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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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6.013 Quiz 1 Solutions, 10/20/2005
Problem 1
a) Apply Integral form of Ampere’s Law:
2π
r
J0r 2
2π rHφ = ∫ ∫ J z (r )rdrdφ → H =
iφ
3R
1
φ =0 r =0
b) The total current going up the inside conductor must return going down in the
outside conductor:
R1
J R2
2π ∫ J z ( r )rdr = 2π R2 K → K = − 0 1 i z
r =0
3R2
Problem 1 Alternate Method:
R1 < r < R2
2π R1
J 0r
J 0 R12
rdrdφ → H φ =
2π rHφ = ∫ ∫
3r
φ =0 r =0 R1
J 0 R12
K z = −H φ ( r = R2 ) = −
3R2
Problem 2
a) The boundary conditions at φ = 0 and φ = α determine the unknowns A and B
Φ (φ = 0) = 0 → B = 0
Φ (φ = α ) = V0 → A =
V0
α
b) The electric field is the negative gradient of the potential. Problem is best done
in cylindrical coordinates:
V
1 ∂Φ(φ )
E = −∇Φ = −
iφ = − 0 iφ
r ∂φ
αr
c) Use the boundary condition on the normal D field at the upper perfect conductor:
σ sf = iφ i[ε 0 E(φ = α + ) − ε E(φ = α − )] =
ε V0
αr
d) Since C=Q/V0 and the total charge is given by
Q = d∫
R2
r = R1
σ sf (r )dr =
⎛R ⎞
εd
ε d ⎛ R2 ⎞
ln ⎜ ⎟
V0 ln ⎜ 2 ⎟ → C =
α
α
R
⎝ 1⎠
⎝ R1 ⎠
Problem 3
8
8
a) By inspection ω = 2π f = 2π ×10 → f = 10 Hz
b) By inspection k = k x 2 + k z 2 = π 1 + 3 = 2π → λ =
2π
= 1m
k
c) c = f λ = 1×10 m/s
8
d) From trigonometry k x = π , k z = π 3, tan(θ i ) =
equal to 30 degrees.
kx
1
π
radians,
→ θi =
=
6
kz
3