Transcript 20. Electric Charge, Force, & Field

26. Magnetism: Force & Field
1.
2.
3.
4.
5.
6.
7.
8.
What is Magnetism?
Magnetic Force & Field
Charged Particles in Magnetic Fields
The Magnetic Force on a Current
Origin of the Magnetic Field
Magnetic Dipoles
Magnetic Matter
Ampere’s law
This ultraviolet image shows delicate loops of millionkelvin ionized gas in the Sun’s atmosphere.
What force shapes the gas into such intricate structures,
and why don’t we see similar things in Earth’s atmosphere?
26.1. What is Magnetism?
Magnets exert (magnetic) forces on each other & materials like irons.
Field description:
A magnet produces a magnetic field B.
Another interacts with B at its position.
Ampere :
Moving charges can produce B.
Magnetic dipoles ~ current loops.
Iron filings align with the magnetic field,
tracing out the field of a bar magnet.
Quantum mechanics:
Intrinsic magnetic moment related to spin.
26.2. Magnetic Force & Field
Magnetic force on a charge q
moving with velocity v in a field B:
FB  q v  B
FB  q v B sin 
[B] = N  s / (C  m) = T (Tesla) = 10,000 G (Gauss)
v  B: greatest F
BEarth ~ 1 G
Bsmall magnet ~ 100 G
BMRI ~ 1 T
Bmagnetar ~ 1011 T
v // B:
GOT IT? 26.1
Figure shows a proton in a magnetic field.
For which of the three proton velocities shown will the magnetic force be greatest?
What will be the direction of the force in all three cases?
 & into paper
Example 26.1. Steering Protons
Figure shows 3 protons entering a 0.10-T magnetic field.
All three are moving at 2.0 Mm/s.
Find the magnetic force on each.
F  q vB
B
F1  q v B ˆi  1.6  1019 C  2.0  106 m / s   0.10 T  ˆi
1
3
2
FB + FE = 0 when v = E/B
 3.2 1014 N ˆi
 32 f N ˆi
F2  0
F3  q v B ˆi  32 f N ˆi
F  q E q vB
velocity selector.
Electromagnetic force
26.3. Charged Particles in Magnetic Fields
FB  v, dx
 WB = 0
For v  B, FB = q v B.
2
v
FB  qvB  m
r
r
v = const
Circular motion
mv
qB
CW about B for q > 0.
GOT IT? 26.2
A uniform magnetic field points out of this page.
Will an electron that’s moving in the plane of the page circle
(a) clockwise or
(b) counterclockwise
as viewed from above the page?
Example 26.2. Mass Spectrometer
A mass spectrometer separates ions according to their ratio of charge to mass.
Such devices are widely used to analyze unknown mixtures,
and to separate isotopes of chemical elements.
Figure below shows ions of charge q & mass m being
first accelerated from rest through a potential difference V,
then entering a region of uniform B pointing out of the page.
Find an expression for the horizontal distance x.
qV 
1
m v2
2
v2
m
qvB
r
x  2r  2
mv
qB
r
2
m
qB
mv
qB
2 qV
m

2
B
2 mV
q
The Cyclotron Frequency
Period of particle in circular orbit in uniform B:
2 m v
2 r

T
v qB
v

2
qB

T
m

2 m
qB
Cyclotron frequency
Motion // B not affected by it
Trajectory in 3-D
Charged particles frozen to B field lines.
Application: The Cyclotron
Whole device in vacuum chamber.
Small V across the dee’s, which alternates
polarities at the cyclotron frequency.
Particle injected at center of gap & spirals out.
E ~ MeVs.
Applications:
Manufacture of radioactive isotopes.
e.g., PET (Positron Emission Tomography).
Higher energies:
Relativity effects  Synchrotron (B also varies)
26.4. The Magnetic Force on a Current
Force on carrier in wire:
f  q vd  B
Force on straight wire of cross section A & length L:
F  n A L q vd  B
 I LB
F out of paper
e moving left deflected upward …
I  n A q vd
Prob 58
Fmag on + charge is also upward
… resulting charge separation
leads to upward force on whole
wire
L  L vˆ d
GOT IT? 26.3
Figure shows a flexible wire passing through a magnetic field that points out of page.
The wire is deflected upward, as shown.
Is the current flowing
(a) to the left or
(b) to the right?
Conceptual Example 26.1. Power Line
A power line runs along Earth’s equator, where B points horizontally from south to north;
the line carries current flowing from west to east.
What’s the direction of the magnetic force on the power line?
F
F  I LB
L
upward
Making the Connection
Earth’s equatorial field strength is 30 T, and the power line carries 500 A.
What’s the magnetic force on a kilometer of the line.
F  I LB

