Chapter 19: Magnetism Magnets Magnets Homework assignment : 18,25,38,45,50 Read Chapter 19...
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Transcript of Chapter 19: Magnetism Magnets Magnets Homework assignment : 18,25,38,45,50 Read Chapter 19...
Chapter 19: MagnetismMagnets
Magnets
Homework assignment : 18,25,38,45,50Read Chapter 19 carefully especially examples.
Magnetic force
Magnets
Magnetic field lines
Magnets
Magnetic field lines
Magnets
NS
Magnetic field lines (cont’d) Electric Field Linesof an Electric Dipole
NS
Magnetic Field Lines of a bar magnet
Magnetic monopole?
Magnets
• Many searches for magnetic monopoles—the existence of which would explain (within framework of QM) the quantization of electric charge (argument of Dirac)
• No monopoles have ever been found:
Perhaps there exist magnetic charges, just like electric charges.
Such an entity would be called a magnetic monopole (having + or magnetic charge).
How can you isolate this magnetic charge?
Try cutting a bar magnet in half:
NS N NS S
Even an individual electron has a magnetic “dipole”!
Magnets Source of magnetic field
What is the source of magnetic fields, if not magnetic charge?
Answer: electric charge in motion!
e.g., current in wire surrounding cylinder (solenoid) produces very similar field to that of bar magnet.
Therefore, understanding source of field generated by bar magnet lies in understanding currents at atomic level within bulk matter.
Orbits of electrons about nuclei
Intrinsic “spin” of electrons (more important effect)
Magnets
Magnetic field of Earth• The geographic North Pole corresponds a magnetic south pole.
• The geographic South Pole corresponds a magnetic north pole.
• The angle between the direction of the magnetic field and the horizontal is called the dip angle. • The difference between true north and, defined as the geographic North Pole, and north indicated by a compass varies from point to point on Earth. This difference is referred to as a magnetic declination.
Magnetic Fields
Magnetic force: Observations
vector product
magnitude:
Magnetic Fields
Magnetic force (Lorentz force)
sinqv
FB SI unit : tesla (T) = Wb/m2
mA
N
m/s C
N
m
Wb][
2TB
right-handrule
Magnetism
Magnetic force (cont’d)
Units of magnetic field
Magnetic Fields
Magnetic force (Lorentz force)
Magnetic force
F
x x x x x x
x x x x x x
x x x x x xv
B
q
v
B
qF = 0
v
B
qF
qvBF max
Magnetic force on a current (straight wire)
Magnetic Force on a Current-Carrying Conductor
sinBIF
Magnetic force on a current (straight wire) (cont’d)
Magnetic Force on a Current-Carrying Conductor
sinBILF
Plane of loop is parallel to the magnetic field
Force and Torque on a Current Loop
=rFsin
Plane of loop : general case
Force and Torque on a Current Loop
Force and Torque on a Current Loop
Plane of loop and magnetic moment
Motor
flip the current direction
Force and Torque on a Current Loop
Case 1: Velocity perpendicular to magnetic field
Motion of Charged Particles in a Magnetic Field
υ perpendicular to B
The particle moves at constant speed υ in a circle in the plane perpendicular to B.
F/m = a provides the acceleration to the center, so
2LF υ B
and
hence
q q Ba a
m m m Rm
RqB
������������� �
v
R
x B
F
Case 1: Velocity perpendicular to magnetic field (cont’d)
Motion of Charged Particles in a Magnetic Field
Case 1: Velocity perpendicular to magnetic field (cont’d)
Motion of Charged Particles in a Magnetic Field
Velocity selector
Case 1: Velocity perpendicular to magnetic field (cont’d)
Motion of Charged Particles in a Magnetic Field
Mass spectrometer
Case 1: Velocity perpendicular to magnetic field (cont’d)
Motion of Charged Particles in a Magnetic Field
Mass spectrometer
Case 1: Velocity perpendicular to magnetic field (cont’d)
Motion of Charged Particles in a Magnetic Field
Mass spectrometer
Case 1: Velocity perpendicular to magnetic field (con’t)
Motion of Charged Particles in a Magnetic Field
Mass spectrometer
L
L L
υ and υ to B. The Lorentz force becomes
F υ B [(υ υ ) B]
Now, the cross product of any two parallel vectors is zero, so
F [υ B] and υ does not contribute to F
q q
q
������������������������������������������
���������������������������������������������������� ����
������������������������������������������
L
.
F is perpendicular to B and to υ , so υ stays constant.
