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Magnetism
Ferromagnetism: Substances that exhibit strong
magnetic properties
ex. iron, nickel, cobalt
neodymium
Nd2Fe14B
Paramagnetism: form of magnetism that occurs only in
the presence of an externally applied
magnetic field; weakly attractive
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examples:
platinum, aluminum, chromium,
oxygen, tungsten
Diamagnetism:
property of an object which causes
it to create a magnetic field in opposition
to an externally applied magnetic field
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examples:
nitrogen, gold, silver,
mercury, diamond
Magnetic Field Around a Straight Wire
r
B=
mo I
2πr
mo = 4p x 10-7 T m/A
I = current
r = distance from wire
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ex. A current of 1.5 A is traveling through a wire from left
to right, as shown below. Determine the magnitude and
direction of the field at a location 0.45 m above the wire.
0.45 m
1.5 A
B=
mo I
2πr
( 4π x 10-7 Tm/A )( 1.5 A )
=
B = 6.7 x 10-7 T
2π ( 0.45 m )
Forces on Current-Carrying Wires
Consider a wire that is
hanging in a magnetic
field created by a
horseshoe magnet:
B
N
Pass current up through the wire:
Direction of force can be found by
a Right-Hand-Rule
S
I
B
N
L
S
I
Magnitude can be found from:
F=BIL
where L = length of
wire in field
N
B
L
L sin x
I
S
What if the wire is at an
angle to the field?
direction of force is
into the paper:
F=L(IxB)
magnitude : F = BIL sin x
direction : right-hand fingers in
direction of I ; curl to B;
thumb points in direction of F
ex. A wire 0.40 m long is placed in a
2.0-T magnetic field at a 60o angle
to the field lines, as shown. A
current of 0.75 A is sent through the
wire. Find the magnitude and
direction of the force on the wire.
F=L(IxB)
I = 0.75 A
60o
B = 2.0 T
magnitude: F = BIL sin x
= ( 2.0 T )( 0.75 A )( 0.40 m ) sin 60
F = 0.52 N
direction: Fingers with I, curl to B :
F = 0.52 N
Forces on Charges Moving in Magnetic Fields
B
A positive charge is
directed up through the
magnetic field at left.
p+
Magnitude of force on particle:
F=qvB
Direction can be found from a RHR
More general:
F=q(vxB)
B
p+
F=q(vxB)
F = qvB sin x in the
direction determined
as shown at right
ex. A proton is sent into a 0.50-T
magnetic field with a speed
of 2.5 x 106 m/s.
(a) Find the magnitude and
direction of the force
exerted on the proton.
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B
p+
F=q(vxB)
magnitude: F = qvB sin x
x = 90o
= ( 1.6021 x 10-19 C )( 2.5 x 106 m/s )( 0.50 T )
F = 2.0 x 10-13 N
direction: proton is deflected to the left
(b) The proton is deflected into
a circular path of radius r.
Find r.
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B
x
Fcent = Fmag
mv2
r
x
x
r
p+
= qvB
mv2 = qvB r
mv = q B r
mv
r =
qB
r = 0.052 m
( 1.67 x 10-27 kg )( 2.5 x 106 m/s )
=
( 1.6021 x 10-19 C )( 0.50 T )
Mass Spectrometer:
Bubble Chamber:
Force Between Two Parallel Current-Carrying Wires
Consider two parallel wires:
I
I
Pass current through each wire in the same direction;
determine the direction of magnetic field generated
by each wire above and below the wires; do the
wires attract or repel?
A video clip
Magnetic Flux
The flow of magnetic field lines through a surface area
B
A
Magnetic
flux
Φ=BA
If field lines are at an angle to the
surface, take the perpendicular
component
Use angle with the normal to
the surface
Φ = A ( B cos x )
More generally:
Φ = AB
= A B cos x
dot product
ex. A loop of wire of radius 4.5 cm is placed in a magnetic
field of strength 1.5 T.
(a) If the plane of the loop is parallel to the field lines,
find the magnetic flux through the loop.
normal to loop
Φ = AB
= A B cos x
= A B cos 90
Φ = 0
90o
B = 1.5 T
ex. A loop of wire of radius 4.5 cm is placed in a magnetic
field of strength 1.5 T.
(a) If the plane of the loop is perpendicular to the field lines,
find the magnetic flux through the loop.
B = 1.5 T
Φ = AB
= A B cos x = A B cos 0
= A B = π r2 B
normal
to loop
= π ( 0.045 m )2 ( 1.5 T )
Φ = 0.0095 T m2
= 0.0095 webers
= 0.0095 Wb
Wilhelm Weber
( 1804 – 1891 )
ex. A loop of wire of radius 4.5 cm is placed in a magnetic
field of strength 1.5 T.
(c) If the plane of the loop is at a 50o angle to the field lines,
find the magnetic flux through the loop.
B = 1.5 T
Φ = AB
= A B cos x
= A B cos 40 = π r2 B cos 40
= π ( 0.045 m )2 ( 1.5 T )( 0.766 )
Φ = 0.0073 T m2
= 0.0073 webers
= 0.0073 Wb
40o
Electromagnetic Induction
_
Consider a loop of wire:
+
I
If an EMF (voltage) is induced in the loop,
current will flow.
