new EM1
-
Upload
cooldude99344 -
Category
Documents
-
view
218 -
download
0
Transcript of new EM1
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 1/24
Chapter 1 Electromagnetic Field Theory
1-1 Electric Fields and Electric Dipoles
Gauss’s law of E : ∫∫∫ ∫∫ ==⋅
→
'1 dvQS d E s
ρ ε ε
∫∫∫ ∫∫∫ =⋅∇ → →
''
1'
vv
theormdivergence
dvdv E ρ ε ε
ρ =⋅∇⇒
→
E and ε0=910
36
1 −×π
( F /m) in
the free space.
For q at→
' R , field point at→
R 2
'
|'|4
→→
→
−=⇒
R R
aq E RR
πε 3|'|4
)'(→→
→→
−
−=
R R
R Rq
πε and
'
'
'
R R
R Ra RR
−
−= .
→
E d!e to a s"ste# of discrete char$es% ∑=
→→
→→→
−
−=
n
k k
k k
R R
R Rq E
1 3|'|
)'(
4
1
πε
&ol!#e so!rce ρ⇒ ∫∫∫ ∫∫∫ →
→∧→
==' ' 3
2'
||4
1'
4
1v v
R dv
R
Rdv
Ra E
ρ
πε
ρ
πε
'!rface so!rce ρ s⇒ ∫∫ ∧→
=' 2
'4
1 s
s R dS R
a E ρ
πε ine so!rce ρl ⇒
∫ ∧→
=' 2
'4
1
l
l R dl R
a E ρ
πε
Eg. how that Coulom!’s law 3
21
2
21
44 R
Rqq
R
qqa F R
πε πε ==
∧→" where
R
Ra R
= " R R = .
(roof)∵ →
= E q F 2
and E r
qS d E 21 4π
ε
==⋅ →→
∫∫ ,
3
1
2
1
44 R
Rq
R
qa E R
πε πε
== ∧→
∴ 2
21
4 R
qqa F R
πε
∧→=
Eg. Determine the electric field intensity of an infinitely long line charge of a
uniform density ρl in air.
('ol.)
∫∫ ∫ ∫ ==⋅→→
s
L
rLE dz Erd S d E 0
2
02
π
π φ
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 2/24
0
2ε
ρ π
LrLE l = ,
r a E l r
02πε
ρ ∧→
=
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 3/24
Eg. Determine the electric field intensity of an infinite planar charge with a
uniform surface charge density ρs.
('ol.) ∫ ==⋅→→
s EA ES S d E 22 ,
0
2ε
ρ A EA s=
<−=
>=
=⇒∧
∧
→
0,2
0,2
0
0
z z
z z
E s
s
ε
ρ
ε
ρ
Eg. # line charge of uniform density ρl in free space forms a semicircle of radius
b. Determine the magnitude and direction of the electric field intensity at the
center of the semicircle. $高考%
('ol.) φ πε
φ ρ sin
4
)(2
0
⋅−=→
b
bd E d l
y ,b
yd b
y E y E l l y
00
0 2sin
4 πε
ρ φ φ
πε
ρ π ∧∧∧→−=−== ∫
Eg. Determine the electric field caused !y spherical cloud of electrons with a
&olume charge density ρ'- ρ( for b R ≤≤0 )!oth ρ( and b are positi&e* and ρ'(
for R+b. $交大電子物理所%
('ol.)
(a) b R ≥
3
03
4bQ
π ρ −= , 2
0
3
0
2
0 34
R
ba
R
Qa E R R
ε
ρ
πε
∧→
−==
(*) b R ≤≤0
E a E R
∧→= , dS aS d R
∧→= , ∫ ∫ ==⋅
→→
i i s s R E dS E S d E 24π
∫∫∫ ∫∫∫ −=−==v v
RdvdvQ 3
00
3
4π ρ ρ ρ ,
0
0
3ε
ρ Ra E R
∧→
−=
Eg. # total charge Q is put on a thin spherical shell of radius b. Determine the
electrical field intensity at an ar!itrary point inside the shell. $台大電研%
('ol.)
