Post on 16-Jan-2016
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Moving Electrical Charge
Magnetic Field
Moving Electrical Charge
BvqFB
BLiFB
dd
BLiFB
1
1
nq
iB
nqVH
The Hall Effect
The net torque on the loop is not zero.
ˆ =iAn B B
The magnetic force on moving charge
Moving Electrical Charge ?B
Hall coefficient
magnetic dipole moment
Chapter 33 The Magnetic Field of a Current
The Electric Field Due to a Charge
2
1 ,
rqE
q
r P r
r
qE ˆ
4
20
q
r P
The Magnetic Field Due to a Motion Charge
v
2
1 ,
rqB
20 sin
4 r
qvB
m/AT104 70
μ0 is the permeability constant.
m/AT104
70
sin , , v3
02
0
4
ˆ
4 r
rvq
r
rvqB
)4
1ˆ
4
1(
30
20 r
rq
r
rqE
q
r P
The Magnetic Field Due to a Motion Charge
v
2
1 ,
rqB sin , , v
30
20
4
ˆ
4 r
rvq
r
rvqB
The Magnetic Field of a Current
ddd
d
d
ddd sis
t
q
t
sqvq
30
20
4
ˆ
4 r
rvdq
r
rvdqBd
30
20 d
4
ˆd
4d
r
rsi
r
rsiB
Biot-Savart law
A straight wire segment
30 sind
4d
r
zriB
30 d
4 r
zdi
2
322
0
)(
d
4 dz
zid
2030 2 2 2
2dd
4 ( )
Lid zB B
z d
2
1220
)4(4
dL
L
d
i
:direction
d
iBdL
2
, 0
Two Parallel Current
Superposition principle
Vector sum
0 11 1
1
ˆ 2
iB n
r
0 22 2
2
ˆ 2
iB n
r
21 BBB
The force exerted by one wire on another
L
dFi
0
212 2
The definition of the ampere
d
iB
2
101
1221 BLiF
d
LiiLBiF
2210
1221
•dib
ia
F dib
ia
F
•
Parallel currents attract, and antiparallel currents repel.
21 ii
AimLmd
NF
1 ,1 ,1
,102When 7
xd
iBy
d
2d 0
xd
axi
)/d(
20
yy dBB
Example
xPd
y
2/
2/
0
2
a
a xd
dx
a
i
2/
2/ln
20
ad
ad
a
i
Example
d
iB
d
2d 0
seccos RRd
d
axi )/d(
20
cosdd BBx
xx dBB
2
0
sec2
)/d(
R
axi
cos
sec2
)/(0
R
adxi
sec2
)/(2
0
R
adxi
tanx R
0 d 2
i
a
d2
0
a
iBx
a
i0R
a
a
i
2tan 10
R
iBRa
R
atRa
2
,2/2
an , 01- a
iBR
2/2,0, 0
dsecd 2Rx
The Magnetic Field of a Current
30
20 d
4
ˆd
4d
r
rsi
r
rsiB
Biot-Savart lawA circular current loop
0d B
r
R
r
siBB z
20
4
dd
szR
iR Rd
)(
1
4
2
0 2322
0
;)(2 2
322
20
zR
iR
R
iBz
2 0,when 0
3
20
2 , If
z
iRBRz
3
20
2
)(
z
RiB
magnetic dipole moment
,cosdd BBz
30
2
z
pB m
304
1 ,
x
pEqdp e
e
The Magnetic Field of a Solenoid
The Magnetic Field of a Solenoid
2322
20
])([2
)d(d
dzR
RzniB
;)(2 2
322
20
zR
iRB
2/
/2- 2322
20
])([
d
2
L
L dzR
zniRB
)])2([
2
])2([
2(
2 21222
122
0
dLR
dL
dLR
dLni
If L>>R,
The field outside the ideal solenoid is zero.
B=μ0ni.
The Magnetic Field of a Solenoid
2322
20
])([2
)d(d
dzR
RzniB
;)(2 2
322
20
zR
iRB
2/
/2- 2322
20
])([
d
2
L
L dzR
zniRB
The field outside the ideal solenoid is zero.
z
B
-L/2 L/2
solenoid for magnetic field
parallel-plate capable for electric field
E B
E=. 0B nI
The Electric Field Due to a Charge
rr
qE ˆ
4
20
The Magnetic Field Due to a Moving Charge
20 ˆ
4 r
rvqB
20 ˆd
4d
r
rsiB
0q
sdE
Gauss’ LawGauss’ Law0 sdB
Ampere’s Law 0d
s
lE ildB 0
Ampere’s Law
loop, Amperean is soughpassingthrcurrent net is i
loop. theboundedby surface
ildB 0
LL
lBlB dcosd
d
20 rr
IL
2
0
0 d2
I
I0
LL
lBlB dcosd
dcosd rl
d
20 rr
IL
2
0
0 d2
I
I0
dcosd rl
'cos''cos
dd
dlBBdl
lBlB
0 0( ) ( ' )2 2 '
I Ird r d
r r
0
d 0 isBs
Ampere’s Law
d
iB
idBsBsB
2
)2(dd
0
0
A straight wire segment
2
2 2322
0
)(
d
4d
L
Ldz
zidBB
d
iBdL
2
,when 0
2122
0
)4(4
dL
L
d
i
Application of Ampere’s law
Long, straight wire (r<R)
2
2
0)2(dR
rirBsB
20
2 R
irB
B
rR
irBsB 0)2(d
Rr
r
iB
2
0
A solenoid
nhisB 0d
From loop2, hB1+ (-h)B2=0
dacdbcab
sB
d
niB 0
loop2B1
B2
ab
sBd Bh
nhiBh 0
B1= B2
A toroid
nir
iNB
iNrBsB
00
0
2
)2(d
Same as a solenoid
The field outside a solenoid
02outB r i
0
0
2out
in
B i r
B ni
0
2out
iB
r
rn2
1
1, 0outout
in
BB
B
0inB ni
ExercisesP768-769 10, 13, 15ProblemsP772 8, 9
A interesting question for interaction force between electric / magnetic field and moving or resting charged particle
23/4/21 zwma@zju.edu.cn 25
A charged particle at rest Electric fieldMoving charged particle Both electric and magnetic field
Whether a charged particle moves or not depends on the chosen frame.
Whether can we conclude thatboth electric and magnetic fieldDepend on the chosen frame?
Einstein’s Postulates: The laws of physics are the same in all inertial reference frames.
Are these two conclusions contradictory?Is it true?
23/4/21 zwma@zju.edu.cn 26
In the frame K（ at rest）
In the frame K’
（ with moving electrons）
S－－ － －－ －
＋＋ ＋ ＋＋ ＋
Ｉ
x
y
r
S－－ － －－ －
＋＋ ＋ ＋＋ ＋
Ｉ
x’
y’
r
In the frame K
In the frame K’
Both electric and magnetic force are zero.
Magnetic force：
Electric force：
Fm We have：
Therefore
)(1
1
)(1
1
2
2
yzz
zyy
xx
vBEE
vBEE
EE
)(1
1
)(1
1
22
22
yzz
zyy
xx
Ec
vBB
Ec
vBB
BB
Transformation equations of E and B:
In different inertial reference frame, the results come from the electric or magnetic field is same.