lesson 2 sources of magnetic fields
TRANSCRIPT
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SOURCES OF
MAGNETIC FIELD
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28.1
Magnetic Field of a Moving Charge
P
Consider
a
charge
q
moving
with
a
constant
velocity
v
.
The
location
of
the
charge
at
any
given
instant
is
called
the
source
point
.
The
point
P
where
the
magnetic
field
due
to
the
charge
is
being
evaluated
is
called
the
field
point
.
r
r
is
the
displacement
vector
from
source
point
to
field
point
.
is
the
angle
between
r
and
v
.
x
y
q
v
+
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28.1
Magnetic Field of a Moving Charge
The
direction
of
the
magnetic
field
at
point
P
is
perpendicular
to
both
r
and
v
and
can
be
determined
using
the
right
hand
rule
.
P
r
B
x
y
q
v
+
0,// FthenBvIf
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28.1
Magnetic Field of a Moving Charge
qB
2
1
rB
vB
sinB0,// FthenBvIf
Experimental
results
about
the
magnitude
of
B
.
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The magnetic field
B
due to a moving charge
q
is
2
sin
4 r
vqB o
w
here:
*104 .7
AmT
o
ATmo 7
1014
28.1
Magnetic Field of a Moving Charge
0,// FthenBvIf
*permeability of free space is themeasure of the amount of resistanceencountered when forming a
magnetic field in a vacuum.
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The magnetic field
B
due to a moving charge
q
is
2
sin
4 r
vqB o
28.1
Magnetic Field of a Moving Charge
2
4 r
rvqB o
w
here:
r
r
r
Vector Magnetic Field
0,// FthenBvIf
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0,// FthenBvIf
28.1 Magnetic Field of a Moving Charge
NOTES:
B = 0 at all points alongthe line through the chargeparallel to the direction ofvelocity. sin = 0
at any distance r from
the charge, B is maximumat all points lying in theplane perpendicular to thedirection of velocity.sin = 1
if the charge is negative,the direction of B isopposite that for a positivecharge.
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2
r
qqkFE
2
2199
)10.0(
)106.1()109( 2
2
m
CF
C
NmE
upwardNFE ,1030.226
Example 1. Two protons move parallel to the x-axis in oppositedirections at the same speed of1.00 x 105m/s. At the instantshown in the figure, where r = 10 cm, find the electric and
magnetic forces (magnitude and direction) on the upper proton.
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2
sin
4 r
qvB o
2
5197
)10.0()/101)(106.1()101(
msmCB
ATm
TB 19106.1
qvBFM
)106.1)(/101)(106.1( 19519 TsmCFM
upwardNFM ,1056.233
Example 1. Two protons move parallel to the x-axis in oppositedirections at the same speed of1.00 x 105m/s. At the instantshown in the figure, where r = 10 cm, find the electric and
magnetic forces (magnitude and direction) on the upper proton.
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Consider a segment dlof a current carrying conductor.
2
sin
4 r
vdQdB o
2
sin
4 r
dt
dldQ
dB o
2
sin
4 r
dldt
dQ
dB o
28.2 Magnetic Field of a Current Element
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28.2 Magnetic Field of a Current Element
The magnetic field due to a current element is
2
sin
4 r
dlIdB o
24 r
rldIBd
o
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Law ofBiot-Savart
The magnetic field ofa current-carrying conductor is
2sin
4 r
dlIB o
24 r
rldI
Bo
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Example 2. Acopper wire carries a steady current of125A. Findthe magnetic field caused by a 1.00-cm segment of this wire at apoint 1.2 m from it ifthe point is (a) point P1straight to the side ofthe segment, and (b) point P
2on a line at 30oto the segment
2
sin
4)(
r
dlIBa o
TB 8107.8
2
7
)2.1(90sin)01.0)(125()101(
mmAB
ATm
TB 8103.4
2
7
)2.1(
30sin)01.0)(125(
)101()( m
mA
Bb ATm
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Consider a conductor of length 2a carrying a current I.
Find the magnetic field Bat point Pat a distancexfrom theconductor on its perpendicular bisector.
