MIT OpenCourseWare
http://ocw.mit.edu
6.013/ESD.013J Electromagnetics and Applications, Fall 2005
Please use the following citation format:
Markus Zahn, Erich Ippen, and David Staelin, 6.013/ESD.013J
Electromagnetics and Applications, Fall 2005. (Massachusetts Institute
of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed
MM DD, YYYY). License: Creative Commons AttributionNoncommercial-Share Alike.
Note: Please use the actual date you accessed this material in your citation.
For more information about citing these materials or our Terms of Use, visit:
http://ocw.mit.edu/terms
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.013 Electromagnetics and Applications
Problem Set #2 SOLUTION
Fall Term 2005
Problem 2.1
a.
By the divergence theorem;
∫ ∇i( ∇ × A )dV = ∫ ( ∇ × A )ida
V
where S encloses V.
S
By Stokes’ theorem:
∫ ( ∇ × A )ida = ∫ Aidl
S'
C
Suppose S’=S is a closed surface
S
S’
C
Or S’ is an open surface with boundary contour C. i.e., S’ is the same as S, except for the
curve C, which makes S’ not a closed surface.
Now consider the limit as CÆ0;
S’
So that S’=S.
If C is 0 then
∫ ( ∇ × A )ida = ∫ Aidl
S'
S
S’
= 0 and
C
∫ ∇i( ∇ × A )dV = ∫ ( ∇ × A )ida = 0
V
S
Since V can be any volume, the argument of the integral must be identically 0:
∇i( ∇ × A ) = 0
b.
A = Ar ir + Aφ iφ + Az iz in the cylindrical geometry.
⎛ 1 ∂Az ∂Aφ ⎞
⎛ ∂Ar ∂Az
∇× A = ⎜
−
−
⎟ ir + ⎜
∂z ⎠
∂r
⎝ ∂z
⎝ r ∂φ
1 ⎛ ∂ ( rAφ ) ∂Ar
⎞
⎜
+
−
i
φ
⎟
∂φ
r ⎜ ∂r
⎠
⎝
⎞
⎟i
⎟z
⎠
1 ∂ ⎡ ⎛ 1 ∂Az ∂Aφ ⎞ ⎤ 1 ∂ ⎛ ∂Ar ∂Az ⎞ ∂ ⎡ 1 ⎛ ∂ ( rAφ ) ∂Ar ⎞ ⎤
⎟⎥
−
−
−
⎢r ⎜
⎟⎥ +
⎜
⎟+ ⎢ ⎜
r ∂r ⎣ ⎝ r ∂φ
∂z ⎠ ⎦ r ∂φ ⎝ ∂z
∂r ⎠ ∂z ⎢ r ⎝⎜ ∂r
∂φ ⎠⎟ ⎥
⎣
⎦
2
2
2
2
2
2
1 ∂ Az 1 ∂Aφ ∂ Aφ 1 ∂ Ar 1 ∂ Az 1 ∂Aφ ∂ Aφ 1 ∂ Ar
=
−
−
+
−
+
+
−
=0
r ∂r ∂φ r ∂z ∂r∂z r ∂φ∂z r ∂r ∂φ r ∂z ∂r ∂z r ∂φ∂z
∇i( ∇ × A ) =
c.
From stokes’ theorem
∫ ( ∇ × ( ∇f ) )ida = ∫ ∇f idl
S
∇f =
Here
C
∂f
∂f
∂f
ix + iy + iz
∂x
∂y
∂z
dl = dxix + dy iy + dziz
,
dl
A
∇f idl =
So
∂f
∂f
∂f
dx + dy + dz = df
∂x
∂y
∂z
∫ ( ∇ × ( ∇f ) )ida = ∫ ∇f idl = ∫ df
S
C
C
S
C
= f
A
=0
A
Since S can be any surface, the argument of the integral must be identically 0:
∇ × ( ∇f ) = 0
d.
∇f =
∂f
1 ∂f
1 ∂f
ir +
iθ +
iφ in spherical coordinate.
