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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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6.013/ESD.013J — Electromagnetics and Applications
Fall 2005
Problem Set 9 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 9.1
A
H=−
B=
I
ı̂φ
2πr
�
0I
− µ2πr
ı̂φ
µI
− 2πr ı̂φ
in free space
in fluid
B
L(h) =
ln ab
[µ0 (l − h) + µ(h + s)]
2π
C
� �
ln ab I 2
1 2 dL(h)
=
[µ − µ0 ]
fx = I
2
dh
4π
D
fx = fg
� �
I 2 ln ab (µ − µ0 )
=⇒ fx = ρm ghπ(b − a ), h =
4π 2 ρm g(b2 − a2 )
2
2
Problem 9.2
A
λ1 = L11 i1 + L12 i2
λ2 = L21 i1 + L22 i2
(L21 = L12 )
The CO-ENERGY:
1
1
L11 i21 + L21 i1 i2 + L22 i22
2
2
1
1
= [L0 + M cos 2θ]i21 + M sin 2θi1 i2 + [L0 − M cos 2θ]i22
2
2
�
Wm
(i1 , i2 , θ) =
so that the torque of electrical origin is
T e = M [sin 2θ(i22 − i21 ) + 2 cos 2θi1 i2 ]
1
Problem Set 9
6.013, Fall 2005
B
i22 − i21 = I 2 [sin2 ωs t − cos2 ωs t] = −I 2 cos 2ωs t
i1 i2 = I 2 sin ωs t cos ωs t =
1
2
I sin 2ωs t
2
Therefore,
T e = M I 2 [− sin 2θ cos 2ωs t + sin 2ωs t cos 2θ]
= M I 2 sin(2ωs t − 2θ).
Substitution of θ = ωm t + γ obtains
T e = −M I 2 sin[2(ωm − ωs )t + 2γ].
For constant torque ωm = ωs :
e
= −M I 2 sin 2γ
Tconstant
C
To determine the equilibrium angle γ0 we write the equation of motion in the steady state
Jθ¨ = T0 − M I 2 sin 2γ0 = 0
1
T0 = M I sin 2γ0 −−−−−−−−−−−−→ γ0 = sin−1
Equilibrium Condition
2
2
�
T0
M I2
�
Equilibrium possible as long as M I 2 > T0 . Now, for the perturbation equation
Jθ¨ = T0 + T � − M I 2 sin(2γ)
J
d2
(ωm t + γ0 + γ � ) = T0 + T � − M I 2 sin(2γ0 + 2γ � )
dt2
d2
�
γ = T � − (2M I 2 cos 2γ0 )γ � .
dt2
The constant terms cancel by the equilibrium condition.
J
D
With T � (t) = I0 u0 (t),
dγ � +
I0
(0 ) =
dt
J
γ � (0+ ) = 0
I0
�
γ (t) =
J
�
2M I 2 cos 2γ0
J
sin
��
2M I 2 cos 2γ0
t
J
�
2
Problem Set 9
6.013, Fall 2005
E
The equilibrium condition is given by
T0 = M I 2 sin 2γ0 =⇒ γ0 =
1
sin−1
2
�
T0
M I2
�
Graphically, we see there are two possible values of γ0 at most. From part D, perturbation motions are stable
if cos 2γ0 > 0, as γ � (t) oscillates with a finite amplitude, and unstable if cos 2γ0 < 0, as then the oscillation
frequency is imaginary and γ � (t) grows to infinite amplitude.
MI 2 sin2γ0
T0
(s)
(u)
π
2
π
γ0
Figure 1: Plot of M I 2 sin 2γ0 versus γ0 showing stable (s) and unstable (u) operating points when a me­
chanical torque T0 is applied. (Image by MIT OpenCourseWare.)
The equilibrium labeled (s) is stable as the resulting torque restores the system to equilibrium with
cos 2γ0 > 0. The point labeled (u) is unstable, as cos 2γ0 is negative.
Problem 9.3
A
When θ = 0, there is no overlap =⇒ A = 0. When θ = π/2, there is complete overlap =⇒ A = πR2 /2
and changes linearly: A(θ) = θR2 . There are 2N − 1 pairs.
C = (2N − 1)θR2
ε0
g
Q = C(θ)V 2
B
1
C(θ)V 2
2
�
∂We� ��
(2N − 1)R2 ε0 V 2
Te =
=
�
∂θ V
2g
We� =
3
Problem Set 9
6.013, Fall 2005
C
J
d2 θ
dθ 1 V 2 (2N − 1)R2 ε0
=
−Kθ
−
B
+
dt2
dt
2
g
V02 (2N − 1)R2 ε0
θ=
2gK
Problem 9.4
A
Ampere’s Law → N i = dHd + xHx
Gauss’ Law → µ0 wbHd = µ0 waHx
Ni
�
=⇒ Hx = �
x + da
b
B
N 2 µ0 aw
�
λ = N µ0 awHx =⇒ L = �
x + da
b
Wm
�
�
1 λ2
1 λ2 da
b +x
=
=
2 L
2 N 2 µ0 aw
C
�
1
λ2
∂Wm ��
=
−
f =−
∂x �λ
2 N 2 µ0 aw
e
D
�
�
V
λ
1 dλ λ da
b +x
I(t) =
+ =
+
R L
R dt
N 2 µ0 aw
E
M
d2 x
1
λ2
= Mg −
2
2
dt
2 N µ0 aw
4