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General Physics (PHY 2140)
Lightning Review
Last lecture:
1. Magnetism
9 Magnetic field
9 Magnetic force on a moving particle
9 Magnetic force on a current
9 Torque on a current loop
9 Motion in a uniform field
Lecture 8
¾
Electricity and Magnetism
1. Magnetism
9 Application of magnetic forces
9 Ampere’s law
2. Induced voltages and induction
9 Magnetic flux
F = qvB sin θ
F = BIl sin θ
τ = NBIA sin θ
r = mv / qB
Review Problem:
How does the aurora borealis (the
Northern Lights) work?
http://www.physics.wayne.edu/~alan/2140Website/Main.htm
Chapter 19-20
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Magnetic Field of the Earth
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The Aurora compared to a CRT
For more info see the aurora home page: http://sec.noaa.gov/pmap/
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19.6 Motion of Charged Particle in magnetic field
Consider positively charge
particle moving in a uniform
magnetic field.
Suppose the initial velocity of
the particle is perpendicular to
the direction of the field.
Then a magnetic force will be
exerted on the particle and
make follow a circular path.
×
×
q
v
×
The magnetic force produces a centripetal acceleration.
×
×
×
×F ×
×
r
Bin
× ×
×
×
F = qvB =
The particle travels on a circular trajectory with a radius:
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
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mv2
r
r=
5
mv
qB
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Application: Mass Spectrometer
Example: Proton moving in uniform magnetic field
A proton is moving in a circular orbit of radius 14 cm in a uniform
uniform
magnetic field of magnitude 0.35 T, directed perpendicular to the
the
velocity of the proton. Find the orbital speed of the proton.
r = 0.14 m
B = 0.35 T
m = 1.67x10-27 kg
q = 1.6 x 10-19 C
r=
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mv
qB
v=
r=
qBr
m
(1.6 ×10
−19
=
)
(1.67 ×10
−27
= 4.7 ×10
6m
(
C ( 0.35T ) 14 ×10−2 m
kg
)
)
mv
qB
s
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See prob. 30 in text
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Review Problem 2
Review Example 1: Flying duck
How does your credit card work?
The stripe on the back of a credit card is a magnetic stripe, often called a
magstripe. The magstripe is made up of tiny iron-based magnetic particles in a
plastic-like film. Each particle is really a tiny bar magnet about 20-millionths of
an inch long.
A duck flying horizontally due north at 15 m/s passes over Atlanta, where
the magnetic field of the Earth is 5.0×10-5 T in a direction 60° below a
horizontal line running north and south. The duck has a positive charge of
4.0×10-8C. What is the magnetic force acting on the duck?
B=5.0 x 10-5 T.
q = 4.0×10-8C
v = 15 m/s
The magstripe can be "written" because the tiny bar magnets can be
magnetized in either a north or south pole direction. The magstripe on the back
of the card is very similar to a piece of cassette tape .
A magstripe reader (you may have seen one hooked to someone's PC at a
bazaar or fair) can understand the information on the three-track stripe.
F = qvB sin θ
θ = 60°
F = qvBsinθ
= (4.0 x10−8 C )(15m / s )(5.0 x10−5 T ) sin 60
= 2.6 x10−13 N
- to the west (into page)
B
v
N
60°
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Review Example 2: Wire in Earth’s B Field
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Ground
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Application: The Velocity Selector
A wire carries a current of 22 A from east to west. Assume that at this location
the magnetic field of the earth is horizontal and directed from south to north,
and has a magnitude of 0.50 x 10-4 T. Find the magnetic force on a 3636-m length
of wire. What happens if the direction of the current is reversed?
reversed?
Felectric = qE
Fmagnetic = qvB
Magnetic force is up…
But the electric force is down.
B=0.50 x 10-4 T.
I = 22 A
l = 36 m
Fmax = BIl
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Fmax = BIl
(
= 0.50 ×10−4 T
Since there is no deflection we
can set these equal to each
other. So we find:
) ( 22 A)( 36m )
qE = qvB
thus we have
v= E/B
= 4.0 ×10−2 N
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Example 2:
19.7 Magnetic Field of a long straight wire
Danish scientist Hans Oersted (1777(1777-1851) discovered
somewhat by accident that an electric current in a wire
deflects a nearby compass needle.