24 b
Q s
π ρ = ,
−=
2
2
2
2
1
1
04 r
dS
r
dS dE s
πε
ρ
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 4/24
α α coscos2
2
2
2
1
1
r
dS
r
dS d ==Ω , 0
coscos40
=
Ω
−Ω
=α α πε
ρ d d dE s
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 5/24
Electric dipole: + pair of e!al *!t opposite char$es -ith separation.
2311
4
)2
()2
(|2
|
2
32/3
2
3
2/32
22/33
Rd R R
Rd R R
d d R R
d R
d R
d R
→→
−−
→→
−
−−
→→
→→
−
→→
⋅+≅⋅−≅
+⋅−=−⋅−=−
2
31|
2|
2
33
R
d R R
d R
→→
−−
→→ ⋅−≅+
−
⋅=
−
⋅≅
+
+−
−
−=⇒
→→→→
→→→→
→→
→→
→→
→→
→
P R R
P R
Rd R
R
d R
R
q
d R
d R
d R
d R
q E
2323333
4
13
4
2
2
2
2
4 πε πε πε
=⋅
−==
→→∧∧∧→θ θ θ θ cossincos Raa z R
+=⇒
∧∧→
θ θ πε
θ sincos24 3
aa R
E R
Eg. #t what &alue of θ does the electric field intensity of a z -directed dipole ha&e
no z -component.
('ol.) )sincos2(4 3
0
θ θ πε
θ
∧∧→
→
+= aa R E r , θ θ θ sincos ∧∧∧ −= aa z r
o z co#ponent 0sinsincoscos2 =⋅−⋅⇒ θ θ θ θ 2tan2 =⇒ θ ⇒! =4. or 12.3
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 6/24
1-, tatic Electric otentials
→→→
∫ ⋅−=−⇒−∇= l d E " " " E
2
112
and ε ρ −=∇ " 2
Electric potential due to a point charge:
R
qdRa
R
qa"
R
R Rπε πε 44 2
=⋅−= ∫ ∞∧∧
)11
(4 12
21 12 R R
q" " "
−=−=πε
Electric potential due to discrete charges:
∑= −
=n
k k
k
R R
q"
1 '4
1
πε
Electric potential due to an electric dipole:
)11
(4 −+
−= R R
q"
πε
5f d 66 R, -e ha7e
+≅
−≅ −
−
+
θ θ cos2
1cos2
1 1
1
R
d R
d R
R
and
−≅
+≅ −
−
−
θ θ cos2
1cos2
1 1
1
R
d R
d R
R
→→
== d q Rqd " ,4
cos2πε θ )(
4 2
" Ra " R
πε
∧→
⋅=⇒
)sincos2(4
3 θ θ
πε θ θ θ
∧∧∧∧→
+=∂∂
−∂∂
−=−∇=⇒ aa R
" a
R
" a" E R R
calar electric potential due to &arious charge distri!utions:
&ol!#e so!rce ρ⇒ ∫∫∫ ='
'4
1
vdv
R"
ρ
πε .
'!rface so!rce ρ s⇒ ∫∫ ='
'41
s
s dS R
" ρ πε
ine so!rce ρl ⇒ ∫ = '4
1dl
R" l ρ
πε
ote: 1. " is a scalar, *!t→
E is a 7ector.
2. " E −∇=→
is 7alid onl" in the static 89 field.
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 7/24
Eg. /!tain a formula for electrical field intensity along the a0is of a uniform line
charge of length L. The uniform line-charge density is ρl . $高考%
('ol.)2
,' L z z z R >−=
∫ − −=
2/
2/0
'
'
4
L
L
l
z z
dz "
πε
ρ ( )( ) 2
,2/
2/ln
4 0
L z
L z
L z l >
−+
=πε
ρ
( )[ ] 2,
2/4 22
0
L z
L z
L z
dz
d" z E l >
−=−=
∧∧→
πε
ρ
Eg. # finite line charge of length L carrying uniform line charge density ρl is
coincident with the x -a0is. Determine V and E in the plane !isecting the line
charge.
('ol.)
−
++
=
+= ∫ − y
L y
L
y #
d#" l
L
L
l ln22
ln24
2
2
0
2/
2/ 22
0πε
ρ
πε
ρ
and " E −∇=→
( )
+=
∧
220 2/
2/
2 y L
L
y y l
πε
ρ
Eg. # charge is distri!uted uniformly o&er an L× L suare plate. Determine V
and E at a point on the a0is perpendicular to the plate and through its center.
