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
!
S
B · ds = −N
d
dt
! 0.125 ! 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),
"
#
! 0.125 ! 0.125
d
3
Vemf = −100
20 cos 10 t
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),
"
#
! 0.125 ! 0.125
d
3
20 cos 10 t
cos x sin 2y dx dy = 0.
Vemf = −100
dt
x=−0.125 y=−0.125
(kV).
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)
!
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 6.14 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 6.16 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=
d
,
σA
Ic =
V
VσA
=
.
R
d
(b)
∂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
and
C=
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
C=
= 2.84 × 10−12 F.
0.5 × 10−2
R=
Problem Set 7
Problem 6
Solution
a)
φ=
!
⃗ =
⃗ dS
B.
S
!
a
2
−a
2
!
a
2
Bz dxdy =
−a
2
!
a
2
−a
2
!
a
2
3x2 (y − a)cos(ωt)dxdy = cos(ωt)
−a
2
!
#a
a
a
3a
a5
a 1"
= 2( )3 (y − a)2 −2 a cos(ωt) = ( )3 [(− )2 − (− )2 ]cos(ωt) = cos(ωt).
2
2 2
2
2
2
2
a
2
−a
2
"
x3
# a2
−a
2
(y − a)dy
Also,
dφ
ωa5
|=
sin(ωt) = 0.314sin(ωt).
dt
4
Sign of emf should be set to positive here, so that the induced current I opposes flux Bz . Hence, emf =
0.314sin(ωt).
b)
! a2 ! a2
!
! a2 ! a2
⃗ =
⃗ dS
By dxdz =
8xzdxdz = 0.
φ=
B.
|emf | = |
S
−a
2
−a
2
−a
2
−a
2
Note that the integral equals zero because xz is an odd function of x/z.
As there is no flux going through the loop, emf = 0.
c)
d
emf = − dφ
dt ; in a DC problem like this, dt = 0, hence emf = 0.
d)
⃗
⃗
⃗
⃗ = dD = ϵ dE => dE = 1 ∇ × B.
⃗
∇×H
dt
dt
dt
µϵ
⃗ = aˆx ( dBz − dBy ) + aˆy ( dBx − dBz ) + aˆz ( dBy − dBx ) = [aˆx (3x2 − 8x) + aˆy (5 − 6x(y − a)) + aˆz 8z]cos(ωt)
∇×B
dy
dz
dz
dx
dx
dy
⃗
1
dE
=
[aˆx (3x2 − 8x) + aˆy (5 − 6x(y − a)) + aˆz 8z]cos(ωt)
dt
µ0 ϵ0
⃗ = 1 [aˆx (3x2 − 8x) + aˆy (5 − 6x(y − a)) + aˆz 8z]sin(ωt) (V /m)
E
µ0 ϵ0 ω
e)
emf m =
$
c
⃗ × B|
⃗ ω=0
V
⃗
⃗ × B).
⃗ dl
(V
%
% aˆx
aˆy
%
%
= % 0 1000
%
% 5z 8xy
aˆz
0
3x2 (y − a)
1
%
%
%
%
% = 1000(3x2 )(y − a)aˆx − 5z(1000)aˆz
%
%
emf m =
+
!
!
a
2
−1000(3x2 )(y − a)|y=a/2,z=0 dx +
−a
2
a
2
2
1000(3x )(y − a)|y=a/2,z=0 dx +
−a
2
!
!
a
2
5z(1000)|x=−a/2,z=0 dy
−a
2
a
2
−5z(1000)|x=a/2,z=0 dy
−a
2
(
)a/2
3a
a 'a/2
3
+ 0 + −1000x (− )
+0
= −1000x (− )
2 −a/2
2 −a/2
&
3
3a a3
a a3
= −1000(− )( ) + 1000(− )( ) = 250a4 = 250(1 × 10− 2)4 = 2.5 (µV )
2 4
2
4
2