Magnetism - Chapter 28
• Historical Overview
– Earliest manifestation was that certain stones, made of material called
magnetite, attracted iron. Known as early as 13th century BC.
– In 1269, Pierre de Maricourt mapped out the directions taken by a
magnetized needle when placed around a permanent magentic sphere.
– Subsequent experiments have shown that all magnets have two poles,
north and south
– Like poles repel, unlike poles attract.
– A Compass is a magnet suspended in the earths magnetic field. Its
north pole points to geographic north or the earth’s magnetic south
pole and vice versa.
– Magnetic poles found in pairs.
– Electricity and Magnetism connected: electric current in a wire can
deflect a compass needle
• The Magnetic Field
– Can describe the interaction of magnetic objects with the idea of a
magnetic field, similar to an electric field to describe the interaction
of charged particles.
– Iron filings sprinkled around a bar magnet trace the magnetic field due
to the bar magnet.
– Figure 28.3 and Figure 28.4, p. 738, in the book are examples of
magnetic fields.
– The magnetic field, B is a vector quantity.
– The magnetic field, B at a certain point can be traced by measuring
the magnetic force F B exerted on an appropriate test particle placed
at that point.
– Charged particle in a magnetic field, B experiences a magnetic force
FB .
– Experiments show that this force is:
∗ Magnitude of the FB proportional to speed and charge, v,q of the
particle.
∗ When a charged particle moves parallel to a magentic filed vector,
the magnetic force on the particle is zero.
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∗ When the velocity vector makes an angle θ with the magnetic
field, the magnetic force acts in a direction perpendicular to both
v, B.
∗ The magnetic force on a negative charge is directed opposite to
the force on a positive charge moving in the same direction.
∗ If the velocity vector makes an angle θ with the magnetic field,
the magnitude of the magnetic force is proportional to sinθ.
∗ See figure 28.2, p. 737 for the right hand rule.
– Magnetic force is more complicated than electric force because it depends on the velocity of the particle and because its direction is perpendicular to both v and B.
–
F B = qv × B.
– If A = (Ax , Ay , Az ) and B = (Bx , By , Bz ), then
A × B = i(Ay Bz − Az By ) − j(Ax Bz − Az Bx ) + k(Ax By − Ay Bx ).
– Look at http://www.phy.syr.edu/courses/java-suite/crosspro.html
– The cross-product or vector product A × B is a vector that is always
perpendicular to the plane defined by the two vectors A and B.
– Definition of the magnetic field at a point in space, compared to the
analgous equation for electric fields:
F e = qE.
– Unit of magnetic field in SI is the tesla (T) where
–
1T = 1N.s/C.m
– Right hand rules for direction of magnetic force - Figure 28.2, p. 737.
– FB = qvBsinθ for magnitude, θ is the angle between v and B
– Discuss the implications of this.
– Differences between electric and magnetic forces on charged particles:
∗ Electric force is always parallel or antiparallel to the electric field,
whereas the magnetic force is perpendicular to the magnetic field.
∗ The electric force on a charged particle is independent of the velocity of the particle but the magnetic force on a charged particle
acts only when the particle is in motion.
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∗ The electric force does work in displacing a charged particle, whereas
the magnetic force associated with a constant magnetic field does
no work when a charged particle is displaced.
∗ This is true because for a small displacement, ds, the work done
by the magnetic force is dW = F B .ds = F B .vdt = 0 since the
magnetic force, F B is a vector perpendicular to the velocity v.
∗ The kinetic energy of a charged particle cannot be altered by a
constant magnetic field alone - can alter the direction of the velocity vector.
• Try the following Quick Quiz An electron moves in the plane of this paper
toward the top of the page. A magnetic field is also in the plane of the page
and directed toward the right. What is the direction of the magnetic force
on the electron? Is it a) toward the top of the page, b) toward the bottom
of the page c) toward the left of the page d) toward the right edge of the
page e) upward out of the page and f) downward into the page?
• If a charged particle moves in a straight line through some region of space,
can you say that the magnetic field in that region is zero?
• Go over Sample Problem 28-1, p. 739.
– An electron in a television picture tube moves toward the front of the
tube with a speed of 8.0 × 106 m/s along the x axis. The neck of the
tube is surrounded by a coil of wire that creates a magnetic field of
magnitude 0.025T, directed at an angle of 60 degrees to the x axis and
lying in the xy plane. Calculate the magnetic force on and acceleration
of the electron.