('ol.)2 L
Q s = ρ , 222 z y y +→ , ∫ −
+−
+++
=
2/
2/
2222
2
0
ln22
ln2
L
L
s dy z y L
z y L
" πε
ρ
+
⋅−
−+
++
= −
2
2
2
1
2
2
2
2
2
0
22
2tan
222
222
ln2
z L
z
L
z
L z
L
L z
L
L
L
Q
πε
+
=−∇= −∧→
2
2
2
1
2
0
22
2tan
z L
z
L
L
Q z " E πε
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 8/24
Eg. # positi&e point charge Q is at the center of a spherical conducting shell of an
inner radius Ri and an outer radius Ro. Determine→
E and V as functions of the
radial distance R. $高考
%
('ol.) R: Ro,0
24
ε π Q R E S d E
s==⋅∫∫
→
, 2
04 R
Q E
πε = , ∫ ∞ =−=
R
R
Q EdR"
04πε
Ri6 R6 Ro, 0= E ,o R
Q
R R" "
00 4πε =
==
R6 Ri, 2
04 R
Q E
πε = , $
R
Q$ EdR" +=+−= ∫
04πε
−+=⇒
−= ioio R R R
Q
" R R
Q
$
111
4
11
4 00 πε πε
Eg. # charge Q is distri!uted uniformly o&er the wall of a circular tu!e of radius
b and height h. Determine→
E and V on its a0is )a* at a point outside the tu!e" )!*
at a point inside the tu!e.
('ol.) ( ) ( )∫ −+
=
−+
= π
ε
ρ
πε
φ ρ 2
0 22
0
22
0
'2
'
'4
''
z z b
bdz
z z b
dz bd d" s s
(a)( ) ( ) ( ) 22
22
00 22
0
ln2'2
'
h z bh z
z bbb
z z b
bdz " s
h s
o
−++−
++=
−+= ∫ ε
ρ
ε
ρ ,
bh
Q s
π ρ
2=
( )
+−
−+=−=
∧∧→
22220
11
2 z bh z b
b z
dz
d" z E s
oε
ρ
(*) ( ) ( )∫ ∫ −++−+=
h
z
s z
s
i z z b
bdz
z z b
bdz
" 22
0
0 22
0 '2
'
'2
'
ε
ρ
ε
ρ
( ) ( ) ( )
−++−⋅++= 2222
2
0
ln2
z hb z h z b z b
b s s ρ
ε
ρ
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 9/24
Eg. Consider two spherical conductors with radii b1 and b, )b,+b1* that are
connected wire. The distance !etween the conductors is &ery large in comparison
to b, so that charges on spherical conductors may !e considered uniformly
distri!uted. # total charge Q is deposited on the spheres. Find )a* the charges onthe two spheres" and )!* the electric field intensities at the sphere surfaces2
('ol.)
(a)20
2
10
1
44 b
Q
b
Q
πε πε = ,
2
1
2
1
b
b
Q
Q= , QQQ =+ 21
;21
2
2
21
1
1bb
bQand Q
bb
bQ
+=
+=
(*) 2
20
2
22
10
1
1
44
8
b
Q E and
b
Qnn
πε πε
== ,1
2
2
1
2
1
2
2
1
b
b
Q
Q
b
b
E
E =
=
Eg. /!tain a formula for the electric field intensity on the a0is of a circular dis3
of radius b that carries uniform surface charge density ρs. $高考%
('ol.) '''' φ d dr r ds = , 22 'r z R +=
( )∫ ∫ +=
π
φ πε
ρ 2
0 0 2/1220
'''
'
4
b s d dr
r z
r " ( )[ ] z b z s −+=
2/122
02ε
ρ
z " z " E
∂
∂−=−∇=
∧→
( )[ ]
( )[ ]
<++−
>+−
= −∧
−∧
0,12
0,1
22/122
0
2/122
0
zb z z z
zb z z z
s
s
ε
ρ ε
ρ
+s z ::1 ( ) 2
22/122
21 z
bb z z −≅+⇒
−,
( )2
0
2
4 z
b z E s
πε
ρ π ∧→
=
<−
>
=∧
∧
04
04
2
0
2
0
z z
Q z
z z
Q z
πε
πε
Eg. 4a3e a two-dimensional s3etch of the euipotential lines and the electricfield lines for an electric dipole.
