Download Magnetic Field

Physics 169
Kitt Peak National Observatory
Monday, March 13, 17
Luis anchordoqui
1
6.1 Magnetic Field
Stationary charges experienced an electric force in an electric field
Moving charges experienced a magnetic force in a magnetic field
~
(electric force)
F~E = q E
~ (magnetic force)
F~B = q~v ⇥ B
More explicitly ☛
~ : Unit = Tesla
Magnetic field B
(T)
1 T = 1 C moving at 1 m/s experiencing 1 N
Common Unit
1 Gauss (G) = 10
4
T ⇠ magnetic field on Earth’s surface
Direction of magnetic force determined from right hand rule
Monday, March 13, 17
2
Right Hand Rule
Monday, March 13, 17
3
Monday, March 13, 17
4
Magnetic Force
Example
What’s the force on
~v = (10ˆ
|
0.1 C charge moving at velocity
6.1 Magnetic Field
~ = ( 3ı̂ + 4k̂) ⇥ 10
20k̂) m/s in a magnetic field B
4
4
N
N
T
For stationary charges, they experienced an electric force in an electric field.
For moving
~ charges, they experienced a magnetic force in a magnetic field.
~ = q~v ⇥ B
F
Mathematically,
FEk̂)
= q⇥
E ((electric
force)
= +0.1
(10ˆ
|
20
3ı̂ +
4k̂) ⇥ 10
FB = qv ⇥ B
5
(magnetic force)
= 10 Direction
(40ı̂of+the60ˆ
| + 30
k̂)determined
N
magnetic
force
from right hand rule.
Note that
~
F~ = q~v ⇥ B
|F~ | = qvB sin ✓
)
☛Bis: Unit
angle
between
Magnetic field
= Tesla
(T)
1T = 1C moving at 1m/s experiencing 1N
✓
~
~v and B
~
✓ = 90 , (i.e. ~v ? B)
Common Unit: 1 Gauss (G) = 10 4 T ⇤ magnetic field on earth’s surface
~
v k B)
Magnetic force is minimum (0) when ✓ = 0 or 180 (i.e. ~
Magnetic force is maximum when
Monday, March 13, 17
Example: What’s the force on a 0.1C charge moving at velocity v = (10ĵ
5
20k̂)ms 1 in a magnetic field B = ( 3î + 4k̂) ⇥ 10 4 T
F = qv ⇥ B
6.2
Motion
APOINT
Point
Charge
in Magnetic
Field
6.2. MOTION of
OF A
CHARGE
IN MAGNETIC
FIELD
75
6.2 Motion of A Point Charge in Magnetic Field
Since Since
F⇥B 6.2
⇥F
⇥v~, therefore
changes
the direction
of the velocity
but not
Motion
of Aonly
Point
Charge
in Magnetic
Field
~v B-field
B ?
its magnitude.
~ -field only changes direction of velocity but not its magnitude
B
Since F⇥B ⇥ ⇥v , therefore B-field only changes the direction of the velocity but not
its magnitude.
~
Generally
☛ ~
FB = q~v ⇥ B = qv? B
⇥B = to
⇥ = qmotion
~ -field
Generally,
q⇥vconsider
B
v B , component ? to B
Only F
need
⇥B consider
⇥ =
We only
need Fto
the
motion
Generally,
= q⇥
v B
q vcircular
B,
We
have
motion
need to consider the motion
component
⇥Wetoonly
B-field.
Magnetic
force
provides
component ⇥ to B-field.
centripetal force on moving charge particles
)
FB
v2
= m
r
v2
We |q|vB
have circular
= m motion. Magnetic
We haveforce
circular
Magnetic
providesmotion.
the r
centripetal
force on the
mv
force provides
moving
charge
) the
r centripetal
= particles.force on the
|q|B
moving charge particles.
r ☛ radius of circular motion
Monday, March 13, 17
FB
v2
= m
r
2
6
Time for moving around one orbit:
Time for moving around one orbit
2 r
2 m
T =
=
Cyclotron Perio
2⇡r
2⇡m
vCyclotron
qBPeriod
T =
=
v
qB
(1) Independent of v
① Independent of v (non-relativistic)
(2) Usem/q
it to
② Use it to measure
(non-relativistic)
measure m/q
Generally, charge
stant velocity mov
ence of constant B
Generally ☛ charged particles with constant velocity
~ -field
moves in helix in presence of constant B
Monday, March 13, 17
7
6.3. HALL EFFECT
76
Note
~ -field
does
work on particles
B
(1)①
B-field
does NO
workNO
on particles.
