```Problem 6.11 The loop shown in P6.11 moves away from a wire carrying a current
I1 = 10 A at a constant velocity u = ŷ7.5 (m/s). If R = 10 Ω and the direction of I2 is
as defined in the figure, find I2 as a function of y0 , the distance between the wire and
the loop. Ignore the internal resistance of the loop.
z
10 cm
R
u
I1 = 10 A
I2
20 cm
u
R
y0
Figure P6.11: Moving loop of Problem 6.11.
Solution: Assume that the wire carrying current I1 is in the same plane as the loop.
The two identical resistors are in series, so I2 = Vemf /2R, where the induced voltage
is due to motion of the loop and is given by Eq. (6.26):
m
Vemf = Vemf
=
Z
(u × B) · dl.
n
C
The magnetic field B is created by the wire carrying I1 . Choosing ẑ to coincide with
the direction of I1 , Eq. (5.30) gives the external magnetic field of a long wire to be
B = φ̂φ
µ0 I1
.
2π r
For positive values of y0 in the y-z plane, ŷ = r̂, so
× B = r̂|u|×
× φ̂φ
u × B = ŷ|u|×
µ0 I1
µ0 I1 u
= ẑ
.
2π r
2π r
Integrating around the four sides of the loop with dl = ẑ dz and the limits of
integration chosen in accordance with the assumed direction of I2 , and recognizing
m , we have
that only the two sides without the resistors contribute to Vemf
¶¯
¶¯
Z 0.2 µ
Z 0 µ
µ0 I1 u ¯¯
µ0 I1 u ¯¯
m
Vemf
=
ẑ
· (ẑ dz)
·
(ẑ
dz)
+
ẑ
2π r ¯r=y0
2π r ¯r=y0 +0.1
0
0.2
µ
¶
1
4π × 10−7 × 10 × 7.5 × 0.2 1
−
=
2π
y0 y0 + 0.1
µ
¶
1
1
= 3 × 10−6
−
(V),
y0 y0 + 0.1
and therefore
µ
¶
m
Vemf
1
1
= 150
−
I2 =
2R
y0 y0 + 0.1
(nA).
6.11
Problem 6.16 The parallel-plate capacitor shown in Fig. P6.16 is filled with a lossy
dielectric material of relative permittivity εr and conductivity σ . The separation
between the plates is d and each plate is of area A. The capacitor is connected to
a time-varying voltage source V (t).
I
A
V(t)
ε, σ
d
Figure P6.16: Parallel-plate capacitor containing a lossy dielectric material (Problem 6.16).
(a) Obtain an expression for Ic , the conduction current flowing between the plates
inside the capacitor, in terms of the given quantities.
(b) Obtain an expression for Id , the displacement current flowing inside the
capacitor.
(c) Based on your expressions for parts (a) and (b), give an equivalent-circuit
representation for the capacitor.
(d) Evaluate the values of the circuit elements for A = 4 cm2 , d = 0.5 cm, εr = 4,
σ = 2.5 (S/m), and V (t) = 10 cos(3π × 103t) (V).
Solution:
(a)
R=
(b)
d
,
σA
Ic =
V
VσA
=
.
R
d
∂D
∂ E εA ∂V
V
,
Id =
· A = εA
=
.
d
∂t
∂t
d ∂t
(c) The conduction current is directly proportional to V , as characteristic of a
resistor, whereas the displacement current varies as ∂ V/∂ t, which is characteristic
E=
of a capacitor. Hence,
R=
d
σA
C=
and
εA
.
d
I
+
ε, σ
V(t)
R
Ic
Id
+
C
V(t)
-
-
Actual Circuit
Equivalent Circuit
Figure P6.16: (a) Equivalent circuit.
(d)
0.5 × 10−2
= 5 Ω,
2.5 × 4 × 10−4
4 × 8.85 × 10−12 × 4 × 10−4
= 2.84 × 10−12 F.