('ol.)
For an electric dipole, 2
04
cos
R
qd "
πε
θ = =constant⇒ θ cosvc R =
E k l d = , -here k is a constant.
++=++
∧∧∧∧∧∧
φ φ θ θ φ θ φ θ θ E a E a E ak d Ra Rd adRa R R R sin
φ θ
φ θ θ
E
d R
E
Rd
E
dR
R
sin
== ,θ θ
θ sincos2 Rd dR = , θ 2sin E c R =
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 10/24
1-5 4agnetic Fields
4agnetic field: % & &
'
−==0 µ µ
, -here (0=4<10 ( A/m) in the free space.
4agnetic flu0 density: ∫ ×
=$
R
R
ad ) & 2
0
4
π
µ
#mpere’s law of ' : ) l ' =⋅∫ d ⇔ * ' =×∇
% * * * & m
o
×∇+=+=×∇
µ
1 or * %
& =−×∇ )(
0 µ ,
) l d ' % &
' =⋅⇒−= ∫
0 µ , and ' ' % ' & r m µ µ χ µ µ 000 )1()( =+=+= ,
-here 01 µ
µ
χ µ =+= mr
Gauss’s law of & : ∫∫ =⋅⇔=⋅∇S
S & & 0d0
∵ 0=⋅∇ & , A∃ f!lfills A & ×∇=
A A A * &2
)( ∇−⋅∇∇=×∇×∇==×∇ µ
>hoose * A A µ −=∇⇒=⋅∇ 20 (ote: ε
ρ −=∇ " 2
is scalar oisson?s e!ation)
∵ '
4
1
'
dv
R
" "
∫∫∫ =
ρ
πε ,∴ '
4 '
dv
R
* A
" ∫∫∫ =
π
µ (+b/m)
4agnetic flu0: ∫∫ ∫ ∫∫ ⋅=⋅×∇=⋅=ΦS $ S
l d AS d AS d & ')( (+b)
6iot-a&art’s law: ∫ ∫ ×
=×
='
32
'
4
'
4$
R
R
Rl d )
R
al d ) &
π
µ
π
µ
∫ ∫∫∫ ==''
4d7'
4$ "
R
l d )
R
* A
π
µ
π
µ , ∵ , - , - , - ×∇+×∇=×∇ )()( ),
∴
∫ ∫ ∫ ∫ ×=
×∇+×∇=×∇=
×∇=×∇=
'
2
''
'4
')1('14
)'(4
'4
$
R
$ $ R
al d ) dl R
dl R
) R
l d ) R
l d ) A &π
µ π
µ π
µ π
µ
ote:
=×∇⇒++=
∂
∂+
∂
∂+
∂
∂=∇
0'''''
l d dz zdy yd# #l d
z z
y y
# #
, and then
32
'
4
)'
(
4 R
Rl d )
R
al d ) &d R ×
=×
=
π
µ
π
µ
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 11/24
Eg. # direct current I flows in a straight wire of length , L. Find the magnetic
flu0 density & at a point located at a distance r from the wire in the !isecting
plane.
('ol.) ')'('' rdz a z z r adz z Rl d r φ =−×=× ,
21
22 )( r z R +=
∫ − +=
+=
L
L
oo
r Lr
)La
r z
rdz ) a &
2223222
)'(
'
4
π
µ
π
µ φ φ
Eg. Find the magnetic flu0 density at the center of a suare loop" with side w
carrying a direct current I .
('ol.)2
. L = ,
2
.r = in this case,
.
) z
...
. )
z & o
o
π
µ
π
µ 22
)2
()2
(2
2
2422
=+
×=
Eg. Find the magnetic flu0 density at a point on the a0is of a circular loop of
radius b that a direct current I .
('ol.) '' φ φ bd al d = , ba z z R r −= ,
2122)( b z R +=
'@'' 2 φ φ d bbzd a Rl d r +=×
2322
22
0 2322
2
)(2'
)(
4 b z
)b z d
b z
b z bz a ) &
or o
+=
+
+= ∫
µ φ
π
µ π
. 5n case of z =0,
b
) z & o
2 µ
=
Eg. Determine the magnetic flu0 density at a point on the a0is of a solenoid with
radius b and length L" and with a current in its N turns of closely wound coil.