Note :
~-field
(2)②
B-field
does NOT
of particles.
doeschange
NOTK.E.
change
K.E.
B
Particle Motion
in Presence
of E-field
& B-field:
Particle
Motion
in Presence
of particles
of E-field & B-field
~Lorentz
~ Lorentz Force
F~ =B q E
+ q~Force
v ⇥B
~ ? B
~
Special Case
: E Case
⇤B
E
Special
~E | = |F~B |
When |F
F = q E + qv
qE = qvB
E
) Ev =
⇥ v=
B
B
When |FE | = |FB |
qE = qvB
)ForCharged
E/B
particles
at vthey=will
will pass
charged particles
movingmoving
at v = E/B,
pass through
the through
crossed
and B fields
displacement.
~without
~vertical
E
i.e.Ecrossed
and B
fields
without vertical displacement
⇥ velocity selector
Applications
Applications
:
• Cyclotron
(Lawrence
• Cyclotron (Lawrence
& Livingston
1934)
velocity selector
& Livingston 1934)
• Measuring e/m
electrons (Thomson
1897)electrons
• for
Measuring
e/m for
(Thomson 1897)
• Mass Spectrometer (Aston 1919)
• Mass Spectrometer (Aston 1919)
6.3
Monday, March 13, 17
Hall Effect
8
6.4
Magnetic Force on Currents
6.3 Magnetic ForceCurrent
on Currents
= many charges moving together
Current = many charges moving together
L L,
Consider
a wire
segment
of length
Consider
a wire
segment,
length
carrying
current
carrying
currenti iininaamagnetic
magnetic field
field.
TotalTotal
magnetic
force
==(q~
vdq⇥v⇥
magnetic
force
(
| {zd
Recall
nqvdd A
Recall
i =i =nqv
A
~⇥ ) · n A L
B
B
⇥ L⌅
}⌅ ) · | {zn⇤ }A
⇤ ⇥
Total number
number ofof
forceforce
on one
on one Total
charge carrier
charge
carrier charge
charge
carrier
carrier
~ = iL
~ ⇥
~⇥ ⇥
) Magnetic force on current F
B
⇥
Magnetic force on current F = iL B
~ = length of current segment
|L|
~ =
⇥ =vector
⇥ = length of current segment; direction =
L
where
L
Vectorwith
of which: |L|
direction of current
{
direction = direction of current
For infinitesimal wire segment d~l ⇥
For an infinitesimal wire segment dl
~ = i dd~lF⇥⇥= B
⇥
dF
i~d⇥l B
Monday, March 13, 17
Example 1: Force on a semicircle current loop
9
For an infinitesimal wire segment d⇥l
Example
dF⇥ = i d⇥l
⇥
B
Force on a semicircle current loop
Example ~
1: Force on a semicircle current loop
~
dl = infinitesimal arc length element ? B
)
)
dl =
⇥ R d✓
dl = Infinitesimal
⇥
arc length element ⇥ B
dF = iRB d✓
dl = R d
dF = iRB d
By symmetry argument ☛ we only need to consider vertical forces
By symmetry argument, we only need to consider vertical forces, dF · sin
ˆ
dF · sin ✓
NetZ force
dF sin
⇡ F =
0
) Net force
F =
dF sin ✓ ˆ
0
Z = iRB sin d
⇡
= iRB
0
F =sin
2iRB
✓ d✓ (downward)
0
F = 2iRB (downward)
Monday, March 13, 17
10
Method 2
Write d~l in ı̂, |ˆ components
d~l =
dl sin ✓ı̂ + dl cos ✓ˆ
|
= Rd✓ (
~ =
B
)
F~ =
=
=
Monday, March 13, 17
(into the page)
~ = i d~l ⇥ B
~
dF
=
)
B k̂
sin ✓ı̂ + cos ✓ˆ
|)
Z
iRB sin ✓ d✓ˆ
|
⇡
iRB cos ✓ı̂