 
  500 A 1.0  103 ˆi m  30  106 ˆj T
 15 kˆ N
upward
c.f. weight of wire is ~ 10 kN

The Hall Effect
e moves to left &
deflected upward
Direction of FB depends on I,
not on sign of charged carriers.
Carriers of both signs are deflected upwards
EH is upward
EH is downward
z
Direction of E due to built-up charges
depends on signs of charged carriers:
Hall effect.
y
x
Steady state, Fz = 0 :

p moves to right &
deflected upward
Hall potential:
Hall coefficient:
RH 
1
nq
q EH  q vd B  0
EH   vd B
VH  vd B h 
I
I
Bh 
B
nqA
nqt
EH
J
 vd 
 RH J
B
nq
26.5. Origin of the Magnetic Field
Biot-Savart Law:
dB 
0 I d L  rˆ
4
r2
0
7
 107 N / A2  10 T  m / A
4
B
0
4
I d L  rˆ
0

 r2
4
Curl of
finger
gives B.
C.f.
Thumb // I
3
d r
E
permeability
constant
J  rˆ
r2
1
4  0
3
d
 r

r
2
rˆ
Example 26.4. Current Loop
Find the magnetic field at an arbitrary point P on the axis
of a circular loop of radius a carrying current I.
dL  r
cos = a / r
B
0
4
I d L  rˆ
 r2
By symmetry, only Bx  0.
d L  rˆ x  cos dL

Bx  0
4

0
2

a
dL
r
Ia
0 I a
0 I a 2
 r3 dL  4  r3  2  a   2 r 3
I a2
 x2  a2 
3/2
Example 26.5. Straight Line
Find the magnetic field produced by an infinitely long straight wire carrying current I.
B
0
4
I d L  rˆ
 r2
d L  rˆ  sin  dL zˆ 

Bz  0
4

y
dL zˆ
r
0
I y

I y
dL
 r3
4
0 I
2 y

lim

1
2
a
x

2 3/2
x
x
B
x


a2
x2  a2
0 I
φˆ
2 d
1
2
y
dx 
 lim
dx

2 3/2
x
x2  a2
a2
x
x
2
a x

1
a2
The Magnetic Force Between Conductors
F  I LB
Field of I1 at I2 :
B1 
0 I1 ˆ
b1
2 d
Force on length L of I2 :
F2  I 2 L  B1 
Force per unit length on I2 :
f2 
F2 0 I1 I 2 ˆ

d
L
2 d
0 I1 I 2 ˆ
Ld
2 d
points toward I1
Hum from electric equipments are vibrations of transformers in response to AC.
Definition: 1 A is the current in two long, parallel wire 1 m apart &
exerting 2107 N per meter of length.
1C is the charge passing in 1s through a wire carrying 1A.
GOT IT? 26.4
A flexible wire is wound into a flat spiral as shown in figure.
If a current flows in the direction shown, will the coil
(a) tighten or
(b) become looser?
No
Does your answer depend on the current direction?
26.6. Magnetic Dipoles
Field on axis of current loop of radius a :
Bx 
0
2
x
I a2
2
a
0 I a 2


x a
2

2 3/2
C.f. electric dipole:
Setting
Far away,
fields look
similar …
… but close in,
they’re different.
k
0
4
N-turn current loop:

E  2k
x3

0 I A
2  x3
p
x3
p   IA
magnetic
dipole
μ  N IA
Detailed calculations show:
•  = I A valid for arbitrary loop.
• Vector behavior of  similar to that of p for r >> a.
Multi-turn loops = electromagnets
Very strong field requires superconducting wires, e.g., MRI scanners.
Orbiting e in atom  .
Detailed calculations show ：
•  = I A valid for arbitrary loop
• Vector behavior of  similar to that of p for r >> a.
Multi-turn loops = electromagnets
Very strong field requires superconducting wires, e.g., MRI scanners.
Orbiting e in atom  .
Application: Magnetic Fields of Earth & Sun
Currents flows in Earth liquid-iron outer core (due to rotation)

dipole field with  = 8.0  1022 Am2 ( direction not exactly S-N )
Field deflects harmful high E solar particles.
Magnetic reversal:
Earth (period ~ millions of yrs) : map for sea-floor spreading over.
Sun (period ~ 11 yrs): coincides with period of sun-spot abundancy.
Dipoles & Monopoles
Some elementary particle theories suggest existence of magnetic monopoles.
But none was ever observed.
Microscopic origin of B:
1. Charged current.
2. Intrinsic spin.
No magnetic charges : B lines always form loops,
either encircling moving charges, or joining the 2 poles of a magnet.
 B  dA  0
Gauss Law
GOT IT? 26.5
The figure shows two fields.
Which could be a magnetic field?
Torque on a Magnetic Dipole
Forces on top & bottom cancel.
Forces on sides also cancel; but give net torque.