('ol.) = &d ')(
.)'(2
23
22
2
0 dz L
/
b z z
)b z
+−
µ
[ ]
+−
−−
+=
++
+−
−== ∫ 22222221220
)(2)(2 b L z
L z
b z
z
L
/)
b z
z
b z L
z L
L
/) &d & oo
L µ µ
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 12/24
#mpere’s law of & : ∫ =⋅⇔=×∇c
) l d & * µ µ A , -here (= (0 in the free space.
Eg. #n infinitely long" straight conductor with a circular cross section of radius b
carries a steady current I . Determine the magnetic flu0 density !oth inside andoutside the conductor. $交大光電所%
('ol.)
(a) 5nside the cond!ctor, br ≤ %
∫ ∫ ====⋅1
2
2
22
0)()(2d
$
oo ) b
r )
b
r r& &rd l & µ
π
π µ π φ
π
=: 22A
b
r) a
o
π
µ φ =
(*) B!tside the cond!ctor%r
) a & ) r&l d &
o
$
o
π
µ µ π φ
22
2
=⇒==⋅∫ Eg. # long line carrying a current I folds !ac3 with semicircular !end of radius
b. Determine magnetic flu0 density at the center point P of the !end. $高考%
('ol.)
21 & & & += , -hereb
) z & o
π
µ
421 ⋅=
,
b
) z & o
4
2
µ =
Eg. # current I flows in the inner conductor of an infinitely long coa0ial line and
returns &ia the outer conductor. The radius of the inner conductor is a" and the
inner and outer radii of the outer conductor are b and c" respecti&ely. Find the
magnetic flu0 density & for all regions and plot & &ersus r . $高考電機技師%
('ol.) ar ≤≤0 , 22
a
r) a &
π
µ φ = , br a ≤≤ ,
r
) a &
π
µ φ
2=
cr b ≤≤ ,r
)
bc
r ca &
π
µ φ
2)(
22
22
−
−=
Eg. Determine the magnetic flu0 density inside an infinitely long solenoid with
air core ha&ing n closely wound turns per unit length and carrying a current I .
('ol.) nL) &L o µ = =: n) & o µ =
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 13/24
Eg. The figure shows an infinitely long solenoid with air core ha&ing n closely
wound turns per unit length. The windings are slanted at an angle
α and carry a
current I . Determine the magnetic flu0 density !oth inside and outside the
solenoid.
('ol.)
>=
<<=
br r
bn) a
r
bn) a
br
&,
sin
2
sin2
0,0
001 α µ
π
α πµ φ ϕ
,
>
<<=
br
br n) z &
o
,0
0,cos
2
α µ ,
21 & & & +=
Eg. Determine the magnetic flu0 density inside a closely wound toroidal coil with
an air core ha&ing N turns and carrying a current I . The toroid has a mean
radius b" and the radius of each turn is a.
('ol.) /) r&l o µ π ==⋅∫ 2dA
r
/) a &a &
o
π
µ φ φ
2 ==
, (ba)6r 6(bCa), 0= & for r 6(ba) and
r :(bCa)
Eg. 7n certain e0periments it is desira!le to ha&e a region of constant magnetic
flu0 density. This can !e created in an off-center cylindrical ca&ity. The uniform
a0ial current density is * z * = . Find the magnitude and direction of & in the
cylindrical ca&ity whose a0is is displaced from that of the conducting part !y a
distance d . 台大電研、清大電研、中原電機]
('ol.) * z * = , ) l & o µ =⋅∫ d
5f no hole eDists,
=
−=
⇒=⇒=
11
111
1
2
111
2
2
22
# *
&
y *
&
* r
& * r &r
o
y
o
#o
o
µ
µ
µ π µ π φ φ
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 14/24
For * in the hole potion,
−=
=
⇒−=22
22
2
2
2
2
2 #
* &
y *
&
*
r
&o
y
o
#o
µ
µ
µ
φ
+t 21 y y = and d # # += 21 021 =+=⇒ # # # & & & , and d
* & & & o y y y
221
µ =+=
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 15/24
1-8 Electromagnetic Forces
9orent force euation: )( &v E q F ×+=
Electric force: E q F e = . 4agnetic force: A×= vq F m
Eg. #n electron is in;ected with an initial &elocity
oo v yv =
into a region where!oth an electric field and a magnetic field & e0ist. Descri!e the motion of the
electron if o E z E = and o & # & = . Discuss the effect of the relati&e magnitude of
E ( and B( on the electron paths in parts.