dF~
0
iRB
hZ
2iRBˆ
|
⇡
sin ✓ d✓ˆ
|+
0
Z
⇡
cos ✓ d✓ı̂
0
i
11
0
=
iRB
sin d ĵ0 +
cos d î 0
0
0
=
2iRB ĵ
=
2iRB ĵ
=
2iRB ĵ
Example 2: Current loop in B-field
Example
2: Current loop in B-field
Example
2: Current
in B-field
Example
2 loop
Current
loop
in B-field
0
For segment2 ☛
For segment2:
For segment2:
For segment2:
ibBibB
sin(90 sin(90
+ ) = ibB+
cos ✓) =
(pointing
downward)
FF22==
ibB cos
✓ (pointing downward)
F2cos
= ibB sin(90
+ )downward)
= ibB cos
F2 = ibB sin(90 + ) = ibB
(pointing
(pointing downward)
For segment4 ☛
For segment4:
For segment4:
For segment4:
F2 = ibB sin(90
) = ibB cos
(pointing upward)
F42 =F2 =ibB
sin(90 ) = ibB
✓)
cos )✓upward)
upward)
F2cos
==
ibBibB
sin(90
= (pointing
ibB cos
(pointing upward)
ibB sin(90
(pointing
Monday, March 13, 17
12
Example 2: Current loop in B-field
For segment 1 ☛
F1 = iaB
For segment 3 ☛
F3 = iaB
) Net force on current loop = 0
But net torque on loop about
O
= ⌧1 + ⌧3
b
b
= iaB · sin ✓ + iaB · sin ✓
2
2
For segment2:
= i |{z}
ab B sin ✓
A = area of loop
F2 = is
ibB
+ )N= turns
ibB cosof wires
(pointing downward)
Suppose loop
a sin(90
coil with
Total torque ☛ ⌧ = N iAB sin ✓
For segment4:
Monday, March 13, 17
F2 = ibB sin(90
13
) = ibB cos
(pointing upward)
Total torque ⇥ = N iAB sin
Define Unit
☛ Unit
vector
area-vector
(usinghand
right
hand rule)
n̂ to represent
Define:
vector
n̂ to represent
the area-vector
(using right
rule)
Then
we the
cantorque
rewrite
torqueasequation as
Then we can
rewrite
equation
B
~⇤
⇤⇥ = N iA n̂
Define:
~⌧ = N iAn̂ ⇥ B
N iA n̂ = ⇤µ = Magnetic dipole moment of loop
⇤
⇤⇥ = ⇤µ B
~ = Magnetic
dipole moment of loop
Define ☛ N iAn̂ = µ
~
~⌧ = µ
~ ⇥B
Monday, March 13, 17
14
7.1
Magnetic Field
⇥
⌅ experiences magnetic force in B-field.
6.4 Sources of Magnetic Field⇧
(
A moving charge
~ -field
experiences
magnetic force in B
⇧
⇤
can generate B-field.
A moving charge
~ -field
can generate B
⇤ due to m
Magnetic field B
µ0 q⇤v r̂
⇤
B=
·
4⇥
r2
where
~ due to moving point charge
Magnetic field B
µ0 = 4⇥
10
7
µ0Permeability
q~v ⇥ r̂
µof
q~v ⇥
~r
0 free
space
(Magnetic const
~
B =
·
=
·
4⇡
r2
4⇡
r3
µ0 qv sin
maximum when =
⇤
|B| =
·
minimum when =
4⇥
r2
Permeability of free space (magnetic constant)
Monday, March 13, 17
µ0 = 4⇡ ⇥ 10
7
T · m/A (or equivalently N/A2 )
15
⇤ at P0 = 0 = B
⇤
B
where
(
µ0 = 4⇥
10
7
Tm/A (N/A2 )
maximum
when ✓ =
Permeability of free
space (Magnetic
constant)
90
µ0 qv sin ✓
~
B =
·
4⇡ ⇤ rµ20 qv sin minimum
maximum
=
90⇥
when when
✓
=
0
/180
|B| =
·
4⇥
r2
minimum when
= 0⇥ /180⇥
~ at P1
~ at P0 = 0 = B
B
⇤ at P0 = 0 = B
⇤ at P1
B
⇤ at P2 < B
⇤ at P3
B
~ at P3
~ at P2 < B
B
However, a single moving charge will NOT generate a steady magnetic field
stationary charges generate steady E-field.
