MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
Final Exam Review: W15D2_3 Coaxial Cable and Power Flow
A coaxial cable consists of two concentric long
hollow cylinders of zero resistance; the inner
has radius a , the outer has radius b , and the
length of both is l , with l >> b , as shown in
the figure. The cable transmits DC power from
a battery to a load. The battery provides an
electromotive force ε between the two
conductors at one end of the cable, and the load
is a resistance R connected between the two
conductors at the other end of the cable. A current I flows down the inner conductor
and back up the outer one. The battery charges the inner conductor to a charge −Q and
the outer conductor to a charge +Q .

(a) Find the direction and magnitude of the electric field E everywhere.
Answer:
Consider a Gaussian surface in the form of a cylinder with radius r and length l, coaxial
with the cylinders. Inside the inner cylinder (r<a) and outside the outer cylinder (r>b) no
charge is enclosed and hence the field is 0. In between the two cylinders (a<r<b) the
charge enclosed by the Gaussian surface is –Q, the total flux through the Gaussian
cylinder is
 
ΦE = 
∫∫ E ⋅ dA = E(2π rl)
Thus, Gauss’s law leads to E(2π rl) =
qenc
, or
ε0

q
Q
E = enc r̂ = −
r̂ (inward) for a < r < b, 0 elsewhere
2π rl
2πε 0 rl

(b) Find the direction and magnitude of the magnetic field B everywhere.
Answer:
Just as with the E field, the enclosed current Ienc in the Ampere’s loop with radius r is
zero inside the inner cylinder (r<a) and outside the outer cylinder (r>b) and hence the
field there is 0. In between the two cylinders (a<r<b) the current enclosed is –I.
Applying Ampere’s law,


∫ B ⋅ d s = B(2π r) = µ I
0 enc
, we obtain

µI
B = − 0 ϕ̂ (clockwise viewing from the left side) for a < r < b, 0 elsewhere
2π r
0
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02

(c) Calculate the Poynting vector S in the cable.
Answer:
For a < r < b , the Poynting vector is
 1  
⎞ ⎛ µ I ⎞ ⎛ QI ⎞
1 ⎛
Q
S=
E×B =
−
r̂ ⎟ × ⎜ − 0 ϕ̂ ⎟ = ⎜ 2 2 ⎟ k̂ (from right to left)
⎜
µ0
µ0 ⎝ 2πε 0 rl ⎠ ⎝ 2π r ⎠ ⎝ 4π ε 0 r l ⎠

On the other hand, for r < a and r > b , we have S = 0 .

(d) By integrating S over appropriate surface, find the power that flows into the coaxial
cable.
Answer:

With dA = (2π r dr )kˆ , the power is
 
P=
∫∫ S ⋅ dA =
S
b 1
⎛ b⎞
QI
QI
(2π rdr) =
ln ⎜ ⎟
∫
2
2
2π ε 0 l ⎝ a ⎠
4π ε 0 l a r
(e) How does your result in (d) compare to the power dissipated in the resistor?
Answer:
Since
 
b
ε = ∫E⋅d s = ∫
a
Q
Q
⎛b⎞
dr =
ln ⎜ ⎟ = IR
2π rlε 0
2π lε 0 ⎝ a ⎠
the charge Q is related to the resistance R by Q =
2π ε 0 lIR
. The above expression for P
ln(b / a)
becomes
⎛ 2π ε 0lIR ⎞ I
⎛ b⎞
P=⎜
ln ⎜ ⎟ = I 2 R
⎟
⎝ ln(b / a) ⎠ 2π ε 0 l ⎝ a ⎠
which is equal to the rate of energy dissipation in a resistor with resistance R.
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