C=
0.5 × 10−2
R=
Problem 6.3 A coil consists of 100 turns of wire wrapped around a square frame
of sides 0.25 m. The coil is centered at the origin with each of its sides parallel to
the x- or y-axis. Find the induced emf across the open-circuited ends of the coil if the
magnetic field is given by
(a) B = ẑ 20e−3t (T)
(b) B = ẑ 20 cos x cos 103t (T)
(c) B = ẑ 20 cos x sin 2y cos 103t (T)
m = 0 V and
Solution: Since the coil is not moving or changing shape, Vemf
tr . From Eq. (6.6),
Vemf = Vemf
Vemf = −N
d
dt
Z
S
B · ds = −N
d
dt
Z 0.125 Z 0.125
−0.125 −0.125
B · (ẑ dx dy),
where N = 100 and the surface normal was chosen to be in the +ẑ direction.
(a) For B = ẑ20e−3t (T),
Vemf = −100
d
(20e−3t (0.25)2 ) = 375e−3t
dt
(V).
(b) For B = ẑ20 cos x cos 103t (T),
µ
¶
Z 0.125 Z 0.125
d
3
20 cos 10 t
Vemf = −100
cos x dx dy = 124.6 sin 103t
dt
x=−0.125 y=−0.125
(c) For B = ẑ20 cos x sin 2y cos 103t (T),
¶
µ
Z 0.125 Z 0.125
d
3
cos x sin 2y dx dy = 0.
20 cos 10 t
Vemf = −100
dt
x=−0.125 y=−0.125
(kV).
Problem 6.4 A stationary conducting loop with an internal resistance of 0.5 Ω is
placed in a time-varying magnetic field. When the loop is closed, a current of 5 A
flows through it. What will the current be if the loop is opened to create a small gap
and a 2-Ω resistor is connected across its open ends?
Solution: Vemf is independent of the resistance which is in the loop. Therefore, when
the loop is intact and the internal resistance is only 0.5 Ω,
Vemf = 5 A × 0.5 Ω = 2.5 V.
When the small gap is created, the total resistance in the loop is infinite and the
current flow is zero. With a 2-Ω resistor in the gap,
I = Vemf /(2 Ω + 0.5 Ω) = 2.5 V/2.5 Ω = 1 (A).
Problem 6.5 A circular-loop TV antenna with 0.02-m2 area is in the presence of a
uniform-amplitude 300-MHz signal. When oriented for maximum response, the loop
develops an emf with a peak value of 30 (mV). What is the peak magnitude of B of
the incident wave?
Solution: TV loop
antennas have one turn. At maximum orientation, Eq. (6.5)
R
evaluates to Φ = B · ds = ±BA for a loop of area A and a uniform magnetic field
with magnitude B = |B|. Since we know the frequency of the field is f = 300 MHz,
we can express B as B = B0 cos (ω t + α0 ) with ω = 2π × 300 × 106 rad/s and α0 an
arbitrary reference phase. From Eq. (6.6),
Vemf = −N
d
dΦ
= −A [B0 cos(ω t + α0 )] = AB0 ω sin(ω t + α0 ).
dt
dt
Vemf is maximum when sin(ω t + α0 ) = 1. Hence,
30 × 10−3 = AB0 ω = 0.02 × B0 × 6π × 108 ,
which yields B0 = 0.8 (nA/m).
Problem 6.6 The square loop shown in Fig. P6.6 is coplanar with a long, straight
wire carrying a current
I(t) = 5 cos(2π × 104t) (A).
(a) Determine the emf induced across a small gap created in the loop.
(b) Determine the direction and magnitude of the current that would flow through
a 4-Ω resistor connected across the gap. The loop has an internal resistance of
1 Ω.
z
10 cm
I(t)
10 cm
5 cm
y
x
Figure P6.6: Loop coplanar with long wire (Problem 6.6).
Solution:
(a) The magnetic field due to the wire is
B = φ̂φ
µ0 I
µ0 I
= −x̂
,
2π r
2π y
where in the plane of the loop, φ̂φ = −x̂ and r = y. The flux passing through the loop
is
¶
Z 15 cm µ
Z
µ0 I
−x̂
· [−x̂ 10 (cm)] dy
Φ = B · ds =
2π y
5 cm
S
µ0 I × 10−1 15
ln
=
2π
5
4π × 10−7 × 5 cos(2π × 104t) × 10−1
=
× 1.1
2π
= 1.1 × 10−7 cos(2π × 104t) (Wb).
Vemf = −
dΦ
= 1.1 × 2π × 104 sin(2π × 104t) × 10−7
dt
= 6.9 × 10−3 sin(2π × 104t) (V).
