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Chapter 28
Magnetic Fields
In this chapter we will cover the following topics:
r
Magnetic field vector B
r
Magnetic force on a moving charge FB
Magnetic field lines
Motion of a moving charge particle in a uniform magnetic field
Magnetic force on a current carrying wire
Magnetic torque on a wire loop
r
Magnetic dipole, magnetic dipole moment µ
Hall effect
Cyclotron particle accelerator
What is the equivalent resistance
between points F and G? Each
Resistor is 8 Ω.
(A) 2.5 Ω
(B) 8.0 Ω
(C) 2.0 Ω
(D) 4.0 Ω
(E) none of these
(28 – 1)
What produces a magnetic field
r
Definition of B
One can generate a magnetic field using one of the
Pass a current through a wire and thus form what is knows
as an "electromagnet".
Use a "permanent" magnet
and its magnitude FB = q v sin φ where φ is the angle between
r
r
r
v and B. In addition FB is perpendiculat to the plave defined
r
r
by v and B. The magnetic force vector is given by the equation:
r
r r
FB = qv × B
Empirically we know that both types of magnets attract
small pieces of iron. Also if supended so that they can
rotate freely they align themselves along the north-south
direction. We can thus say that these magnets create in
r
the surrounding space a "magnetic field" B which
r
manifests itself by exerting a magnetic force FB .
SI unit of B : The defining equation is FB = q v sin φ
If we shoot a particle with charge q = 1 C at right
r
angles (φ = 90°) to B with speed v = 1m/s and the
We will use the magnetic force to define precicely
r
the magnetic field vector B.
The Vector Product of two Vectors
r
r r r
r
The vector product c = a × b of the vectors a and b
r
is a vector c
r
The magnitude of c is given by the equation:
c = ab sin φ
r
The direction of c is perpendicular to the plane P defined
r
r
by the vectors a and b
r
The sense of the vector c is given by the right hand rule:
r
r
a. Place the vectors a and b tail to tail
r
b. Rotate a in the plane P along the shortest angle
r
so that it coincides with b
c. Rotate the fingers of the right hand in the same direction
r
d. The thumb of the right hand gives the sense of c
The vector product of two vectors is also known as
r
r r
FB = qv × B
r
The magnetic field vector is defined in terms of the force FB it
r
exerts on a charge q which moves with velocity v . We inject
r
the charge q in a region where we wish to determine B at
random directions, trying to scan all the possible directions.
r
There is one direction for which the force FB on q is zero. This
r
r
direction is parallel with B . For all other directions FB is not zero
following methods:
(28 – 2)
FB = q vB sin φ
magnetic force FB = 1 N, then B = 1 tesla
(28 – 3)
r
r r
The Vector Product c = a ×b in terms of Vector Components
r
r
r
a = a x iˆ + a y ˆj + az kˆ , b = b x iˆ + by ˆj + bz kˆ , c = c x iˆ + c y ˆj + cz kˆ
r
The vector components of vector c are given by the equations:
cx = a y bz − az by , c y = a z bx − ax bz , c z = a xby − a y bx
Note : Those familiar with the use of determinants can use the expression
$i
$j k$
r r
a × b = ax a y a z
bx
by
bz
Note : The order of the two vectors in the cross product is important
r r
r r
b × a = − a ×b
(
)
the "cross" product
(28 – 4)
(28 – 5)
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Magnetic Field Lines : In analogy with the electric field lines we
introduce the concept of magnetic field lines which help visualize
r
the magnetic field vector B without using equations.
r
The relation between the magnetic field lines and B are:
r
1. At any point P the magnetic field vector B is tangent to the magnetic field lines
Magnetic field lines of a permanent magnet
The magnetic field lines of a permanent magnet are shown
in the figure. The lines pass through the body of the magnet
and form closed loops. This is in contrast to the electric field
lines that originate and terminate on elecric charges.
The closed magnetic field lines enter one point of the magnet
and exit at the other end. The end of the magnet from which
r
BP
P
magnetic field line
the lines emerge is known as the north pole of the magnet.
The other end where the lines enter is called the south pole
of the magnet. The two poles of the magnet cannot be
separated. Together they form what is known as a
"magnetic dipole"
r
2. The magnitude of the magnetic field vector B is proportional
to the density of the magnetic field lines
r
BQ
r
BP
BP > BQ
Q
P
magnetic field lines
(28 – 6)
(28 – 7)
r
FE
hitt
(28 – 8)
Cathode
r
r
FE = qE
The current in the 5.0 Ω resistor in the circuit
shown is:
(1)0.42A (2) 0.67A (3) 1.5A (4) 2.4A (5) 3.0A
Anode
r
FB
r
r r
FB = qv × B
Discovery of the electron : A cathode ray tube is shwon in the figure. Electrons are
emitted from a hot filament known as the "cathode". They are accelerated by a
voltage V applied between the cathode and a second electrode known as the "anode".