– Draw the picture
– Use the equation FB = |q|vBsinθ.
– Plugging in the numbers we get FB = 2.8 × 10−14 N.
– What direction is v × B?
– Because the charge is negative, FB is in the negative z direction.
– Once we have determined the force, a = FB /me .
– Plugging in the numbers, a = 3.1 × 1016 m/s2 .
• Motion of a charged Particle in a Uniform Magnetic Field
– See http://lectureonline.cl.msu.edu/ mmp/kap21/cd533capp.htm
– See Fig 28-8, p. 742.
– Special case of a positively charged particle moving in a uniform magnetic field when the initial velocity is perpendicular to the field.
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– Particle moves in a circle: from figure 28.8 and
– Use Newton’s second law
–
X
F = FB = ma,
–
qvB =
–
r=
mv ∗ ∗2
r
mv
qB
– Radius of circular path proportional to linear momentum of the particle
– Angular speed ω =
v
r
qB
m
2πr
= 2πm
v
qB
=
– Period of motion T =
– Angular speed = cyclotron frequency
– If the velocity vector is at an initial angle to the magnetic field B, the
path is a helix.
– Try This i) A charged particle is moving perpendicular to a magnetic
field in a circle of radius r. The magnitude of the magnetic field is
increased. Compared with the initial radius of the circular path, is
the radius of the new path a) smaller, b) larger or c) equal in size?
ii) An identical particle enters the field with v perpendicular to B,
but with a higher speed v than the first particle. Compared with the
radius of the circle for the first particle in the same magnetic field, is
the radius of the circle for the second particle a) smaller, b) larger or
c) equal in size?
– Example 28.2, p. 743. Example 28-3, p. 747.
• Applications Involving Charged Particles moving in a Magnetic Field
– A charge moving with velocity v in the presence of an electric and
magnetic field E, B experiences a total force, called the Lorentz force:
–
F = qE + qv × B.
– Instruments such as mass spectrometers and cyclotrons, synchrotrons
use this equation: read about these in the text,p. 748.
• Magnetic force on a Current carrying Conductor
– A magnetic force is exerted on a single charged particle when it moves
through an external magnetic field.
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– A current carrying wire experiences a magnetic force when placed in
an external magnetic field.
– Magnetic force on the wire is due to the sum of all the magnetic forces
on the individual charged particles in the wire.
– Magnetic force on the wire is transmitted to the ”bulk” of the wire
through collisions with the atoms making up the wire.
– Consider a segment of wire of length l, cross-sectional area, A, carrying
a current, I in a uniform external magnetic field, B:
– Magnetic force on a charge q moving with drift velocity vd is qv d × B.
– Volume of the segment is Al, number of charges in this segment is nAl,
where n is the number of charges per unit volume.
– Total magnetic force on the wire of length l is
F B = (qvd ) × BnAl
– Noting that I = nqvd A, we have
F B = Il × B,
where l is a vector in the direction of the current I.
– Now consider an arbitrarily shaped wire of uniform cross-section in an
external magnetic field, the force is
dF B = Ids × B
where ds is a vector representing length segment.
– Now to obtain the total magnetic force F B in a length of the wire
between arbitrary points a, b, integrate this equation between these
points a, b:
Z
FB = I
b
a
ds × B.
– Problem 28-6, p. 751.
• Torque on a current loop placed in a uniform magnetic field
– Now know the force extered on a current carrying conductor in an
external magnetic field.
– If this wire is a loop, a torque is generated - great practical value in
design of motors and generators.
– Consider a rectangular loop carrying a current I in the presence of a
uniform magnetic field in the plane of the loop - se fig 28-20, 28-21, p.
752.
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– The sides of the loop parallel to the magnetic field experience no force.
Non-zero forces are exerted on the sides of the loop perpendicular to
the field.
– On each side the magnitude of the force is IaB, (why?) where a is the
length of the side carrying current I in a magnetic field B.
– However, the direction of the force on each side of the loop perpendicular to the magnetic field is opposite so that if the loop is pivoted
so that it can rotate, it will rotate about an axis perpendicular to its
plane. This is the torque.
– The magnitude of the torque, τmax is
b
b
τmax = (IaB) + (IaB) = IabB
2
2
where the moment arm about the axis is b/2.
– Because the area of the loop is A = ab, the magnitude of the torque is
τmax = IAB
– Torque occurs only when the field B is parallel to the plane of the
loop.
