# 6.641 �� Electromagnetic Fields, Forces, and Motion ```MIT OpenCourseWare
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6.641
�� Electromagnetic Fields, Forces, and Motion
Spring 2005
6.641 — Electromagnetic Fields, Forces, and Motion
Spring 2005
Problem Set 8 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 8.1
A
From Fig. 6P.1 we see the geometric relations
r � = r, θ � = θ − Ωt, z � = z, t� = t
There is also a set of back transformations
r = r � , θ = θ � + Ωt, z = z � , t = t�
(1)
B
Using the chain rule for partial derivatives
�
�� � �
�� � �
�� � �
�� �
∂Ψ
∂r
∂Ψ
∂θ
∂Ψ
∂z
∂Ψ
∂t
∂Ψ
+
+
+
=
∂t�
∂r
∂t�
∂θ
∂t�
∂z
∂t�
∂t
∂t�
From (1) we learn that
∂r
∂θ
∂z
∂t
= 0, � = Ω, � = 0, � = 1
∂t�
∂t
∂t
∂t
Hence,
∂Ψ
∂Ψ
∂Ψ
=
+Ω
∂t�
∂t
∂θ
We note that the remaining partial derivatives of Ψ are
∂Ψ ∂Ψ
∂Ψ ∂Ψ
∂Ψ
∂Ψ
=
, � =
, � =
�
∂r
∂r ∂θ
∂θ ∂z
∂z
Problem 8.2
A
In the frame rotating with the cylinder
Ē � (r � ) =
K&macr;
ir
r�
H̄ � = 0, B̄ � = &micro;0 H̄ � = 0
But then since r �
= r, v̄r (r) = rω īθ
Ē = Ē � − v̄r &times; B̄ � = Ē � =
K
īr
r
1
Problem Set 8
V =
�
b
6.641, Spring 2005
Ē &middot; d&macr;l =
a
�
b
a
K
b
dr = K ln( )
r
a
V 1−
→
→
&macr; = V� � 1 −
E
i = Ē � = � b � � ir
b r r
r
ln a
ln a
The surface charge density is then
→
−
&macr; � = ε0 V 1 = σa
σa� = ir &middot; ε0 E
ln ab a
→
−
&macr; � = ε0 V 1 = σb
σb� = − ir &middot; ε0 E
ln ab b
B
J&macr; = J&macr;� + v̄r ρ�
But in this problem we have only surface currents and charges
K̄ = K̄ � + v̄r σ � = v̄r σ �
aωε0 V −
→
→ ωε0 V −
&macr;
� b � iθ =
K(a)
=
iθ
a ln a
ln ab
bωε0 V →
ωε0 V −
→
−
&macr;
K(b)
=−
i =−
iθ
b θ
b ln a
ln ab
C
→
&macr; = − ωε0 V −
H
iz
b
ln a
D
&macr; =H
&macr; � + v̄r &times; D̄� = v̄r &times; D̄�
H
�
�
�
�
ε
V
1
→ −
−
→
0
�
&macr; =rω
H
i
i
&times;
θ
r
ln ab r �
→
&macr; = − ωε0 V −
H
iz
ln ab
This result checks with the calculation of part (c).
Problem 8.3
A
We assume the simple magnetic ﬁeld
�
→
−
− Di i3 0 &lt; x1 &lt; x
&macr;
H=
0
x &lt; x1
λ(x) =
�
&micro;0 W x
i
B̄ &middot; d&macr;a =
D
2
Problem Set 8
6.641, Spring 2005
B
&micro;0 W x
λ(x, i)
=
i
D
Since the system is linear
L(x) =
W � (i, x) =
1 &micro;0 W x 2
1
L(x)i2 =
i
2
2 D
C
fe =
�
∂Wm
1 &micro;0 W 2
=
i
∂x
2 D
(2)
D
The mechanical equation is
M
d2 x
dx
1 &micro;0 W 2
+B
=
i
dt2
dt
2 D
(3)
The electrical circuit equation is
�
�
d &micro;0 W x
dλ
=
i = V0
dt
dt
D
(4)
E
From (3) we learn that
dx
&micro;0 W 2
=
i = constant
dt
2BD
while from (4) with i constant, we learn that
&micro;0 W i dx
= V0
D dt
Solving these two simultaneously
�
�1
DV02 3
dx
=
dt
2&micro;0 W B
F
From (2)
i=
�
2BD dx
=
&micro;0 W dt
�
D
&micro;0 W
� 32
1
1
(2B) 3 V03
G
As in part A,
�
−
→
− Di i3 0 &lt; x1 &lt; x
H̄ =
0
x &lt; x1
3
Problem Set 8
6.641, Spring 2005
H
The surface current K̄ is
K̄ = −
i(t)
ī2
D
The force on the short is
� &macr;
�
&macr; �
&macr; &times; &micro;0 H1 + &micro;0 H2
F&macr; = J&macr; &times; B̄dv = DW K
2
&micro;0 W 2 −
→
=
i (t) i1