22 yxr
22)sin(
yx
x
dydlLet
sin4
2r
dlIB o
28.3 Magnetic Field of a Straight Current-Carrying Conductor
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sin4
2r
dlIB o
a
ao
yx
x
yx
dyIB 22224
a
a
o
yx
xdyIB
23
224
28.3 Magnetic Field of a Straight Current-Carrying Conductor
22
2
4 axx
aIB o
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If the length 2a>>x, theconductor can be considered tobe infinitely long.
Ifa>>x, xa
aI
Bo 2
4
28.3 Magnetic Field of a Straight Current-Carrying Conductor
22
24 axx
aIB o
rIB o
2
x
IB
o
2
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Example 3. The figure shows an end-on view oftwo longparallel straight wires perpendicular to the xy-plane, eachcarrying a current of10 A but in opposite directions. Findthe magnitude and direction ofthe magnetic field at pointsP1, P2, and P3.
x
y
0
10cm
20cm
-10cm
-30cm
I1
=1
0A
P1 P2 P3I2
=1
0A
B2
B1
r
IB
o
2For P1:
mAxB
ATm
20.010)102( 71
jTxB 101 51
m
AxB
ATm
40.0
10)102(
7
2
jTxB 105 62
jTxjTxBnet 10510165
jTxBnet 1056
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Example 3. The figure shows an end-on view oftwo longparallel straight wires perpendicular to the xy-plane, eachcarrying a current of10 A but in opposite directions. Findthe magnitude and direction ofthe magnetic field at pointsP1, P2, and P3.
r
IB
o
2For P2:
mAxB
ATm
10.010)102( 71
jTxB 102 51
m
AxB
ATm
10.0
10)102(
7
2
jTxB 102 52
jTxjTxBnet 10210255
jTxBnet 1045
x
y
0
10cm
20cm
-10cm
-30cm
I1
=1
0A
P1 P2 P3I2
=1
0A
B1
B2
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Example 3. The figure shows an end-on view oftwo longparallel straight wires perpendicular to the xy-plane, eachcarrying a current of10 A but in opposite directions. Findthe magnitude and direction ofthe magnetic field at pointsP1, P2, and P3.
r
IB
o
2For P3:
mAxB
ATm
30.010)102( 71
jTxB 1067.6 61
m
AxB
ATm
10.0
10)102(
7
2
jTxB 102 52
jTxjTxBnet 1021067.656
jTxBnet 1033.15
x
y
0
10cm
20cm
-10cm
-30cm
I1
=1
0A
P1 P2 P3I2
=1
0A
B2
B1
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Consider two long and straight parallel current-carryingwires. Determine the magnitude and direction ofthe magnetic
force they exert on each other.
28.4 Magnetic Force Between Parallel Conductors
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I1 I2
B1
B2
F12
F21
28.4 Magnetic Force Between Parallel Conductors
Force per unit length exerted
by wire 1 on wire 2 (F12)
r
1212 LBIF
r
ILIF o
2
1212
r
IIx
L
FATm 21712 )102(
Force per unit length exertedby wire 2 on wire 1 (F
21
)
2121 LBIF
r
ILIF o
2
2121
r
IIx
L
FATm 21721 )102(
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I1 I2
B1B2
F12F
21
28.4 Magnetic Force Between Parallel Conductors
If the currents in the conductors are in opposite directions:
rx
Two parallel wires with currents in the same direction attracteach other.
Two parallel wires with currents in opposite directions repel
each other.
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Example 4. Find the magnitude and direction of the Fnetper unitlength on wire 3 due to wires 1 and 2.
x
y
8 cm
I3 = 20 A
6 cm
I2 = 5 AI1 = 10 A
F13
F23
Fnet
mN
ATm
m
AA
L
F
4
713
104
10.0
)20)(10()102(
mN
ATm
m
AA
L
F
4
723
1033.3
06.0
)20)(5()102(
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x
y
8 cm
I3 = 20 A
6 cm
I2 = 5 AI1 = 10 A
F13
F23
Fnet
mN
mNx
L
F
4
4
102.3
0)8.0)(104(
mN
mN
mNy
L
F
5
44
103.9
1033.3)6.0)(104(
mN
mN
mNnet
L
F
4
2524
1033.3
)103.9()102.3(
xaboveo
2.16
00032.0
000093.0tan
1
Example 4. Find the magnitude and direction of the Fnetper unitlength on wire 3 due to wires 1 and 2.