('ol.) )( &v E et
vm ×+−=
∂∂
,
=−=
+−=
=
⇒
−−=
∂
∂
−=∂
∂
=∂
∂
⇒
=
=
oo
o
o
z
o
o
o
o
o y
#
yoo z
zo y
#
o
o
&
m
etv
R
Ev
&
Et
&
Evv
v
v & E
m
e
t
v
v &m
e
t
v
t
v
& # &
E z E
00
0
Esin)(
cos)(
0
)(
0
ω ω
ω
(5fo
oo &
E v ≠ )
22222
22
2 )()()(
),cos1(
sin
0
ooo
o
o
o
oo
o
o
o
o
o
cc zt
&
E y
&
E vct
c z
t &
E t
c y
#
ω ω
ω ω
ω ω
=++−⇒
−=−−=
+=
=
⇒
4agnetic force due to & and I :
& )d &d dt
dq &
dt
d dqdF &vq F mm
×=×=×=⇒×= , ∴ ∫ ×=$
m &d ) F
Eg. Determine the force per unit length !etween two infinitely long parallel
conducting wires carrying currents I 1 and I , in the same direction. The wires are
separated !y a distance d . $清大電研%
('ol.) )(' 12212 & z ) F ×= ,d ) ) y F
d ) # &
π µ
π µ
2'
2 210
1210
12 −=⇒−=
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 16/24
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 17/24
Eg. Calculate the force per unit length on each of three euidistant" infinitely
long" parallel wires d apart" each carrying a current of I in the same direction.
pecify the direction of the force.
('ol.)
d
) y )& y & ) z -
d
) # & # & ) ) ) )
π
µ
π
µ
2
3,
2
330cos2,
2
0
222
0
122321 −=−=×−======
Eg. The !ar ##’" ser&es as a conducting path for the current I in two &ery long
parallel lines. The lines ha&e a radius b and are spaced at a distance d apart. Find
the direction and the magnitude of the magnetic force on the !ar. $中山物理所%
('ol.)
)
11
(4 0
yd y
)
z & −+−= π
µ
, dy yd =
)1ln(2
)11
(4
2
0
2
0 −−==⇒−
+−=×=⇒ ∫ −
b
d ) # F d F dy
yd y
) # & )d F d
bd
b π
µ
π
µ
#pplication of the electric forces: he epaper
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 18/24
calar electric potential function: ∫∫∫ ='
'4
1
"
dv R
" ρ
πε
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 19/24
<ector magnetic potential function: ∫∫∫ ='
'4
"
dv R
* A
π
µ
=etarded potentials:
∫∫∫ −
='
')(
4
1),(
"
dv R
v Rt t R"
ρ
πε , ')/(4
),('
dv R
v Rt * t R A" ∫∫∫
−=
π µ
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 20/24
1-> Faraday’s 9aw and 4agnetic Dipoles
Faraday’s law: t
& E
∂∂
−=×∇
or ∫ ∫∫ ⋅∂∂
−=⋅$ S
S d t
&d E
∵ 0)()( =∂∂+×∇⇒×∇∂∂−=∂∂−=×∇t
A E At t
& E
∴ ∃" f!lfillst
A" E "
t
A E
∂
∂−−∇=⇒−∇=
∂
∂+
ote: 5n static field% " E −∇= , *!t in ti#e7ar"in$ field%t
A" E
∂∂
−−∇=
Emf : ∫ ⋅= l d E " . 4agnetic flu0: ∫∫ ⋅=ΦS
S d & , ∫ ∫∫ ⋅∂∂
−=⋅$ S
S d t
&d E
⇒
dt
d
"
Φ
−=
4otional emf : ∫ ⋅×=$
l d &v" )(' ("olt )
l d &vl d E " E &vq
F &vq F
$
mmm
m
⋅×=⋅−=⇒−≡×=⇒×= ∫ ∫ )('
Eg. # circular loop of N turns of conducting wire line in the xy-plane with its
center at the origin of a magnetic field specified !y t br & z & ω π sin)2cos(0= "
where b is the radius of the loop and ω is the angular freuency. Find the emf
induced in the loop.