2
(5)
(6)
I
∂E2 −
&micro;0 di −
∂ B̄
→
→
i3 = −
=
i3
∂x1
∂t
D dt
�
�
→
&macr; = &micro;0 x di + C −
i2
E
D dt
�
�
&micro;0 x di V (t) −
→
i2
=
−
D dt
W
� &times; Ē =
J
Choosing a contour with the right leg in the moving short, the left leg ﬁxed at x1 = 0
�
�
d
→
−
→� −
E &middot; dl = −
B̄ &middot; dā
dt S
C
Since E � = 0 in the short and we are only considering quasistatic ﬁelds
�
dx
∂H0
→
−
→� −
+ W &micro;0 H0
E &middot; d l = V (t) = W x&micro;0
∂t
dt
�
�
d &micro;0 W x
=
i(t)
dt
D
(7)
K
n̄ &times; (Ē b ) = Vn B̄ b
Here
&micro;0 i −
dx b
→
−
→
, B̄ = −
i3
n̄ = i1 , Vn =
dt
D
�
�
�
�
&micro;0 x di V (t) −
dx &micro;0 −
→
→
Ēb =
−
i2 = −
i i2
D dt
W
dt D
�
�
d &micro;0 xW i
&micro;0 xW di &micro;0 iW dx
+
=
V (t) =
D dt
D dt
dt
D
(8)
L
Equations (6) and (2) are identical. Equations (8), (7), and (4) are identical if V (t) = V0 . Since we used (2)
and (4) to solve the ﬁrst part we would get the same answer using (6) and (7) in the second part.
4
Problem Set 8
6.641, Spring 2005
M
Since
di
dt
=0
Ē2 (x) = −
V (t) −
V0 −
→
→
iy = − iy
w
W
Problem 8.4
A
The electric ﬁeld in the moving laminations is
−
→
−
→�
−
→
J
i −
J
→
=
=
iz
E� =
σ
σ
σA
The electric ﬁeld in the stationary frame is
�
�
i
−
→ −
→�
−
→
E = E − V̄ &times; B̄ =
+ rωBy iz
σA
&micro;0 N i
By = −
S
�
�
2D &micro;0 2DrωN
V =
i
−
σA
S
Now we have the V − i characteristic of the device. The device is in series with an inductance and a load
resistor Rt = RL + Rint .
�
�
2D &micro;0 2DrN
−
ω i+
=0
Rt +
σA
S
S
dt
B
Let
R1 = Rt +
i = I0 e−
Pd =
If
R1
L
2D 2D&micro;0 rN ω
−
,L =
σA
S
S
t
I 2 � R1 �2
i2
= 0 e− L t
RL
RL
R1 = Rt +
2D 2D&micro;0 rN ω
−
&lt;0
σA
S
the power delivered is unbounded as t → ∞.
C
As the current becomes large, the electrical nonlinearity of the magnetic circuit will limit the exponential
growth and determine a level of stable steady state operation (see Fig. 6.4.12).
5
Problem Set 8
6.641, Spring 2005
Problem 8.5
A
The armature circuit equation is
vA = Ra ia + GIf ω
(9)
The equation of motion is
J
dω
= GIf ia
dt
Which may be integrated to yield
�
G t
ω(t) =
ia (t)
J −∞
(10)
Combining (10) with (9)
(GIf )2
vA = Ra ia +
Jr
�
t
ia (t)
−∞
We recognize that
C=
Jr
(GIf )2
C=
Jr
(0.5)
=
2
(GIf )
(1.5)2 (1)
B
Problem 8.6
1
dif
+ if (Rr + Rf − Gω) +
(Lr + Lf )
dt
C
�
if dt = 0
d2 if
(Rr + Rf − Gω) dif
1
+
+
if = 0
dt2
Lr + Lf
dt
(Lr + Lf )C
if = Iest
(Rr + Rf − Gω)
1
s+
=0
(Lr + Lf )
(Lr + Lf )C
��
� 12
�2
1
Rr + Rf − Gω
Rr + Rf − Gω
−
s=−
&plusmn;
2(Lr + Lf )
2(Lr + Lf )
(Lr + Lf )C
s2 +
A
Self excited if −Gω + Rr + Rf &lt; 0 ⇒ ω &gt;
Rr +Rf
G
6
Problem Set 8
6.641, Spring 2005
B
dc self-excitation
�
Rr + Rf − Gω
2(Lr + Lf )
�2
−
1
4(Lr + Lf )
&gt; 0, C &gt;
(Lr + Lf )C
[Rr + Rf − Gω]2
�2
−
1
4(Lr + Lf )
&lt;0⇒C&lt;
(Lr + Lf )C
[Rr + Rf − Gω]2
ac self-excitation
�
Rr + Rf − Gω
2(Lr + Lf )
C
Frequency ω0 =
�
1
(Lr +Lf )C
−
�
Rr +Rf −Gω
2(Lr +Lf )
�2 � 12
7
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