Download UNIT 3

DEPARTMENT OF PHYSICS AND ASTRONOMY
Electricity and Magnetism
Course PA113 – Unit 3
PA113/Unit 3
UNIT 3 – Introductory Lecture
 The Magnetic Field
– Chapter 28
 Sources of the Magnetic Field
– Chapter 29
PA113/Unit 3
Importance of Magnetic Fields
 Practical Uses
– Electric motors, Loud speakers, Navigation
(Earth’s magnetic field)
 In Experimental Physics
– Mass spectrometers, Particle accelerators,
Plasma confinement
 In the Universe
– Stars (e.g. the Sun), Interstellar space,
Intergalactic structure, Jets
PA113/Unit 3
Importance of Magnetic Fields
PA113/Unit 3
Importance of Magnetic Fields
 Units – SI Tesla (T) = (N C-1)/(m s-1) or N A-1 m -1
– 1 Gauss (G) = 10-4 T
 Examples
–
–
–
–
–
–
–
Terrestrial B field ~ 4x10-5 T
Solenoid ~ 10-3 T
Permanent magnet ~ 10-1 T
Atomic interactions ~ 10 T
Superconducting magnet ~ 102 T
White dwarfs ~ 102 - 103 T
Neutron stars < 108 T
PA113/Unit 3
Ch28 – The Magnetic Field
 28-1 Force exerted by a Magnetic Field
 28-2 Motion of a point charge in a
Magnetic Field
 28-3 Torques on current loops and
magnets
 28-4 The Hall Effect
PA113/Unit 3
Vector Notation
 The DOT product
  
C  A  B  C  AB cos 
 The CROSS product
  
C  A  B  C  AB sin 
PA113/Unit 3
28-1 The Force Exerted by a
Magnetic Field
 Key Concept – Magnetic fields apply a
force to moving charges
F  qv  B
Current element
PA113/Unit 3
dF  I dl  B
28-1 The Force Exerted by a
Magnetic Field
PA113/Unit 3
Representation of Magnetic Field
 Like electric field, can be represented by field
lines
– Field direction indicated by direction of lines
– Field strength indicated by density of lines
 But, unlike electric field
– Magnetic field lines perpendicular to force
– No isolated magnetic poles, so no points in space
where field lines begin or end
PA113/Unit 3
28-2 Motion of a Point Charge in a
Magnetic Field
 Key Concept – Force is perpendicular to
field direction and velocity
 Therefore, magnetic fields do no work
on particles
 There is no change in magnitude of
velocity, just direction
PA113/Unit 3
Motion of a Point Charge in a
Magnetic Field
PA113/Unit 3
28-2 Motion of a Point Charge in a
Magnetic Field
 Radius of circular orbit
 Cyclotron period
2m
T
qB
 Cyclotron frequency
PA113/Unit 3
mv
r
qB
1
qB
f  
T 2m
28-3 Torques on Current Loops
and Magnets
 Key concept – a current loop
experiences no net force in a uniform B
field but does experience a torque
PA113/Unit 3
28-3 Torques on Current Loops
and Magnets
F1  F2  IaB
F b sin
  NIAB
IaBb
IAB
sin
sin
2
τ  μB
Magnetic dipole moment
μ  NIAn
PA113/Unit 3
Potential Energy of a Magnetic
Dipole in a Magnetic Field
 Potential energy
 Work done….. dW  d   B sin d
dU  dW  B sin d
Integrate
U   B cos  Uo
U  B cos  μ  B
PA113/Unit 3
Zero at
θ = 90o
28-4 The Hall Effect
PA113/Unit 3
Vh = vdBw
Ch29 – Sources of the Magnetic
Field
 29-1 The Magnetic Field of moving point
charges
 29-2 The Magnetic Field of Currents
– Biot-Savart Law
 29-3 Gauss’ Law for Magnetism
 29-4 Ampère’s Law
 29-5 Magnetism in matter
PA113/Unit 3
29-1 The Magnetic Field of Moving
Point Charges
 Point charge q moving with velocity v
produces a field B at point P
o q v  rˆ
B
2
4 r
μo= permeability of free space
μo= 4 x 10-7 T·m·A-1
PA113/Unit 3
29-2 The Magnetic Field of
Currents: The Biot-Savart Law
 Key concept – current as a series of
moving charges – replace qv by Idl
o Id l  rˆ
dB 
2
4 r
Add each element to
get total B field
PA113/Unit 3
29-3 Gauss’ Law for Magnetism
 Key concept – The net flux of magnetic
field lines through a closed surface is
zero (i.e. no magnetic monopoles)
m, net   BndA  0
s
Magnetic flux
PA113/Unit 3
29-3 Gauss’ Law for Magnetism
Electric dipole
PA113/Unit 3
Magnetic dipole
(or current loop)
29-4 Ampère’s Law
 Key concept – like Gauss’ law for electric
field, uses symmetry to calculate B field
around a closed curve C
B.
d
l


oIc

c
N.B. This version assumes the currents are steady
PA113/Unit 3
29-5 Magnetism in Matter
 Magnetization, M = m Bapp/0
 m is the magnetic susceptibility
 Paramagnetic
– M in same direction as B, dipoles weakly add to B
field (small +ve m )
 Diamagnetic
– M in opposite direction to B, dipoles weakly
oppose B field (small -ve m )
 Ferromagnetic
– Large +ve m, dipoles strongly add to B-field. Can
result in permanent magnetic field in material.
PA113/Unit 3
End of lecture 1
PA113/Unit 3
DEPARTMENT OF PHYSICS AND ASTRONOMY
Electricity and Magnetism
Course 113 – Unit 3
PA113/Unit 3
UNIT 3 – Problem solving Lecture
 The Magnetic Field
– Chapter 28
 Sources of the Magnetic Field
– Chapter 29
PA113/Unit 3
Problem Solving




