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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science 6.013 Electromagnetics and Applications
Problem Set #1 SOLUTION
Fall Term 2005 Problem 1.1
b.
T=
s
Q1Q2 Mg
=
where sin θ = ,
2
2l
cos θ 4πε 0 s sin θ
Q⎞
⎛
⎜ Q1 = Q2 = ⎟
2⎠
⎝
Mg4πε 0 ( 2l sin θ ) sin θ
Q1Q2 Q 2
2
= 1 , sin θ tan θ =
=
Q1Q2 cos θ
16πε 0l 2 Mg 64πε 0l 2
Mg
2
c.
i+
d ( −q )
dq
, v
=0⇒i =
dt
dt
iR = R
dq
= − Rq0ω sin (ωt )
dt
Problem 1.2
∫
H ids = ∫ J i
da , H ≈
b.
Ampere’s integral law
c.
NΦ NBSb µ0 N 2 a
=
L= =
=
i
i
i
2
d.
v=L
Cb
Sb
λ
di
dv
, i = C , v ( t = 0 ) = V ⇒ i = I sin (ωt ) , ω =
dt
dt
At t = 0 , v ( t = 0 ) = V = LI ω ⇒ I =
I =V
e.
Ni
2π a
V
C
= V
Lω
L
1
LC
1
1
; Note : LI 2 = CV 2
2
2
C
ω
1
=
sin (ωt
) , f =
L
2π 2π LC
1
1
C
Mv 2 = CV 2 ⇒ v = V
M
2
2
1
2
1
CV = Mgh ⇒ h =
CV 2
2
2Mg
f.
L = 0.1mH , I = 2000 A, v = 707m / s, h = 255m
Problem 1.3
m
qE0t 2
qE0
t 2
d 2
z
, vz 0
t = z0 = 0
=
qE
⇒
z
=
+
v
t
+
z
=
0
0
z0
2m
2m
dt 2
1
m
d 2x
x
= 0 ⇒ x = v0t ⇒ t =
2
v0
dt
z=
qE0x 2
qE0 L2
,
z
x
=
L
=
h
=
(
)
2mv0 2
2mv0 2
Problem 1.4
(
b. The Lorentz force law F =q E + v × B
)
In the steady state F = 0 , so: E = −v × B
⎧⎪ v y iy ; postive charge carries
v =⎨
, B = B0 iz
⎪⎩−v y iy ; negative charge carries
⎧⎪ v y B0 ix
; postive charge carries
E=⎨
⎪⎩−v y B0 ix ; negative charge carries
c.
VH = Φ ( x = d ) − Φ ( x = 0 ) = − ∫ Ex dx = ∫ Ex dx
d
0
0
d
⎧ v y B0
d ; positive charge carriers
VH = ⎨
⎩−v y B0 d ; negative charge carriers
d. As seen in part c, different polarity charge carriers have opposite polarity voltage, so the
answer is an indubitable “Yes!”.
Problem 1.5
a. As the line currents have infinite extent in the z direction the magnetic field has no
dependence on the z coordinate.
The magnetic field of a z-directed line current at the origin is: H =
I
2π r
iφ
Convert cylindrical coordinates to Cartesian coordinates and move the line current to
( 0, d / 2 ) , the magnetic field is
H=
⎛ ⎛
d⎞
⎞
⎜ − ⎜ y − 2 ⎟ ix + xiy ⎟
⎛
⎠
⎠
d ⎞ ⎞⎝ ⎝
⎛
2π ⎜ x 2 + ⎜ y − ⎟ ⎟
⎜
⎝
2 ⎠ ⎟⎠
⎝
I
2
Moving the line current to
H=
( 0, −d / 2 )
gives the magnetic field as
⎛ ⎛
d⎞
⎞
⎜ − ⎜ y + 2 ⎟ ix + xiy ⎟
⎛
⎠
⎠
d ⎞ ⎞⎝ ⎝
⎛
2π ⎜ x 2 + ⎜ y + ⎟ ⎟
⎜
⎝
2 ⎠ ⎟⎠
⎝
I
2
2
The total magnetic field due to the two line currents is
H total =
⎛ ⎛
d⎞
⎞
I2
⎛ ⎛
d⎞
⎞
− y + ⎟ ix + xiy ⎟
⎜ − ⎜ y− 2 ⎟ ix + xiy ⎟ +
2 ⎜ ⎜
2⎠
⎛
⎛
⎠
⎠
⎠
d ⎞ ⎞⎝ ⎝
d ⎞ ⎞⎝ ⎝
⎛
⎛
2π ⎜ x 2 + ⎜ y + ⎟ ⎟
2π ⎜ x 2 + ⎜y− ⎟ ⎟
⎜
⎟
⎜
⎟
2⎠ ⎠
2⎠ ⎠
⎝
⎝
⎝
⎝
I1
2
b. The force density on a line current (force per length) is F = I × B .
At
( 0, d / 2 )
I 2 ix
2π d
the magnetic field is: H = − F = I1 × µ0 H 2 = −
µ0 I1 I 2
iy
2π d
d
d
I2
2
2
.
H x ( x, y = 0 ) =
−
2
⎛ 2 ⎛ d ⎞ ⎞
⎛ 2 ⎛ d ⎞2 ⎞
2π ⎜ x + ⎜ ⎟ ⎟ 2π ⎜ x + ⎜ ⎟ ⎟
⎜
⎜
⎝ 2⎠ ⎟⎠
⎝ 2 ⎠ ⎟⎠
⎝
⎝
I1
c.
When I1 / I 2 = 1 , H x ( x, y = 0 ) = 0
H y ( x, y = 0 ) =
I1 x I2 x
,
+
2
⎛ 2 ⎛ d ⎞ ⎞
⎛ 2 ⎛ d ⎞ 2 ⎞
2π ⎜ x + ⎜ ⎟ ⎟ 2π ⎜ x + ⎜ ⎟ ⎟
⎜
⎜
⎝ 2⎠ ⎟⎠
⎝ 2 ⎠ ⎟⎠
⎝
⎝
When I1 / I 2 = −1 , H y ( x, y = 0 ) = 0
3