Maxwell’s Equations - جامعة نزوى · 2014. 5. 26. · Maxwell’s Equations for...
Transcript of Maxwell’s Equations - جامعة نزوى · 2014. 5. 26. · Maxwell’s Equations for...
Chapter 5
Maxwell’s Equations for Time-Varying Fields
Figure 5-17: Loop of Problem 5.2.
Solution
• Since the single-turn loop is not moving or changing shape
with time, .
VVtr
emf
m
emf emfV and V 0 Therefore ,
sd .1
s
tr
emfemf
t
B
RRR
VI
V
If we take the surface normal to be +𝒛 , then the right hand rule
gives positive flowing current to be in the +𝝓 direction.
(A) cossin 00 t
R
ABtB
tR
AI
,where A is the area of the loop.
(a)
A, 𝜔 and R are positive quantities. At t = 0, cos𝜔t = 1
so 𝐼 < 0 and the current is flowing in the −𝝓 direction
(so as to produce an induced magnetic field that
opposes 𝐵).
(b)
At 𝜔 t =𝜋/4, cos𝜔t = 2/2 so 𝐼 < 0 and the current is still flowing
in the − 𝝓 direction.
(c)
At 𝜔 t =𝜋/2, cos𝜔t = 0 so 𝐼 = 0 .There is no current
flowing in either direction.
Solution:
Since the loop is not moving or changing shape
with time,
VVtr
emf
m
emf emfV and V 0
𝑉𝑒𝑚𝑓 = −𝑁𝑑
𝑑𝑡 𝐵. 𝑑𝑠 𝑠
= −N𝑑
𝑑𝑡 𝐵. (𝒛 𝑑𝑥𝑑𝑦)
0.125
−0.125
0.125
−0.125
where N = 100 and the surface normal was chosen to be in the + 𝒛
direction.
For
= −100𝑑
𝑑𝑡(20 𝑐𝑜𝑠103𝑡 (sin 0.125 ×
180
𝜋− sin −0.125 ×
180
𝜋)(0.125 − (−0.125)
= 124.67 sin 103t (kV)
Note: 𝑑
𝑑𝑡𝑐𝑜𝑠103𝑡 = −103𝑠𝑖𝑛103𝑡
, sin −𝜃 = −sin (𝜃)
To convert
from radians
to degrees
Not included in your text-book
Vemf is independent of the resistance which is in the loop. Therefore, when the
loop is intact and the internal resistance is only 0.5Ω.
𝑉𝑒𝑚𝑓 = 𝐼𝑅 = 5 × 0.5 =× 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
𝑅 =
2.5
0.5 + 2 = 1 (A)
Solution:
Solution:
Φ = 𝐵. 𝑑𝑠 = ±𝐵𝐴
𝐵 = 𝐵
Frequency of the field f =300MHz
𝐵 = 𝐵0cos (𝜔𝑡 + 𝛼0)
𝜔=2𝜋𝑓 = 2𝜋 × 300 × 106 𝑟𝑎𝑑/𝑠
𝑉𝑒𝑚𝑓 = 𝐴𝐵0𝜔
∴ 𝐵0 =𝑉𝑒𝑚𝑓
𝐴𝜔=
20×10−3
10−2×6𝜋×108= 1.06 (𝑛𝐴/𝑚)
= −𝐴𝑑
𝑑𝑡𝐵0cos (𝜔𝑡 + 𝛼0)
= 𝐴𝜔𝐵0sin (𝜔𝑡 + 𝛼0)
Vemf is maximum when sin 𝜔𝑡 + 𝛼0 = 1
𝑉𝑒𝑚𝑓 = −𝑁𝑑Φ
𝑑𝑡= −A
𝑑𝐵
𝑑𝑡
Not included in your text-book
solution
𝜔 =7200×2𝜋
60= 240𝜋 rad/s
𝐴 = 5 × 10 × 10−4 = 5 × 10−3 𝑚2
𝑉𝑒𝑚𝑓 = 𝐴𝜔𝐵0𝑠𝑖𝑛𝜔𝑡
it can be seen that the peak voltage is when 𝑠𝑖𝑛𝜔𝑡 = 1
V) ( 85.18105240105 63
0 BAVpeak
emf
5.5 A square loop is coplanar with a long, straight wire carrying
a current
)( )4102cos(5.2)( Atti
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- resistor connected across the gap. The loop has
an internal resistance of 1-.
i(t)
10 cm
10 cm
5 cm
z
y
x
For a long wire carrying
current, recall that:
r
Io
2φβ ˆ
l
→
→ →
At t = 0, B is a maximum, it points in 𝒙 -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: Not included in your text-book
(1) The counter-clockwise circulating current in
a solenoid is increasing at a rate of 8.39 A/s.
The cross-sectional area of the solenoid is 3.14159 cm2,
and there are 163 turns on its18.4 cm length.
