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Chapter 9

MAXWELL'S EQUATIONS

Part Two

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conducting bar is connected via flexible leads to a pair of rails in a

magnetic field B = 6 cos 10t ax mWb/m2 as in Figure 9.20. If the z-axis is

the equilibrium position of the bar and its velocity is 2 cos 10t ay m/s, find

the voltage induced in it.

2

105

0 0

emf

6cos(10 ) mWb/m

2cos(10 ) m/s

. 6cos(10 ) . 6cos(10 ) (5)( 10)

30( 10)cos(10 )

x

y

emf

y

x x

motional transformer

t

u t

dV

dt

B dS t dydz t y

y t

B a

a

a a

4

2

sin(10 )2cos(10 )

5

sin(10 )30( 10)cos(10 ) 30 10 cos(10 )

5

6sin(10 ) 300 cos(10 )

. . ( ).

6cos(10 ) mWb/m , 60sin(10 )

2cos(10 ) m/s

emf

x x

y

dyu

dt

ty udt t dt

ty t t

t t

dV E dl d u dl

dt

dt t

dt

u t

B

S B

BB a a

a

, dl=dzx zdS dydz a a

5

105 5

2

0 0 0

2

2

. ( ).

60sin(10 ) 12cos (10 )

60sin(10 ) (5)(10+y) 12cos (10 ) (5)

=300(10+y)sin(10 ) 60cos (10 )

sin(10 )2cos(10 )

5

emf

y

emf

dV d u dl

dt

t dydz t dz

t t

V t t

dyu

dt

ty udt t dt

BS B

6

A car travels at 120 km/hr. If the earth's magnetic field is 4.3 X 10"5 Wb/m2, find

the induced voltage in the car bumper of length 1.6 m. Assume that the angle

between the earth magnetic field and the normal to the car is 65°.

5

( ). | || | sin

120,000 = ( )(4.3 10 )sin(25) (1.6) 0.96 mV

3600

emfV u dl u B

B

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As portrayed in Figure 9.21, a bar magnet is thrust toward the center of a coil of 10

turns and resistance 15 fl. If the magnetic flux through the coil changes from 0.45

Wb to 0.64 Wb in 0.02 s, what is the magnitude and direction (as viewed from the

side near the magnet) of the induced current?

0.64 0.45(10) 95 V

0.02 0

956.33 A

15

emf

emf

dV N

dt

VI

R

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- It is not an electric current of moving charges ,

but a time-varying electric field .

-This provides a mechanism whereby a time-varying electric

field can create a magnetic field, complementary to

Faraday's law, in which a time-varying magnetic field can

produce an electric field.

- "mutual embrace" of electric and magnetic fields can produce

propagating electromagnetic waves. This would not be

possible without the displacement current.

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Types of current:

1) Conduction current: through conductor J=σE

2) Convection current: through isolator J=ρvu

3) Displacement current: is due to time varying field.

Simple example(RC circuit)

- For resistor the conduction current model is valid(JR= σRER)

- But the conduction current doesn’t characterized the

capacitor current for ideal capacitor (perfect insulator σc=0)

- The capacitor current depends on charging and discharging

of the capacitor plates.

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c c c

c c

c

A( ) , Q=CV(t)=C E (t) d= E (t) d= AE (t)

E (t) (t)( )

( ) (t)

c

c

c

dQi t

dt d

d dDi t A A

dt dt

i t dD

A dt

c (t) (displacement current density)D

dDJ

dt

The displacement current doesn't exist under static conditions.

cv

(t) , J= E+tot

dDJ J u

dt

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c

Ampere's law// . ( ).

(t

.

)to

t

t

otIen H dl H dS J dS

dDH J J

dt

- Without the term Jd, electromagnetic wave propagation (radio or TV waves,

for example) would be impossible.

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For non ideal capacitor:

D

D

J E

dDJ E

dt

J

J

- At low frequencies, Jd is usually neglected compared with J.

However, at radio frequencies, the two terms are comparable

D

conduction

good conductor (J negligible)

good insulator (J negligible)

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In free space E=20cos(wt-50x) ay V/m , find:

(a) Jd (b) H

y

d d

x y z

(a) 20 sin( 50 ) A/m

(b) H=J+ J J

H

20 sin( 50 )

20 sin( 50 )

cos( 520

d o o

y yz x z x

x zo

zo

z o

dD dEJ t x

dt dt

dH dHdH dH dH dH

dy dz dz dx dx dy

dH dHt x

dz dx

dHt x

dx

tH

a

a a a

z

0 )0.4 cos( 50 )

50

0.4 cos( 50 ) A/m

o

o

xt x

H t x

a

14

A 50-V voltage generator at 20 MHz is connected to the plates of an air dielectric

parallel plate capacitor with plate area 2.8 cm2 and separation distance 0.2 mm.

Find the maximum value of displacement current density and displacement

current.

6

4

46

,max 3

,max 2

,max 4

50cos( ) 50cos(2 20 10 )

2.8 10

(50cos( ))50 sin( )

(2.8 10 )50(2 20 10 ) 0.0778 A

0.2 10

0.0778277.7 A/m

2.8 10

d

od

d

d

V t t

A

dV A d t AI C t

dt d dt d

I

IJ

A

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The ratio JIJd (conduction current density to displacement current density) is very

important at high frequencies. Calculate the ratio at 1 GHz for:

- sea water (µ=µo, ε = 8εo, σ = 25 S/m)

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255.55 ,good conductor

(2 10 )(81 )D o

J

J

16

A conductor with cross-sectional area of 10 cm2 carries a conduction current

0.2 sinl09t mA. Given that σ = 2.5 X 106 S/m and εr = 6, calculate the magnitude of

the displacement current density.

311

6

9 11 2

0.2 10| | 0.2 8 10

2.5 10

| | (2 10 )(6 )(8 10 ) 26.7 pA/m

c

D

D o

J E E

dDJ j E

dt

J E

Maxwell’s equations in final forms

enc v

cenc

: D d S Q dv

: B d S 0

(Non existence of isolated magneticcharge)

Gauss’s Law

Gauss’s Law for magnetic

Faraday's Lawd B.dS

E d Ldt

:

Ampere's Law:dD (t)

H d L I J

.dS

dt

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Maxwell’s Equations in final forms

vD

B 0

dBE

dt

dDH J

dt