Read the book!!!!!
Look at some examples
Try out some questions
Draw a diagram – include vector nature
of the field (r and v or dl )
PA113/Unit 3
You must know how to…

Calculate force on a moving charge
–

Understand the properties of a dipole
–

Or current element
Torque and magnetic moment
Calculate the B field using
1. The Biot-Savart law
2. Ampère’s Law

Understand Gauss’ Law for Magnetism
PA113/Unit 3
29-2 Example – the Biot-Savart
Law applied to a current loop
PA113/Unit 3
Field due to a current loop
ˆ
I
d
l

r
ˆ

o Id
l

r

o
B
ddB

22
44
 rr
o Idl
dB 
2
2
4 x  R
PA113/Unit 3
Field due to a current loop

R
dBx  dB sin   dB
2
2
 x R




o Idl
dB 
2
2
4 x  R
o Idl
dBx 
2
2
4 x  R
PA113/Unit 3
R
x R
2
2
Field due to a current loop
o Idl
dBx 
2
2
4 x  R
R
x R
2
2
 oo IRIR
Bx   dBx  
dl
3
/
2
3
/
2

2 2
44x2x2 RR

2πR
o 2R 2 I
Bx 
3/ 2
2
2
4 x  R 
PA113/Unit 3
Magnetic field lines of 2 loops
PA113/Unit 3
Many loops – a solenoid
PA113/Unit 3
The B field in a very long solenoid
Can use the Biot-Savart Law or Ampère’s Law
Length L
N turns
n = N/L
Radius R
di=nIdx
Field in a very long solenoid: B =0nI
PA113/Unit 3
Current I
Field around and inside a wire
Classic example of the use of Ampère’s Law
o Ir
B
2
2 R
PA113/Unit 3
 B.d l   I
o c
c
o 2 I
B
4 R
Direction of field around a wire
PA113/Unit 3
End of lecture 2
PA113/Unit 3
DEPARTMENT OF PHYSICS AND ASTRONOMY
Electricity and Magnetism
Course 113 – Unit 3
PA113/Unit 3
UNIT 3 – Follow-up Lecture
 The Magnetic Field
– Chapter 28
 Sources of the Magnetic Field
– Chapter 29
PA113/Unit 3
Ch28 – The Magnetic Field
 28-1 Force exerted by a Magnetic Field
 28-2 Motion of a point charge in a
Magnetic Field
 28-3 Torques on current loops and
magnets
 28-4 The Hall Effect
PA113/Unit 3
28-1 The Force Exerted by a
Magnetic Field
 Key Concept – Magnetic fields apply a
force to moving charges
F  qv  B
Current element
PA113/Unit 3
dF  I dl  B
28-2 Motion of a Point Charge in a
Magnetic Field
 Radius of circular orbit
 Cyclotron period
2m
T
qB
 Cyclotron frequency
PA113/Unit 3
mv
r
qB
1
qB
f  
T 2m
28-3 Torques on Current Loops
and Magnets
F1  F2  IaB
F b sin
  NIAB
IaBb
IAB
sin
sin
2
τ  μB
Magnetic dipole moment
μ  NIAn
PA113/Unit 3
Ch29 – Sources of the Magnetic
Field
 29-1 The Magnetic Field of moving point
charges
 29-2 The Magnetic Field of Currents
– Biot-Savart Law
 29-3 Gauss’ Law for Magnetism
 29-4 Ampère’s Law
 29-5 Magnetism in matter
PA113/Unit 3
29-2 The Magnetic Field of
Currents: The Biot-Savart Law
 Key concept – current as a series of
moving charges – replace qv by Idl
o Id l  rˆ
dB 
2
4 r
Add each element to
get total B field
PA113/Unit 3
29-3 Gauss’ Law for Magnetism
 Key concept – The net flux of magnetic
field lines through a closed surface is
zero (i.e. no magnetic monopoles)
m, net   BndA  0
s
Magnetic flux
PA113/Unit 3
29-4 Ampère’s Law
 Key concept – like Gauss’ law for electric
field, uses symmetry to calculate B field
around a closed curve C
B.
d
l


oIc

c
N.B. This version assumes the currents are steady
PA113/Unit 3
Example
PA113/Unit 3
Field of a tightly wound toroid
o NI
B
, ( a  r  b)
2 r
B  0, r  a
B  0, r  b
If b-a < r then B varies little – principle of fusion reactors
PA113/Unit 3
Why use fusion?
• Chemical reaction C+02  CO2 (e.g. Coal)
goes at ~700 K and gives ~107 J kg-1
• Fission, such as U235 + n  Ba143 + Kr91 + 2n
goes at ~103 K and gives ~1012 J kg-1
• Fusion, such as in the Sun, H2 + H3  He4 + n
goes at ~108 K and gives ~1014 J kg-1
PA113/Unit 3
Conditions required
PA113/Unit 3
Typical Fusion Reaction Chains
The Sun
PA113/Unit 3
The laboratory
Tokamak Fusion Test Reactor
Operated from 1982 – 1997
Max Temp = 510 million K; Max power = 10.7 MW
PA113/Unit 3
Reactor Results
PA113/Unit 3
End of lecture 3
PA113/Unit 3
Definition of the Ampère
 Force between 2 straight parallel
conducting wires
d F2  Id l2  B1
PA113/Unit 3