What is the magnitude of the induced 𝑉𝑒𝑚𝑓
produced by the increasing current?
dt
dI
l
N
dt
dB
l
NIB 00 ,
dt
dI
l
AN
dt
BAdN
dt
dNVemf 0
2)(
• Thus, the induced 𝑉𝑒𝑚𝑓 is
mVV
x
xxV
dt
dI
l
ANV
emf
emf
emf
4783.0
104.18
)39.8)(1014159.3()163(1042
427
2
0
• (2) In the previous question,
1. The 𝑉𝑒𝑚𝑓 tries to keep the current in the solenoid flowing in the
counter-clockwise direction.
2. The 𝑉𝑒𝑚𝑓 does not effect the current in the solenoid.
3. Not enough information is given to determine the effect
of the 𝑉𝑒𝑚𝑓.
4. By the right hand rule, the 𝑉𝑒𝑚𝑓 produces magnetic fields in
a direction perpendicular to the prevailing magnetic field.
5. The 𝑉𝑒𝑚𝑓 attempts to move the current in the solenoid in
the clockwise direction. Choose the correct statement
• Explanation:
From Ohm's law and Faraday's law, the current in magnitude is :
dt
d
RR
VI
1
where Φ is the magnetic flux through the loop.
We know the sign of the rate of change of the magnetic flux is
changed when the magnet is withdrawn upward, which, according to
the equation the direction of the current is also changed.
From Lenz's law, we know when the magnet is moved down toward
the loop, the current in the loop is counterclockwise as viewed from
above.
book-Not included in your text Problem:
In the Figure shown, the north pole of the
magnet is First moved down toward the loop
of wire, then is withdrawn upward.
As viewed from above, the induced current
in the loop is:
1. For both cases clockwise with increasing magnitude.
2. For both cases counterclockwise with de-creasing magnitude.
3. For both cases counterclockwise with in-creasing magnitude.
4. For both cases clockwise with decreasing magnitude.
5. First clockwise, then counterclockwise.
6. First counterclockwise, then clockwise.
Choose the correct statement
• Explanation:
As the current is increasing in the counter-
clockwise direction, by Lenz's law, the 𝑉𝑒𝑚𝑓 will
attempt to retard the current, which establishes
an 𝑉𝑒𝑚𝑓 that tries to counter the flow of the
current, which in this case would be in the
clockwise direction.
solution
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rad/s 660
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r ˆu
bar. on the
point any for velocity theis u , ).(12
4
0
5.0
24
0
05.0
4
0
5.0
4m
emf
1
2
V
rdrrrr
drrzr
whereldBu
r
m
emfVV
Note that 𝝓 × 𝒛 = 𝒓
Figure 5-22: Rotating cylinder in a magnetic field (Problem 5.9).
2ˆ10560
12002ˆˆ 2
ru
l
BuV0
12 ld . )(
dzz . 6ˆ)2ˆ0.1
0
r
V 77.31.01212.ˆ12
1.0
0
1.0
0
zzdzz
solution
Problem 5.10 The electromagnetic generator shown in
Fig. 5-12 is connected to an electric light bulb with a
resistance of 100 W. If the loop area is 0.1 m2 and it
rotates at 3600 revolutions per minute in a uniform
magnetic flux density, B0 = 0.2 T, determine the amplitude
of the current generated in the light bulb.
)sin(
)sin(V
: isgenerator acby generated voltagesinusodial the
00
00emf
ctV
ctBA
mA 75.4 A1075.4
R
VI
V
3-
0
100
54.7
54.72.060
236001.0
00
BAV
solution
a) 𝑅 =𝑑
𝜎𝐴 , 𝐼𝑐 =
𝑉
𝑅=
𝑉𝜎𝐴
𝑑
b) 𝐸 =𝑉
𝑑 , 𝐼𝑑 =
𝜕𝐷
𝜕𝑡 . 𝐴 = 𝐴
𝜕𝜀𝐸
𝜕𝑡
= 𝜀𝐴𝜕𝐸
𝜕𝑡= 𝜀𝐴
𝜕
𝜕𝑡
𝑉
𝑑
= 𝜀𝐴
𝑑 𝜕𝑉
𝜕𝑡
C) The conduction current Ic directly proportional
to voltage V , as characteristic of a resistor,
while the displacement current 𝐼𝑑 varies as 𝜕𝑉
𝜕𝑡
which is a characteristic of a capacitor:
𝑅 =𝑑
𝜎𝐴 and 𝐶 =
𝜀𝐴
𝑑
d) 𝑅 =0.5×10−2
2.5×2 ×10−4= 10Ω
𝜀 = 𝜀𝑟𝜀0
𝐶 =4×8.85×10−12×2×10−4
0.5×10−2
= 1.42 × 10−12 𝐹