('ol.) )sin()12
(G
2)sin()2
cos( 0
2
00 t &
brdr z t
b
r & z
b
ω π
π π ω
π −=⋅=Φ ∫
)cos()12
(G
0
2 t &b /
dt
/d " ω ω
π
π −−=
Φ−=
Eg. # metal !ar slides o&er a pair of conducting rails in a uniform magnetic field
0 & z & = with a constant &elocity . )a* Determine the open-circuit &oltage V ( that
appears across terminals 1 and ,. )!* #ssuming that a resistance R is connected!etween the two terminals" find the electric power dissipated in R. eglect the
electric resistance of the metal !ar and of the conducting rails. $交大電子物理
所%
('ol). (a)
∫ −=⋅×=−='1
'200210 )()( hv&dl y & z v #" " " (" )
(*)
R
hv& R
R
hv& R ) P e
2
0202 )()( === (+ )
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 21/24
Eg. The circuit in Fig. is situated in a magnetic field )3
210cos(3 #t z & π π −=
!" . #ssuming R'1>?" find the current # . $中山物理所%
('ol.) ∫ ⋅−×=Φ −
6.0
06 )2.0(10)
3210cos(3 d# #t π π
)10cos()6.03
210cos(4 9 t t
dt
d " π π π −−×=
Φ−= ,
)2.010sin(6.12
π π −== t R
" i
Eg. # conducting sliding !ar oscillates o&er two parallel conducting rails in a
sinusoidally &arying magnetic field )cos( t z & ω = " . The position of the sliding
!ar is gi&en !y x '(.5>)1-cosω$ *" and the rails are terminated in a resistance
R'(.,?. Find # .
('ol.) ).0(2.0cos #t −⋅=Φ ω , #=0.3(1cos0t ), dt
d
Ri
Φ−= 1
)cos21(sin.1 t t i ω ω ω +⋅=⇒
Eg. The Faraday dis3 generator consists of a circular metal dis3 rotating with a
constant angular &elocity @ in a uniform and constant magnetic field of flu0
density 0 & z & = that is parallel to the a0is of rotation. 6rush contacts are the
open-circuit &oltage of the generator if the radius of the dis3 is b.
('ol.) ∫ ∫ ∫ ==⋅×=⋅×=2
)()()(2
00
0
4
3 00
b &rdr &dr a & z r al d &v"
br
ω ω ω φ
(" )
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 22/24
4agnetic dipole moment: m ' m z )S z = , -here S is the area of the loop that carries
) and m= )S .
<ector potential of a magnetic dipole: 2
0
4
R
am A R
π
µ ×=
, -here
( )θ θ π
µ θ sincos2
4 3 aa
R
m A & R
o +=×∇=
Eg. For the small rectangular loop with sides a and b that carries a current I .
Find the &ector magnetic potential A at a distant point P ) x " y" z *. #nd determine
the magnetic flu0 density & and A . $交大光電所%
('ol.)
2
0
4
R
am
A
R
π
µ ×
=
, -here m= )ab, ( )θ θ π
µ θ sincos24 3 aa R
m
A & R
o
+=×∇=
4agnetiation &ector:
"
m
%
"
k
k
" ∆=
∑∆
=
→∆
1
0li#
( A/m), -here k m is the
#a$netic dipole #o#ent of an ato#.