The electrons pass through a hole in the anode and they form a narrow beam. They
hit the fluorescent coating of the right wall of the cathode ray tube where they produce
a spot of light. J.J. Thomson in 1897 used a version of this tube to investigate the nature
of the particle beam that caused the fluorescent spot. He applied constant electric and
magnetic fields in the tube region to the right of the anode. With the fields oriented as
r
r
shown in the figure the electric force FE and the magnetic force FB have opposite
directions. By adjusting B and E Thomson was able to have a zero net force
r
v
r
F
C
. Br
electron
.
r
r=
mv
qB
ω=
qB
m
Motion of a charged particle in a uniform magnetic field
(also known as cyclotron motion)
A particle of mass m and charge q when injected with a speed
r
v at right angles to a uniform magnetic field B, follows a
circular orbit, with uniform speed. The centripetal force
required for such motion is provided by the magnetic force
r
r r
FB = qv × B
r=
mv⊥
qB
T=
2π m
qB
Helical paths
We now consider the motion of
a charged in a uniform magnetic
r
field B when its initial velocity
r
r
v forms and angle φ with B.
r
We decompose v into two
components.
The circular orbit of radius r for an electron is shown in the figure. The magnetic force
One component ( v
v2
mv
2π r 2π mv 2π m
FB = q vB = ma = m → r =
. The period T =
=
=
r
v
q Bv
qB
qB
v = v cos φ
The particle executes two independent motions.
r
One is the cyclotron motion is in the plane perpendicular to B we have
qB
qB
1
. The angular frequency ω = 2π f =
=
T 2π m
m
Note 1 : The cyclotron period does not depend on the speed v. All particles of the
same mass complete their circular orbit during the same time T regardless of speed
analyzed in the previous page. Its radius r =
speed v . The combination of the two motions results in a helical path (see fig.b)
Note 2 : Fast particles move on larger radius circular orbits, while slower particles move
on smaller radius orbits. All orbits have the same period T
(28 – 9)
The pitch p of the helix is given by: p = Tv =
The corresponding frequency f =
)
r
r
parallel to B and the other ( v⊥ ) perpendicular to B (see fig.a)
v⊥ = v sin φ
mv⊥
2π m
. Its period T =
qB
qB
r
The second motion is along the direction of B and it is linear motion with constant
2π mv cos φ
qB
(28 – 10)
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Magnetic force on a current carrying wire.
Consider a wire of length L which carries a current i as shown in
Two isolated conducting spheres are separated by a large
distance. Sphere 1 has a radius of R and an initial charge 2Q
while sphere 2 has a radius of 3R and an initial charge 8Q. A
very thin copper wire is now connected to the spheres.
Charge flows between the spheres until they reach the same
electrical potential. How much charge will be transferred
from sphere 2 to sphere 1? (Note that the charge transferred
can be positive, negative or zero .)
(1) Q/2
(2) 2Q
(3) –Q/3
(5) None of these
the figure. A uniform magnetic field B is present in the vicinity
r
of the wire. Experimentaly it was found that a force FB is
r
r
exerted by B on the wire, and that FB is perpendicular
r
FB
to the wire. The magnetic force on the wire is the vector sum
r
of all the magnetic forces exerted by B on the electrons that
constitute i. The total charge q that flows through the wire
in time t is given by:
L
q = it = i
Here vd is the drift velocity of the electrons
vd
(4) -3Q
in the wire.
The magnetic force FB = qvd B sin 90° = i
L
vd B = iLB
vd
FB = iLB
(28 – 11)
(28 – 12)
Magnetic force on a straight wire in a uniform
magnetic field.
Top view
i
.
magnitude is equal to the wire length L and
Magnetic torque on a current loop
Consider the rectangular loop in fig.a with sides of lengths a and b which carries
a current i. The loop is placed in a magnetic field so that the normal nˆ to the loop
r
forms an angle θ with B. The magnitude of the magnetic force on sides 1 and 3 is:
The magnetic force magnitude FB = iLB sin φ
φ
r
dL
r
r r
dFB = idL × B
r
r r
FB = i ∫ dL × B
Magnetic force on a wire of arbitrary shape
placed in a non - uniform magnetic field.
F1 = F3 = iaB sin 90° = iaB. The magnetic force on sides 2 and 4 is:
In this case we divide the wire into elements of
length dL which can be considered as straight.
F2 = F4 = ibB sin(90 − θ ) = ibB cos θ . These forces cancel in pairs and thus Fnet = 0
The torque about the loop center C of F2 and F4 is zero because both forces pass
The magnetic force on each element is:
r
r r
dFB = idL × B The net magnetic force on the
r
r r
wire is given by the integral: FB = i ∫ dL × B
Magnetic dipole moment :
The torque of a coil that has N loops exerted
r
through point C. The moment arm for F1 and F3 is equal to (b / 2) sin θ . The two
torques tend to rotare the loop in the same (clockwise) direction and thus add up.