A single moving charge will NOT generate a steady magnetic field
steady currents
generate steady B-field.
~-field
stationary charges generate ☛
steady E
steady currents
Monday, March 13, 17
~ -field
generate ☛
steady B
16
Principle of Superposition
field
at point P can be obtained
7.1.Magnetic
MAGNETIC
FIELD
82
by integrating contribution from individual current segments
)
µ0 dq ~v ⇥ r̂
~
dB =
·
field at 4⇡
point P can
r2 be
Magnetic
obtained by integrating the contribution from individual current segments.
d~s
(Principle
Superposition)
~v = dq ·
=
Note ofdq
dt
i d~s
µ0 i d~s ⇥ r̂
~
☛ Biot-Savart Law
dB =
·
2
dq ⇥v r̂ r
0
⇥ = µ4⇡
dB
·
4 circuit
r2
For current around a whole
Z
Z
d⇥s
µ0 id~s ⇥ r̂
Notice: dq ⇥v = dq ·
= i d⇥s
~
~
dt B =
dB =
·
4⇡
r2
⇥ =
dB
µ0 i d⇥s r̂
·
Biot-Savart
Law
entire
circuit
entire
circuit
4
r2
Biot-Savart
Lawcircuit:
is to magnetic field as
For current
around a whole
ˆ
ˆ
Coulomb’s Law
is
to
electric
µ0 ifield
d⇥s r̂
⇥ =
⇥ =
B
dB
·
4
r2
~
Basic element ofentire
E -field ➢
entire Electric charges dq
circuit
circuit
~ -field ➢ Current element i d~s
Basic element ofB
Biot-Savart Law is to magnetic field as
Monday, March
13, 17 is to electric field.
Coulomb’s
Law
Basic element of E-field: Electric charges dq
Basic element of B-field: Current element i d⇥s
17
Basic element of B-field: Current element i d⇥s
Example 1 : Magnetic field due to straight current segment
Example 1
)
Magnetic field due to straight current segment
|d~s ⇥ r̂| = dz sin
= dz sin(⇡
)
(Trigonometric identity)
d
d · dz
p
= dz ·
=
r
d2 + z 2
Monday, March 13, 17
18
µ0 idz d
µ0 i
d
dB =
· 2 ·
=
· 2
dz
2
3/2
4⇡
r
r
4⇡ (d + z )
)
B =
Z
L/2
L/2
µ0 id
dB =
4⇡
Z
µ0 i
z
B =
· 2
4⇡d (z + d2 )1/2
+L/2
L/2
dz
(d2 + z 2 )3/2
+L/2
L/2
µ0 i
L
B =
·
⌘1/2
4⇡d ⇣ L2
2
4 +d
Limiting Cases
When L
d (B-field due to long wire)
⇣ L2
⌘ 1/2
⇣ L2 ⌘ 1/2
2
+ d2
⇡
=
4
4
L
µ0 i
2⇡d direction of B-field determine from right-hand screw rule
Recall E =
2⇡✏0 d for an infinite long line of charge
)
B =
Monday, March 13, 17
19
Recall : E =
⇥
for an infinite long line of charge.