above plane
beneath plane
Fside  a I B
1
2
1
 b a I B sin 
2
 side  b Fside sin 
  2 side  b a I B sin 
  B sin 
Torque on dipole:
τ  μB
Associated energy: U  μ  B
Application: Electric Motors
Rotating loop
CW
Commutator
CCW
Brushes
Battery
Always CCW
Example 26.6. Hybrid Car Motor
Toyota’s Prius gas-electric hybrid car uses a 50-kW electric motor that
develops a maximum torque of 400 Nm.
Suppose you want to produce this same torque in the motor just discussed,
with a 950-turn rectangular coil of wire measuring 30 cm by 20 cm,
in a uniform field of 50 mT.
How much current does the motor need?
τ  μB
Max torque at sin = 1 :
Rectangular coil :

I
950-turn coil
 max
N AB
 max   B
N I A
 max  N I A B

400 N  m
 140 A
950
0.30
m

0.20
m
0.050
T
 


26.7. Magnetic Matter
Ferromagnetism:
e.g., Fe, Ni, Co, & their alloys.
Material divided into magnetic domains in which ’s are aligned.
T > TC : orientations of  in different domains random
 <  >  B (paramagnetic)
T < TC : orientations of  in different domains aligned
 <  >  0 even when B = 0 (ferromagnetic)
Atomic
current loops
are CCW.
No cancellation on
boundaries:
net I on surface.
Adjacent loops cancel
 // Bapplied
Field inside dipole > Bapplied
Distant field similar.
Internal field opposite.
Field inside dipole < Eapplied
p // Eapplied
Paramagnetism :
Materials with randomly oriented permanent  and very weak - interaction.
<>= B
with  > 0
 = magnetic susceptibility
Diamagnetism :
Materials with no intrinsic .
<>= B
( induced  )
with  < 0
26.8. Ampere’s law
Field around long wire carrying current I :
B


0 I
φˆ
2 r
B  dr  
2
0
from Biot-Savart law
0 I
r d
2 r
 0 I
True for arbitrary closed paths & steady currents:
Bdr = 0

B  dr  0 I enclosed
Ampere’s law
c.f. Gauss’ law
net field from ALL sources
Example 26.7. Solar Currents
The long dimension of the rectangular loop in figure is 400 Mm,
and B near the loop has a constant magnitude of 2 mT.
Estimate the total current enclosed by the rectangle.
B  dr on the short segment
 B  dr = 0 there.
B // dr on the long segment 

rectangle
B  dr  2 B L
 2  2  103 T  400  106 m 
 1.6  106 T  m
I
2BL
0
1.6  106 T  m
 1.3 1012 A

7
4   10 T  m / A
Direction: into the page
3D: around equator
GOT IT? 26.6
The figure shows three parallel wires carrying current of the same magnitude I,
but in one of them the current is opposite to that of the other two.
If

B  dr  0

around loop 2,
B  dr
0
(a)
what is
around loop 1, and
A
(b)
which current is the opposite one?
Using Ampere’s Law
STRATEGY 26.1 Ampère’s Law :
Base on symmetry, choose the amperian loop such that B is either // or  to it.
Example 26.8. Outside & Inside a Wire
A long, straight wire of radius R carries a current I distributed uniformly over its cross section.
Find the magnetic field
(a)
outside and
(b)
inside the wire.
By symmetry, B is azimuthal.
Amperian loop is a circle.

(a)
(b)
B  dr  2  r B
I enclosed  I
I enclosed
 r2
I
 R2
B
0 I
2 r
B
0 I r
2  R2
Example 26.9. Current Sheet
An infinite flat sheet carries current out of this page.
The current is distributed uniformly along the sheet, with current per unit width given by JS .
Find the magnetic field of this sheet.
By symmetry, B is // to sheet &  I.
Amperian loop is a rectangle.

Far out, B lines
nearly circular
I enclosed  J S L
B
Close in, B lines
~ infinite sheet
B  dr  2 B L
1
0 J S
2
Fields of Simple Current Distributions
For arbitrary distributions:
Far away from any loop ~ dipole.
Very near any wire ~ infinite straight wire.
Very near any current sheet ~ infinite flat sheet.
Solenoids
Solenoid: a long, tightly wounded coil.
n turns per unit length.
n L loops in L.
encircled current = nLI.

B  dr  B L
B  0 n I
I enclosed  n L I
solenoid field
Boutside = 0
Application: magnetic switches, valves, ….
Table 26.1. Fields of Ssome Simple Charge & Current Distributions
Field(r)
Q
E
J
r 3
Dipole
Dipole
r 2
Point /
spherical
NA
r 1
Line
Line
const
Sheet
Sheet
B
Example 26.10. MRI Scanner
The solenoid used in an MRI scanner is 2.4 m long & 95 cm in diameter.
It’s wounded from superconducting wire 2.0 mm in diameter,
with adjacent turns separated by an insulating layer of negligible thickness.
Find the current that will produce a 1.5-T magnetic field inside the solenoid.
wire diameter = 1/500 m
I
B
0 n
n = 500 turns/m
 2.4 kA

1.5 T
 4 107 T  m / A 500 m1 
Field of a solenoid is very similar to that of a bar magnet.