– If the uniform magnetic field makes an angle θ with a line perpendicular to the plane of the loop, then τ = IABsinθ.
– A vector expression for this is
τ = IA × B,
where A is a vector perpendicular to the plane of the loop and has a
magnitude equal to the area of the loop.
– Right hand rule to determine the sense of A: four fingers of the right
hand curled in the direction of the loop current, thumb points in the
direction of A.
– The magnetic dipole moment µ is defined as
µ = IA.
– SI unit of the magnetic dipole moment is the A.m−2 , ampere-metersquared.
– Thus torque is
τ = µ × B.
– What happens if a coil consists of N turns of wire, each having the
same area and carrying the same current?
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• Go over example 28.7, 28.8, p. 753
• The Biot-Savart Law
– Investigated placing an object in an existing magnetic field.
– Investigated force on a moving charge/current carrying wire when
placed in a magnetic field.
– Now an electric current in a wire also deflects a magnetic field; so the
electric current acts as a source of the magnetic field.
– The Biot-Savart law gives an expression for the magentic field due
to an infinitesimally small element of current that is part of a larger
current distribution.
– Consider the magnetic field dB at point P created by an inifitessimal
length ds of wire:
∗ dB is perpendicular to both ds (which is in the direction of the
current) and to the unit vector r̂ directed from the element toward
P
∗ The magnitude of dB is inversely proportional to r 2 , where r is
the distance from the element to P
∗ The magitude of dB is proportional to the current I and to the
length ds of the element
∗ The magnitude of dB is proportional to sinθ, where θ is the angle
between ds and r̂.
– In other words:
Ids × r̂
r2
, where km is a constant = 10−7 T.m/A in SI units.
dB = km
– km = µ0 /4π, where µ0 is another constant called the permeability of
free space.
– Thus the Biot-Savart law can be written
dB =
µ0 Ids × r̂
.
4π r 2
– The Biot-Savart law only gives the magnetic field due to the current
element Ids. To find the total magnetic field, need to integrate the
above equation over the length of the entire conductor.
– Important differences between and similarities between Biot-Savart
law and electric field due to a charge distribution E = ke rq2 r̂:
∗ Current element Ids produces a magnetic field and the charge
element dq produces an electric field.
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∗ Magnitude of the magnetic field varies as the inverse square of the
distance from the current element as does the magnitude of the
electric field due to a charge element.
∗ Electric field due to a charge element is radial.
∗ Magnetic field due to a current element is perpendicular to both
the current element and the radius vector: right hand rule.
– Magnitude of the magnetic field at a distance r from the wire is
B=
µ0 I
.
2πr
– Go over the proof of this - p. 767.
– Magnetic field due to a Current in a Circular Arc of Wire. See fig.
29-6, p. 767. Then
dB =
µ0 ids
µ0 idssin90
=
.
2
4π R
4π R2
– Now add up over all values of φ,
µ0 i Z φ
dφ,
B=
4πR 0
which is
B=
µ0 iφ
.
4πR
– φ should be in radians not degrees.
– The magnitude of the magnetic field at the center of a full circle of
current is
µ0
B= .
i
– Try Problems, 29-1, 29-2, p. 769-770.
• Magnetic Force Between Two parallel Conductors
– Because a current in a conductor sets up its own magnetic field and
since we know the magnetic force that acts on a current carrying conductor in an external magnetic field,
– Two current carrying conductors exert magnetic forces on each other.
– Two wires, separated by a distance a (a >> radius of wires), carrying
currents I1 , I2 in the same direction.
– Can determine the force on one wire due to the magnetic field set up
by the current in the other wire.
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– Wire 2, carrying current I2 , sets up magnetic field B 2 at the position
of wire 1.
– Thus the force on wire 1, due to the magnetic field set up there by the
current in wire 2 is
F1 = I1 lB2 = I1 l(
lµ0 I1 I2
µ0 I2
)=
.
2πa
2πa
– Can rewrite this expression in terms of the force per unit length as
F1 l = µ0 I1 I2
2πa
– Direction of F 1 is downward, toward wire 2, because l × B 2 is downward.
– The force, F 2 on wire 2 is equal in magnitude and opposite in direction
to F 1 .
– Thus, the magnetic force per unit length exerted by each long current
carrying wire on the other is
µ0 I1 I2
F
=
.
l
2πa
– Parallel conductors carrying currents in the same direction attract
each other but parallel conductors carrying conductors in opposite
directions repel each other.