Faraday: Anytime the magnetic flux that is passing
through the loop changes, an EMF is induced
Faraday's Law
of Induction:
ΔΦ
EMF = - N
Δt
Δ (BA)
EMF = - N
Δt
N = number of coils
negative sign refers
to Lenz's Law
ex. A loop of wire of radius 0.36 m is in a 4.5-T magnetic
field. If the loop is taken out of the field in 0.50 s, what
is the magnitude of the induced EMF in the loop?
EMF = - N Δ (BA)
Δt
=
=
A ΔB
ΔB = 4.5 T
Δt
( 0.40715 m2 )( 4.5 T )
EMF = 3.6 V
Δt = 0.50 s
A = πr2 = π (0.36 m)2
= 0.40715 m2
N=1
0.50 s
Use Lenz's Law: If a bar magnet is brought near a loop
of wire, in what direction will current be induced?
S
N
N
S
I
Consider loop of wire in
magnetic field
Pull loop out of field;
in what direction will current be
induced in the loop?
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B
Consider loop of wire in
magnetic field
Pull loop out of field;
in what direction will current be
induced in the loop?
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Current induced to try to
keep the field in the loop
Induced current will be
clockwise around loop
I
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x
ex. A loop of wire in the shape of a square
with sides of 12.0 cm is in a magnetic
field of strength 24.5 T, as shown at right.
The wire has a resistance of 1.5 Ω.
B = 24.5 T
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12.0 cm
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ex. A loop of wire in the shape of a square
with sides of 12.0 cm is in a magnetic
field of strength 24.5 T, as shown at right.
The wire has a resistance of 1.5 Ω.
(a) If the loop is removed from the field
in 0.30 s, find the magnitude and
direction of the induced current
in the loop.
Δ (BA)
EMF = - N
Δt
=
=
EMF = 1.2 V
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A ΔB
Δt
( 0.120 m )2 ( 24.5 T )
0.30 s
B = 24.5 T
= 1.176 V
12.0 cm
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ex. A loop of wire in the shape of a square
with sides of 12.0 cm is in a magnetic
field of strength 24.5 T, as shown at right.
The wire has a resistance of 1.5 Ω.
(a)
magnitude of EMF = 1.2 V
direction of current in loop = ?
Field inside loop is decreasing;
how will loop respond?
B = 24.5 T
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I
Current is induced to keep field in loop
Direction of current is clockwise around loop
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(b) Find the force necessary to pull
the wire from the field.
force to pull wire = force that field
exerts on the wire
F = BIL
I =
V
R
B = 24.5 T
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R = 1.5 Ω
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x
I = ?
=
1.176 V
1.5 Ω
= 0.784 A
I
x
x
x
F = BIL
= ( 24.5 T )( 0.784 A )( 0.12 m )
F = 2.3 N
EMF = 1.176 V
ex. A rod of zero resistance and length
x
L = 0.45 m slides to the right on two
zero-resistance wires. The wires are
x
joined by a 12.5 Ω resistor, and the
x
system is placed in a 4.8-T magnetic field.
(a) Find the speed v necessary to produce x
a current in the resistor of 0.36 A.
x
L = 0.45 m
R = 12.5 Ω
I = 0.36 A
Can find voltage in loop
V = I R = ( 0.36 A )( 12.5 Ω ) = 4.5 V
EMF = 4.5 V
B = 4.8 T
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Lx
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B = 4.8 T
v
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EMF = 4.5 V
Δ (BA)
EMF = - N
Δt
EMF =
EMF =
v =
B ΔA
Δt
=
B L Δd
EMF
BL
v = 2.1 m/s
Δt
=
B = 4.8 T
L = 0.45 m
Δ (BA)
=
Δt
B Δ( L d )
Δt
= BLv
4.5 V
( 4.8 T )( 0.45 m )
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v
EMF = B L v
EMF induced by
moving a wire of
length L at a
speed v through a
B field
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(b) Find the direction of the current
through the resistor
Use Lenz’s Law
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?
(b) Find the direction of the current
through the resistor
Use Lenz’s Law
As rod moves to the right, field
through loop (into the page) increases
How does loop respond?
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Field will be set up to oppose the increase; field will be
generated that is out of the page
(b) Find the direction of the current
through the resistor
Use Lenz’s Law
As rod moves to the right, field
through loop (into the page) increases
How does loop respond?
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I
Field will be set up to oppose the increase; field will be
generated that is out of the page
Current is induced that flows down through the resistor
B = 4.8 T
(c) How would your answers to (a) and (b)
x
change if the rod were moved to the
left instead of to the right?
x
Answer to (a) would be unchanged,
since direction of v did not play a
role
x
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v
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Would direction of induced current also be the same?
B = 4.8 T
(c) How would your answers to (a) and (b)
x
change if the rod were moved to the
left instead of to the right?
x
Answer to (a) would be unchanged,
since direction of v did not play a
role
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Would direction of induced current also be the same?
As rod is moved to the left, the field inside loop decreases;
loop responds by generating field into page to keep it
Current is induced that moves up through the resistor