∂r
r ∂θ
r sin θ ∂φ
⎡ ⎛
1 ∂f ⎞
⎛ 1 ∂f
∂⎜
⎢ ∂ ⎜ sin θ
⎟
r sin θ ∂φ ⎠
1 ⎢ ⎝
r ∂θ
∇ × ( ∇f ) =
− ⎝
∂θ
∂φ
r sin θ ⎢
⎢
⎣
⎡ ⎛ 1 ∂f
∂ r
1 ⎢ ⎜⎝ r ∂θ
+ ⎢
r⎢
∂r
⎢⎣
1
= 2
r sin θ
⎡
⎛
1 ∂f ⎞ ⎤
⎞⎤
⎛ ∂f ⎞
∂⎜ ⎟ ∂⎜r
⎢
⎟⎥
⎟⎥
⎠⎥ i + 1 ⎢ 1
⎝ ∂r ⎠ − ⎝ r sin θ ∂φ ⎠ ⎥ i
⎥ r r ⎢sin θ ∂φ
⎥θ
∂φ
⎥
⎢
⎥
⎦
⎣
⎦
⎞
⎛ ∂f ⎞ ⎤
⎟ ∂ ⎜ ⎟ ⎥
⎠ − ⎝ ∂r ⎠ ⎥ i
φ
∂θ ⎥
⎥⎦
⎛ ∂2 f
1 ⎛ ∂2 f
1 ⎛ ∂2 f
∂2 f ⎞
∂2 f ⎞
∂2 f ⎞
−
−
−
⎜
⎟ ir + ⎜
⎟ iθ + ⎜
⎟ iφ = 0
r ⎝ ∂r ∂φ ∂r ∂φ ⎠
r ⎝ ∂r ∂θ ∂r ∂θ ⎠
⎝ ∂θ∂φ ∂θ∂φ ⎠
Problem 2.2
a. The total z directed current on the cylinder is:
2
I 0 = ∫ J z ( r )2π rdr = ∫
R1
0
b.
R1
0
R J πR2
⎛ r ⎞
J 2π
J 0 ⎜ ⎟ 2π rdr = 0 2 r 4
1 = 0 1
0
4R1
2
⎝ R1 ⎠
By Ampere’s integral law
∫
C
H ids = ∫ J ida
S
⎧ r
J 0 2π 4 r J 0 2π 4
r' =
r , r < R1
⎪ ∫0 J z ( r ')2π r ' dr ' =
H φ ( r ) 2π r = ⎨
0 4R12
4R12
⎪I ,
R1 < r < R2
⎩ 0
⎧ J0 3
r , r < R1
2
⎪
⎪ 4R1
H φ ( r ) = ⎨
2
⎪ J 0 R1 , R < r < R
1
2
⎪⎩ 4r
J 0 R12
4R2
c.
The current density at r = R2 is: K z = −H φ ( r = R2 ) = −
d.
J 0π R12
= −I 0
The total current on cylinder of r = R2 is: I (r = R2 ) = 2π R2 K z = −
2
Problem 2.3
4
a.
The total charge on the sphere is: q =
b. By Gauss’ law in free space
∫
R1
0
⎛ r ⎞
4πρ 0 R13
ρ0 ⎜ ⎟ 4π r 2 dr =
7
⎝ R1 ⎠
1
∫∫ E ida = ε ∫
s
0
V
ρ dV
⎧ 1 r ⎛ r ' ⎞4
4πρ 0 r 7
2
⎪
, r < R
1
ρ0 ⎜ ⎟ 4π r ' dr ' =
7ε 0 R14
⎪ ε 0 ∫0 ⎝ R1 ⎠
2
4π r Er = ⎨
4πρ 0 R13
⎪1
,
q
=
R1 < r < R2
⎪ε
7ε 0
⎩ 0
⎧ ρ 0 r 5
, r < R1
⎪
4
⎪ 7ε 0 R1 Er = ⎨
3
⎪ ρ0 R1 ,R
< r < R2
⎪⎩ 7ε 0 r 2 1
c.
The surface charge on the sphere of radius R2 is:
σ s = −ε 0 Er ( r = R2 ) = −
P0 R13
7R2 2
d. The total charge at r = R2 is: q ( r = R2 ) = σ s 4π R2 = −
2
4πρ0 R13
= −q
7
Problem 2.4
i
a.
Hφ =
b.
v (t ) =
2π r
, λ f = µ0 N ∫
z +l
z
dz ∫
R+h
R
⎡ µ Nl ⎛ h ⎞ ⎤
Hφ dr = ⎢ 0 ln ⎜ 1+ ⎟ ⎥ i
⎝ R ⎠⎦
⎣ 2π
dλf
⎡ µ Nl ⎛ h ⎞ ⎤
= ⎢ 0 ln ⎜ 1+ ⎟ ⎥ ω I 0 cos (ωt )
dt
⎝ R ⎠⎦
⎣ 2π
with the values given in the problem statement v ( t ) =
dλf
dt
= 1.35 cos (120π t ) mV
c. The coil should be placed vertically centered on the wire so that half the coil flux is positive
and half is negative so that no net flux links the coil.
•
I
×
.