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28.5 Magnetic Field of a Circular Current Loop
Consider a circular current loop ofradius a. Use the LawofBiot and Savart to find the magnetic field at point Pon the
axis ofthe loop.
24 r
dlIdB o
2222
2
)(4
cos4
ax
a
ax
dlI
r
dlI
dB
o
o
x
2222
2
)(4
sin4
ax
x
ax
dlI
r
dlIdB
o
oy
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MAGNETIC FIELD OF A CIRCULAR CURRENT LOOP
0yB
a
ox dl
ax
IaB
2
022
2
3
)(4
xBB
)2()(4 2
322
aax
IaB ox
23
)(2 22
2
ax
INaB o
Where N is the number ofturns
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MAGNETIC FIELD AT THE CENTER OF A CIRCULARCURRENT LOOP
23
)(222
2
axIaB o
Ifx=0, or at the center ofthe loop
a
IB
o
2
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Example 5.
(a)
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Example 5.
(b)
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MAGNETIC FIELD OF OTHER CONDUCTORS
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2
sin
4 r
dlIdB o
(a)
27 )05.0(90sin)0011.0)(10(101
mmAxdB
ATm
TxBd7
1040.4
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(b)
27 )149.0(7.19sin)0011.0)(10(101
mmAxdB
ATm
TxBd 81067.1
mmmr 149.014.005.0 22 o
7.19tan1451
(c) 00 becausedB
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For the 12-A wire:
Txm
mAxBd cm
cm
ATm 8
2
85.2
7 1079.8)08.0(
)0015.0)(12(101
For the 24-A wire:
Txm
mAxBd cm
cm
ATm 7
2
85.2
7 1076.1)08.0(
)0015.0)(24(101
x
TxBd net81079.8
x
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eastTxm
Ax
r
IB
ATmo ,1091.2
50.5
800)102(
2
57
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I1=25A
B1
B2
(a)
40cm -x
I2=75A
x
21 BB
2
2
1
1
22 r
I
r
I oo
xcm
A
x
A
40
7525
3
40
x
xcm
cmx 10
Bnet = 0 at a point 10cm from I1and 30 cm from I2.
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I1=25A
B1
B2
(b)
40cm
I2=75A
x
21 BB
2
2
1
1
22 r
I
r
I oo
xcm
A
x
A
40
7525
3
40
x
xcm
cmx 20
Bnet = 0 at a point 20cm from I1and 60 cm from I2.
x
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Txm
Ax
r
IB
ATmo 57 1011.1
045.0
50.2)102(
2
I
-
v
)1011.1)(/106)(106.1( 5419 TxsmxCxqvBF
NxF19
1007.1
same direction as the current
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21 BB
2
2
1
1
22 r
I
r
I oo
2
1
1
2 rr
I
I
(a)
Amm
AI 2)50.0(
5.1
62
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upwardTx
m
Ax
r
IB
ATmo
Q
67
1
11 104.2
5.0
6)102(
2
(b)
downwardTxm
Ax
r
IB
ATmo
Q
67
2
22 1067.2
5.1
2)102(
2
upwardTxBnetQ6
1013.2
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(c)
TxTxTxBmm
mm
x
6
16.07
18.06 103.1)105()102(
B2 B1
Txm
AxB
ATm
S
67
1 1026.0
6)102(
Txm
AxB
ATm
S
77
2 1058.0
2)102(
TxTxTxBmm
mm
y
6
18.07
16.06 106.1)105()102(
TxBBB yxnetS622
1006.2
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FBD:
x
mg
F
6o
T
F
6o
mg
T
I Ir
mxmr 31036.8)6sin04.0(2
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FBD:
x
mg
F
6o
T
F
6o
mg
T
I I
FT 6sin mgT 6cos
mgIl
mgIlB
mgF r
Io
26tan
l
mgrI
o
2
)6(tan
AI 2.23
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Due to segment a:
a
IB oa
4
Due to segment b:
b
IB ob
4
x
b
I
a
IB oonet
44
ba
IB onet
11
4
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Due to vertical segment: 00sin4 2
1 rdlI
B o
Due to horizontal segment: 0180sin4 2
2 rdlI
B o
RIB o
83
Due to quarter-circle:
R
IB onet
8