2⇤ 0 d
Example
2 :2
A circular
current loop
Example
A circular
current loop
~ 1 at
s1 generating a magnetic field dB
For every current element id~
there is an opposite current element
~ 1 sin ↵ =
so that ☛ dB
Monday, March 13, 17
P
~2
id~s2 generating dB
~ 2 sin ↵
dB
20
~ -field needs to be considered at point
Only vertical component of B
P
* d~s ? r̂
z}|{
µ0 i ds sin 90
dB =
·
4⇡
r2
Z
~ -field at point P
) B
B =
dB cos
| {z ↵}
vertical
component
around
circuit
)
B =
Z
2⇡
0
µ0 i cos ↵
· |{z}
ds
2
4⇡r
µ0 i R
=
·
4⇡ r3
)
µ0 iR2
B =
2(R2 + z 2 )3/2
Monday, March 13, 17
µ0 iR2
B =
2r3
2Z
Rd✓
2⇡
d✓
0
☛ direction of B-field determined
from right-hand screw rule
21
Limiting Cases
① B-field at center of loop
② For z
z = 0
R
1. MAGNETIC FIELD
(3) A current arc:
B =
)
⇣
µ0 i
B =
2R
µ0 iR2
2z 3 1 +
R2
z2
⌘3/2
µ0 iR2
1
⇡
/ 3
2z 3
z
85
p
Recall E-field for an electric dipole ☛ E, =
4⇡✏0 x3
ˆ
) A circular current loop is also called a magnetic
dipole
B Z=
B =
dB cos
⇤⇥ ⌅
dB
cos
| {z ↵} z = 0
around
z = 0 )
= 0 here.
around circuit circuit
↵ = 0 here
R⇥ = length of arc
⌅⇤ ⇥
R✓z =}|length
{ of arc
③ Arc current
ZR✓
B =
Monday, March 13, 17
µ0 i✓
4⇡R
µ0R
i
µ
i
0
·
ds⌅
= = · 3 · 3 |{z}
ds
⇤⇥
4⇤
r
4⇡ |{z}
r ⇤⇥0 ⌅
0 Rd
Rd✓
R =R
r =r
when ↵ = 0
when = 0
µ0 i ⇥
B =
4⇤R
ˆ
22
=
Example 3
Magnetic field of a solenoid
B =
µ0 i R
·
·
4⇤ ⇤⇥r3⌅
R=r
when = 0
0
ds⌅
⇤⇥
Rd
µ0 i ⇥
4⇤R
Solenoid can produce a strong and uniform magnetic field inside its coils
Example 3 : Magnetic field of a solenoid
Solenoid is used to produce a strong and uniform magnetic field inside its
coils.
Consider a solenoid of length L consisting of N turns of wire.
N
Consider
solenoid
of length
consisting
of N
Define: n =aNumber
of turns
per unit L
length
=
L
Consider B-field at distance d from the
center
Monday, March
13, of
17 the solenoid:
For a segment of length dz, number of
current turns = ndz
turns of wire
23
Define
N
Consider a solenoid of length L consisting of N turns of wire.L
N
~
Define:
n
=
Number
of
turns
per
unit
length
=
Consider B -field at distance d from center Lof solenoid
For segment of length dz
number of current turns = n dz
n = Number of turns per unit length =
)
Consider B-field at distance d from the
center
of the solenoid:
= n i dz
Total
current
For a segment of length dz, number of
current turns = ndz
Total current = ni dz
Using result from one coil in Example 2
~ -field from coils of length dz at distance z from center
we get B
µ0 (ni dz)R2
dB =
2r3
Monday, March 13, 17
24
µ0 (ni dz)R2
dB =
2r3
p
However r =
R2 + (z
)
B =
Z
+L/2
However rL/2
=
B =
"
(Integrating over
entire solenoid)
R2 + (z d)2
⇧dB
µ0 niR2
=
2
d)2
Z
+L/2
L/2
dz
[R2 + (z d)2 ]3/2
L
2
L
2
#
+
d
ˆ d+L/2
µ0 ni
r
r
+
(Integrating
over
⌘dB
⇣
⌘2 the
2
2
B 2= ⇣ L
L
entire
R + 2 L/2
+d
R2 + solenoid)
d
2
ˆ
µ0 niR2 +L/2
dz
=
2
along negative
2 z direction
d)2 ]3/2
L/2 [R + (z
L
2
L
2
⇥
+d
d
µ0 ni ⇤
⌅
⇧
B =
+⇧
Ideal Solenoid L
L
L
R2
R2 + ( 2 + d)2
R2 + ( 2 d)2
µ along
ni negative z direction
then
B =
Ideal Solenoid
:
)
Monday, March 13, 17
0
[1 + 1]
2
B = µ0 ni
L⇥R
directionµof
0 niB-field determined from
then B =
[1 + 1]
2
right-hand screw rule
direction of B-field determined
B = µ0 ni ; from right-hand screw rule
25
µ0 ni
Monday, March 13, 17
26
Monday, March 13, 17
27