– This is used to define the ampere:
– If two long parallel wires, 1m apart, each carrying the same current
and the force per unit length on each wire is 2×10−7 M/m, the current
in each wire is defined to be 1A.
– Use this to define Coulomb, the unit of charge:
– If a conductor carries a steady current of 1A, the quantity of charge
that flows through a cross-section of the conductor in 1 second is 1 C.
– Try This: A loose spiral spring is hung from the ceiling and a large
current element is sent through it. Do the coils a) move closer together,
b) move further apart or c) not move at all?
• Amperes Law
– Iron filings sprinkled in a plane around a wire. When the wire carries
no current, iron filings point in the same direction (Earth’s magnetic
field). When the wire carries a steady current, ironf filings form circles
around wire. Thus a current carrying conductor produces a magnetic
field.
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– Direction of B is consistent with the right hand rule.
– B is proportional to the current and inversely proportional to the
distance from the wire.
– Gauss’s Law: a relationship between electric charge and the electric
field it produces.
– Now investigate the relation between a current and the magnetic field
it produces:
∗ B.ds for a small element ds is Bds because the vectors are parallel
to each other.
∗ Thus
I
I
µ0 I
= µ0 I.
B.ds = B ds =
2πr
∗ Ampere’s law: when a steady current passes through the area
surrounded by an arbitrary closed path, then
Z
B.ds = µ0 I.
∗ Amperes law valid only for steady currents.
∗ Magnetic Field Outside and Inside a Long Straight Wire with
Current, p. 773-774.
∗ Example 29.3 - p. 775.
– The Magnetic Field of a Solenoid
∗ A solenoid is a long wire wound in the form of a helix. Can produce a reasonably uniform magnetic field throughout the volume
enclosed by the solenoid, except close to its ends.
∗ Magnetic field is similar to that of a bar magnet - Figure 29-18,
29-19, p. 776.
∗ For an ideal solenoid, field outside is negligible and field inside is
uniform.
∗ Applying Ampere’s law,
N
I,
l
where n = N/l is the number of turns per unit length.
∗ Magnetic field of a Toroid.
B = µ0
– Current Carrying Coil as a Magnetic Dipole.
∗ Saw that a coil placed in an external magnetic field B, is acted on
by a torque given by
τ = µ × B,
where µ is the magnetic moment of the coil and is NIA, where N
is the number of turns, I the current and A the area of the coil.
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∗ But what about the magnetc field exerted by the coil at a point
in the surrounding space?
∗ Its
µ0 iR2
,
B(z) =
2(R2 + z 2 )3/2
where R is the radius of the circular loop, z is the distance of the
point in question from the center of the loop. The direction of
the magnetic field is in the same direction as the magnetic dipole
moment of the loop - find this by the right hand rule.
∗ Go over the proof of this - p.. 779.
– What causes certain materials to exhibit strong magnetic properties?
– Consider the Bohr model of the atom, in which electrons are assumed
to move in circular orbits about the much more massive nucleus.
– Electrons have a charge 1.6 × 10−19 C and circle the atom once every
10−16 s.
– Orbiting electron is equivalent to a current of about 1.6 × 10−3 A.
– Each orbiting electron is therefore a current loop with a corresponding
magnetic moment.
– In most substances, the magnetic moment of one electron in an atom
is canceled by that of another electron in the atom orbiting in the
opposite direction.
– Magnetic effect produced by the orbital motion of the electron is either
zero or very small for most materials.
– Electron also has an intrinsic angular momentum, called spin which
contributes to its magnetic moment.
– Compare the spin of earth about its axis to its rotation around the
sun.
– In atoms or or ions containing multiple electrons, many electrons are
paired up with spins in opposite directions leading to cancellation of
spin magnetic moment.
– But in an atom with an odd number of electrons, there is at least one
”unpaired” electron and a spin magnetic moment.
– This leads to various types of magnetic behaviour.
– Ferromagnetic: atoms with spin magnetic moments that tend to align
parellel to each other even in a weak external magnetic field, eg. iron,
cobalt, nickel.
– Once the moments are aligned, the substance remains aligned after
the external field is removed.
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– Ferromagnetic materials contain domains within which all magnetic
moments are aligned.
– domains range from 10−12 − 10−8 m3 in volume containing some 1017 −
1021 atoms.
– Boundaries between domains having different spin orientations are
called domain walls.
– Extent to which a ferromagnetic substance retains its magentism: hard
or soft.
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