×∇−×∇=∇×=
×=
R
% %
Rdv
R % dv
R
a % Ad
oo Ro ''1
4')
1('
4'
4
2
π
µ
π
µ
π
µ
∫∫∫ ∫∫∫
×∇−
×∇=⇒
' '
'4
''
4 " "
oo dv R
% dv
R
% A
π
µ
π
µ ( ''d'
'
dS F v F s"
×−=×∇
∫∫ ∫∫∫ )
∫∫ ∫∫∫ ×+
×∇= 'd
'
4'd
'
4'
S R
a % v
R
% no
"
o
π
µ
π
µ
∴ 9a$neti@ation 7ol!#e c!rrent densit"% % * m ×∇= ( A/m2)
9a$neti@ation s!rface c!rrent densit"% nms a % * ×= ( A/m)
Eui&alent 4agnetiation Charge Densities:
'd)'(
4
1'd
'
4
1
''
v
R
% S
R
a % "
" S
n
m ∫∫∫ ∫∫ ⋅∇
−+⋅
=
π π
(ote:
(∫∫∫ ∫∫
⋅∇−+
⋅=
'
''
4
1'
'
4
1
" o
n
o
dv R
P dS
R
a P "
πε πε )
Hefine the #a$neti@ation s!rface char$e densit" as nms a % ⋅= ρ and the
#a$neti@ation 7ol!#e char$e densit" as % m ⋅−∇= ρ
Eg. # circular rod of magnetic material with permea!ility ! is inserted coa0ially
in the long solenoid. The radius a of the rod is less than the inner radius b of the
solenoid. The solenoid’s winding has n turns per unit length and carries a
current I . )a* Find the &alues of &
" '
" and %
inside the solenoid for r Aa and
for aAr Ab. )!* Bhat are the eui&alent magnetiation current densities % m and % ms
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 23/24
for the magnetied rod2 $清大電研%
('ol.) (a) r 6 a% n) z ' = , n) z & µ = , n) z ' &
% )1(00
−=−= µ
µ
µ ,
a 6 r 6 b% n) z '
=, n) z &
o
µ = , 0
= %
(*) 0' =×∇= % * m , n) an) a z a % * r nms )1()1)((00
−=−×=×= µ
µ
µ
µ φ
Eg. # ferromagnetic sphere of radius b is uniformly magnetied with a
magnetiation 0 % z % = . )a* Determine the eui&alent magnetiation current
densities m * and ms * . )!* Determine the magnetic flu0 density at the center of
the sphere. $台大電研%
('ol.) (a) 0' =×∇= % * m , θ θ θ φ θ sin)sincos( o Ro Rms % aa % aa * =×−=
(*) θ µ θ θ µ 3
23
2
2
sin2
)(2
)sin)(d( oomso % zb
bb * z &d == , 000
3
32dsin
2 % z % z & oo µ θ θ µ π == ∫
.
Eg. Determine the magnetic flu0 density on the a0is of a uniformly magnetied
circular cylinder of a magnetic material. The cylinder has a radius b" length L
and a0ial magnetiation 0 % z % = . $台大物研%
('ol.) 0' =×∇= % * m " 00)( % aa % z a % * r nms φ =×=×= " 'dz * d) ms=
[ ]
+− −−+=+−= ∫ 222200
0 23
22
2
00
)(2
)'(2
'
b L z
L z
b z
z % z
b z z
dz b % z & L µ µ
7/23/2019 new EM1
http://slidepdf.com/reader/full/new-em1 24/24
Eg. # cylindrical !ar magnet of radius b and length L has a uniform
magnetiation 0 % z % = along its a0is. se the eui&alent magnetiation charge
density concept to determine the magnetic flu0 density at an ar!itrary distant
point. $交大電子所
%
('ol.)
−=⋅=
side.all
bottom %
to %
a % nms
,0
,
,
0
0
ρ " ρ#=0 in the interior re$ion
0
22 % bbq msm π ρ π == )11
(4
),,(−
−+
=⇒ R R
q z y #" m
mπ
( A)
2
2
2 4
cos)(
4
cos
R
L % b
R
Lq om
π
θ π
π
θ
== 24
cos
R
% 1
π
θ
= , -here 0
2
L% b % 1 π =
)sincos2(4 3
θ θ π
µ µ θ aa
R
% " & R
1 omo +=∇−= (1 )
>onsider an infinitel" lon$ solenoid -ith n t!rns per !nit len$th aro!nd to create a
#a$netic fieldE a 7olta$e " 1= dt d n Φ− is ind!ced !nit len$th, -hich opposes the
c!rrent chan$e. o-er P 1=" 1 ) per !nit len$th #!st *e s!pplied to o7erco#e this
ind!ced 7olta$e in order to increase the c!rrent to ) . he -orI per !nit 7ol!#e
re!ired to prod!ce a final #a$netic fl!D densit" &f is + 1= ∫ - &
'd&0
.