
5
�
(0 · îz ) · (r dr dφîz )
��
�
0
Equating LHS and RHS:
�
J0 r 2 π r < b
2πrHφ =
J0 b2 π a > r > b
−
→
H =
r
Problem Set 1
6.641, Spring 2005
z
a
b
region
(r>a)
region
(a>r>b)
region
(b>r)
Figure 6: Ampere’s continuity condition for Problem 1.3 Part C (Image by MIT OpenCourseWare)
Hφ (r = a− ) = −Kz
J0 b2
J0 b2
= −Kz =
2a
2a
Problem 1.4
A
We can simply add the fields of the two point charges. Start with the field of a point charge q at origin and
let SR be sphere of radius R centered at the origin. By Gauss:
�
�
→ →
−
ε0 E · d−
a = ρdV
SR
→
In this case ρ = δ(−
r )q, so RHS is
�
� � �
→
ρdV =
δ(−
r )qdxdydz = q
LHS is
�
−
→ →
ε0 Er · d−
a = (ε0
SR
Er
����
)(surface area of Sr )
symmetry again
= 4πr 2 ε0 Er
Equate LHS and RHS
4πr 2 ε0 Er = q
−
→
E =
q
îr
4πr 2 ε0
Convert to cartesian: Any point is given by
−
→
r = x(r, θ, φ)îx + y(r, θ, φ)îy + z(r, θ, φ)îz
By spherical coordinates
x = r sin θ cos φ
y = r sin θ sin φ
6
Problem Set 1
6.641, Spring 2005
z = r cos θ
→
−
r = r sin θ cos φîx + r sin θ sin φîy + r cos θ îz
îr � line formed by varying r and fixing φ and θ
r̄ = ri¯r
Thus,
so,
i¯r = sin θ cos φiˆx + sin θ sin φîy + cos θ îz
x
y
z
iˆy + �
iˆz
=�
iˆx + �
2
2
2
2
2
2
2
x + y 2 + z 2
x +y +z
x +y +z
−
→
E =
q
îr
4πε0 (x2 + y 2 + z 2 )
→
−
→
−
→
−
E total = E 1 + E 2
→
−
→
−
→
−
−
→
E 1 is just E with y → y − d2 . E 2 is just E with y → y + d2 . Problem has y = 0
(i)
⎡
⎤ �
d
−
→
→
−
x
q
z
2
ˆiy ⎦ ·
ˆix + �
ˆiz − �
E total = E 1 = ⎣ �
2
2
2
2
d
d
d
4πε0 (x +
x2 + 4 + z 2
x2 + 4 + z 2
x2 + 4 + z 2
�
�
q
d
xîx − îy + zˆiz
=
3
d2
2
2
2
2
4πε0 (x + 4 + z )
(ii)
→
−
−
→
−
→
E total = E 1 + E 2
xîx + z îz
=q
� �2
3
2πε0 (x2 + d2 + z 2 ) 2
B
(iii)
→
−
−
→
−
→
E total = E 1 + E 2
−dq îy
=
�
�3
2
4πε0 x2 + d4 + z 2 2
−
→
−
→
F = q1 E
−
→
E doesn’t include field of q
(i)
−
→
F = 0, by Newton’s third law a body cannot exert a net force on itself.
(ii)
−
→
−
→
→
−
d
F = q1 E = q E 2 (x = 0, y = , z = 0)
2
q 2 i¯y
q 2 i¯y
=
=
4πε0 (d2 )
4πε0 d2
(iii)
−
→
q 2 i¯y
F =−
4πε0 d2
7
d2
4
+ z2 )