Thus, there is a circular motion with , while there is
a constant υ . The result is a helical motion.
mR
qB
������������������������������������������
��������������
Case 2: General case
Motion of Charged Particles in a Magnetic Field
υ at any angle to B.
Begin by separating the two components of υ into
//
//
Case 2: General case (cont’d)
Motion of Charged Particles in a Magnetic Field
Since the magnetic field does not exertforce on a charge that travels in its direction,the component of velocity in the magneticfield direction does not change.
Exercises
Exercise 1
If a proton moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what is the speed of the proton and the frequency of motion?
v
r
x x
x x
1 m
qBf
2
kg
TCf
27
19
1067.12
4.0106.1
HzHzf 68 101.61067.128.6
4.06.1
Hzf 6101.6
2 m
qBrv
kg
mTCv
27
19
1067.1
21.04.0106.1
s
m
s
mv 68 101.81067.1
21.04.06.1
s
mv 6101.8
Exercises
Exercise 2
Example of the force on a fast moving proton due to the earth’s magnetic field. (Already we know we can neglect gravity, but can we neglect magnetism?)
Let v = 107 m/s moving North.
What is the direction and magnitude of F?
Take B = 0.5x10-4 T and v B to get maximum effect.
T) 105.0(m/s) 10106.1( 4719 CqvBF
NFM 17108 (a very fast-moving proton)
V/m) 100(C) 106.1( 19 qEFE
NFE 17106.1
B
N
F
v
vxB is into the paper (west). Check with globe
Magnetic field due to a long straight wire
Iron filings
Magnetic Field of a Long Straight Wire and Ampere’s Law
r
IB
2
0Magnetic fieldby a long wire
x10-7 T m/A
permeability of free space
right-handrule 2
Ampere’s (circular) law : A circular path
Ampere’s Law
• Consider any circular path of radius R centered on the wire carrying current I. • Evaluate the scalar product B·s around this path. • Note that B and s are parallel at all points along the path. • Also the magnitude of B is constant on this path. So the sum of all the s terms around the circle is
0
2
IB
r
On substitution for BAmpere’s circuital law (valid for any closed path)
s
s
s
s
)2(// rBsBsB iiii
IrBsBsB iiii 0// )2(
amount of currentthat penetrates theloop
Two parallel wires
Force Between Parallel Conductors
At a distance a from the wire with current I1 the magnetic field due to the wire is given by
2 12F L BI �������������������������� ��
d
II
L
F
2
2102
Ld
II
d
ILILBIF 21010
2122 22
d
IB 10
1 2
d
Force per unit length
Two parallel wires (cont’d)
Force Between Parallel Conductors
Parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other.
d d
Definition of ampere
Force Between Parallel Conductors
2 12F L BI �������������������������� ��
The chosen definition is that for d = L = 1m, The ampere is made to be such that F2 = 2×10−7 N when I1=I2=1 ampere
This choice does two things (1) it makes the ampere (and also the volt) have very convenient magnitudes for every day life and (2) it fixes the size of μ0 = 4π×10−7. Note ε0 = 1/(μ0c2). All the other units follow almost automatically.
d
IB 10
1 2
Ld
IIF 210
2 2
d
Magnetic field by a current loop
Magnetic Fields of Current Loops and Solenoids
x1
x2
1) The segment x1 produces a magnetic fieldmagnitude B1 at the center of the loop, directed out of the page.2) The segment x2 produces a magnetic fieldmagnitude B2 at the center of the loop, directed out of the page. The magnitude of B1 and B2 are the same.
R
IB
20
R
The magnitude of the magneticfield at the center of a circularloop carrying current I
R
INB
20
The magnitude of the magneticfield at the center of N circularloops carrying current I
• If d << L, the B field is to first order contained within the solenoid, in the axial direction, and of constant magnitude. In this limit, we can calculate the field using Ampere's law.
L• A solenoid is defined by a current i flowing through
a wire that is wrapped n turns per unit length on cylinder of radius d and length L. d
Magnetic Fields of Current Loops and Solenoids
Magnetic field by a solenoid
NIBsBsB iiii 0//
Inside the solenoid, B is constant and outsideit is zero in this approximation. Apply Ampere’slaw to the rectangular loop represented by bluedashed lines.
N
nnIIN
B 00
number of turn per unit length
InR
IB 0
0 vs.2
2/322
20
2 Rx
RIB
The magnetic field of a solenoid is essentially identical to that of a barmagnet
Magnetic field by a solenoid (cont’d)
solenoid bar magnet
A mystery of :
R
x
P
I
B field at point P:
In a solenoid, the B field at its axis:
nIdxRx
RnIB 02/322
20
2