The net torque τ = τ 1 +τ 3 =(iabB / 2)sin θ + (iabB / 2)sin θ = iabB sin θ = iAB sin θ
r
r
τ = µ×B
by a uniform magnetic field B and carrries a
r r
U = −µ ⋅ B
R L
L
R
current i is given by the equation: τ = NiAB
r
We define a new vector µ associated with the coil
U = −µ B
the coil.
The magnitude of the magnetic dipole moment µ = NiA
Its direction is perpendicular to the plane of the coil
r
The sense of µ is defined by the right hand rule. We curl the fingers of the right hand
in the direction of the current. The thumb gives us the sense. The torque can
r
r
expressed in the form: τ = µ B sin θ where θ is the angle between µ and B.
r r r
In vector form: τ = µ × B
r r
The potential energy of the coil is: U = − µ B cos θ = − µ ⋅ B
L
R
pushes them to the right (labeled "R") side of the strip. This
accumulates negative charge on the R-side and leaves the left
side (labeled "L") of the strip positively charged. As a result
r
of the accumulated charge, an electric field E is generated as
shown in the figure so that the electric force balances the magnetic
force on the moving charges. FE = FB → eE = evd B →
U has a minimum value of − µ B for θ = 0 (position of stable equilibrium)
U has a maximum value of µ B for θ = 180° (position of unstable equilibrium)
Note : For both positions the net torque τ = 0
The Hall effect
In 1879 Edwin Hall carried out an experiment in which
he was able to determine that conduction in metals is due
to the motion of negative charges (electrons). He was also
able to determine the concentration n of the electrons.
He used a strip of copper of width d and thickness l. He passed
a current i along the length of the strip and applied a magnetic
field B perpendicular to the strip as shown in the figure. In the
r
r
presence of B the electrons experience a magnetic force FB that
which is known as the magnetic dipole moment of
U = µB
(28 – 13)
C
has a direction that coincides with that of the current.
r
dF
Side view
CFnet = 0
If we assume the more general case for which the
r
magnetic field B froms and angle φ with the wire
the magnetic force equation can be written in vector
r
r r
r
form as: FB = iL × B Here L is a vector whose
r
r r
FB = iL × B
r
B
τ net = iAB sin θ
(28 – 14)
E = vd B (eqs.1). From chapter 26 we have: J = nevd →
vd =
J
i
i
=
=
ne Ane ldne
(eqs.2)
(28 – 15)
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R L
L
R
E = vd B (eqs.1).
vd = i / ldne
(28 – 17)
The cyclotron particle accelerator
The cyclotron accelerator consissts of two hollow
(eqs.2)
Hall measured the potential difference V between the left and
the right side of the metal strip. V = Ed (eqs.3)
conductors in the shape of the letter dee (these are
known as the "dees" of the cyclotron. Between the
two dees an oscillator of frequency f osc creates an
We substitute E from eqs.3 and vd from eqs.(2) into eqs.1 and get:
V
i
Bi
=B
→
n=
(eqs.4)
d
ldne
V le
Fig. a and b were drawn assuming that the carriers are electrons.
In this case if we define V = VL − VR we get a positive value.
L
R
If we assume that the current is due to the motion
of positive charges (see fig.c) then positive charges accumulate on
the R-side and negative charges on the L-side and thus V = VL − VR
is now a negative number.
By determining the polarity of the voltage develops between the
left and right hand side of the strip Hall was able to prove that
current was composed of moving electrons. From the value of V
(28 – 16)
using equation 4 he was able to determine the concentration of
the negative charge carriers.
oscillating electric field E that exists only in the gap
r=
mv
eB
f =
eB
2π m
between the two dees. At the same time a constant
magnetic field B is applied perpendicular to the
plane of the dees.
In the figure we show a cyclotron accelerator for protons. The protons follow circular
mv
eB
and rotate with the same frequency f =
. If the cyclotron
eB
2π m
frequency matches the oscillator frequency thne the protons during their trip through
orbits of radius r =
the gap between the dees are accelerated by the electric field that exists in the gap.
The faster protons travel on increasingly larger radius orbits. Thus the electric
field changes the speed of the protons while the magnetic field changes only
the direction of their velocity and forces them to move on circular (cyclotron) orbits.
Hitt
Four identical light bulbs A, B, C and D are connected in series
to a constant voltage source. A wire is then connected
across B. What is the brightness of A relative to its former
brightness?
(1) 0.44
(2) 0.56
none of these
(3) 1.78
(4) 2.25
(5)
4