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

V 70725.0109

21018ˆ.ˆ1018

ˆ . )103(ˆ)6ˆ(

rad/s 660

1802

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