(b)
Iind =
Vemf
6.9 × 10−3
=
sin(2π × 104t) = 1.38 sin(2π × 104t) (mA).
4+1
5
At t = 0, B is a maximum, it points in −x̂-direction, and since it varies as
cos(2π × 104t), it is decreasing. Hence, the induced current has to be CCW when
looking down on the loop, as shown in the figure.
Problem 6.7 The rectangular conducting loop shown in Fig. P6.7 rotates at 6,000
revolutions per minute in a uniform magnetic flux density given by
B = ŷ 50 (mT).
Determine the current induced in the loop if its internal resistance is 0.5 Ω.
z
2c
m
B
3 cm
B
y
φ(t)
ω
x
Figure P6.7: Rotating loop in a magnetic field (Problem 6.7).
Solution:
Φ=
Z
S
B · dS = ŷ 50 × 10−3 · ŷ(2 × 3 × 10−4 ) cos φ (t) = 3 × 10−5 cos φ (t),
2π × 6 × 103
60
Φ = 3 × 10−5 cos(200π t) (Wb),
dΦ
Vemf = −
= 3 × 10−5 × 200π sin(200π t) = 18.85 × 10−3 sin(200π t) (V),
dt
Vemf
Iind =
= 37.7 sin(200π t) (mA).
0.5
φ (t) = ω t =
The direction of the current is CW (if looking at it along −x̂-direction) when the loop
is in the first quadrant (0 ≤ φ ≤ π /2). The current reverses direction in the second
Problem 6.14 The plates of a parallel-plate capacitor have areas of 10 cm2 each
and are separated by 2 cm. The capacitor is filled with a dielectric material with
ε = 4ε0 , and the voltage across it is given by V (t) = 30 cos 2π × 106t (V). Find the
displacement current.
Solution: Since the voltage is of the form given by Eq. (6.46) with V0 = 30 V and
ω = 2π × 106 rad/s, the displacement current is given by Eq. (6.49):
εA
V0 ω sin ω t
d
4 × 8.854 × 10−12 × 10 × 10−4
=−
× 30 × 2π × 106 sin(2π × 106t)
2 × 10−2
= −0.33 sin (2π × 106t) (mA).
Id = −
Problem 6.15 A coaxial capacitor of length l = 6 cm uses an insulating dielectric
material with εr = 9. The radii of the cylindrical conductors are 0.5 cm and 1 cm. If
the voltage applied across the capacitor is
V (t) = 50 sin(120π t) (V)
what is the displacement current?
Solution:
l
Id
r
+
V(t)
2a
2b
-
Figure P6.15:
To find the displacement current, we need to know E in the dielectric space between
the cylindrical conductors. From Eqs. (4.114) and (4.115),
Q
,
2πε rl
µ ¶
b
Q
V=
ln
.
2πε l
a
E = −r̂
Hence,
E = −r̂
D = εE
72.1
50 sin(120π t)
V
¡ b ¢ = −r̂
= −r̂
sin(120π t) (V/m),
r ln 2
r
r ln a
= εr ε0 E
= −r̂ 9 × 8.85 × 10−12 ×
72.1
sin(120π t)
r
5.75 × 10−9
sin(120π t) (C/m2 ).
r
The displacement current flows between the conductors through an imaginary
cylindrical surface of length l and radius r. The current flowing from the outer
conductor to the inner conductor along −r̂ crosses surface S where
= −r̂
S = −r̂ 2π rl.
Hence,
∂D
∂
Id =
· S = −r̂
∂t
∂t
µ
¶
5.75 × 10−9
sin(120π t) · (−r̂ 2π rl)
r
= 5.75 × 10−9 × 120π × 2π l cos(120π t)
= 0.82 cos(120π t) (µ A).
Alternatively, since the coaxial capacitor is lossless, its displacement current has to
be equal to the conduction current flowing through the wires connected to the voltage
sources. The capacitance of a coaxial capacitor is given by (4.116) as
C=
The current is
I =C
2πε l
¡ ¢.
ln ba
2πε l
dV
= ¡ b ¢ [120π × 50 cos(120π t)] = 0.82 cos(120π t) (µ A),
dt
ln a
which is the same answer we obtained before.
```