In 1820, he performed a simple experiment with many
compasses that clearly showed the presence of a
magnetic field around a wire carrying a current.
Consider the velocity selector. The electric field between the plates of the
velocity selector is 950 V/m, and the magnetic field in the velocity selector has
a magnitude of 0.930 T directed at right angles to the electric field. Calculate
the speed of an ion that passes undeflected through the velocity selector.
E = 950 V/m
B = 0.93 T
v=?
v=
E 950 V/m
=
= 1021 m/s
B
0.93 T
I=0
I
v = E/B
aligned with
earth’s field
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The passage of a steady current in a wire produces a
magnetic field around the wire.
„
B=
Field form concentric lines around the wire
Direction of the field given by the right hand rule.
If the wire is grasped in the right hand with the thumb in the
direction of the current, the fingers will curl in the direction of
the field.
„
Magnitude of the field
B=
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Magnitude of the field
Magnetic Field due to Currents
„
aligned in a
circular pattern
μo I
2π r
I
r
B
μo called the permeability of free space
I
μo = 4π ×10−7 Tm / A
μo I
2π r
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Ampere’s Law
Consider a case where B is constant and uniform.
Consider a circular path surrounding a current, divided
in segments Δl, Ampere showed that the sum of the
products of the field by the length of the segment is
equal to μo times the current.
∑ B Δl = μ I
∑ B Δl = B ∑ Δl = B 2π r = μ I
o enc
Andre-Marie Ampere
Then one finds:
I
o enc
r
B =
Δl
B
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μo I
2π r
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l
1
19.8 Magnetic Force between two parallel conductors
B2
2
F1
I1
d
I2
B2 =
μo I 2
2π d
μIIl
⎡μ I ⎤
F1 = B2 I1l = ⎢ o 2 ⎥ I1l = o 1 2
2π d
⎣ 2π d ⎦
F1 μo I1I 2
=
2π d
l
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Force per
unit length
( Attractive )
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Definition of the SI unit Ampere
F1 μo I1I 2
=
2π d
l
Example 1: Levitating a wire
Used to define the SI unit of current called
Ampere.
If two long, parallel wires 1 m apart carry the same current, and
and the
magnetic force per unit length on each wire is 2x10-7 N/m, then the
current is defined to be 1 A.
Two wires, each having a weight per units length of 1.0x10-4 N/m, are
strung parallel to one another above the surface of the Earth, one
one
directly above the other. The wires are aligned northnorth-south. When their
distance of separation is 0.10 m what must be the current in each
each in
order for the lower wire to levitate the upper wire. (Assume the two
wires carry the same current).
l
1
2
d
I1
I2
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F1
1
I1
B2
mg/l
2
l
d
I2
Two wires, each having a weight per
units length of 1.0x10-4 N/m, are strung
parallel to one another above the surface
of the Earth, one directly above the
other. The wires are aligned north-south.
When their distance of separation is 0.10
m what must be the current in each in
order for the lower wire to levitate the
upper wire. (Assume the two wires carry
the same current).
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Example 2: magnetic field between the wires
The two wires in the figure below carry currents of 3.00A and 5.00A
5.00A in
the direction indicated (into the page). Find the direction and magnitude
of the magnetic field at a point midway between the wires.
5.00 A
3.00 A
Weight of wire per unit
length:
mg/l = 1.0x10-4 N/m
Wire separation:
d=0.1 m
I1 = I2
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F1 mg μo I 2
=
=
2π d
l
l
1.0 ×10−4 N / m =
X
(
X
Bi =
μo Ii
2π d
20.0 cm
)( )
4π ×10−7 Tm A I 2
( 2π )( 0.10m)
Bnet =
I = 7.1 A
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μo
2π i0.1m
( 5.00 A − 3.00 A) = 4 ×10−6 T
Upwards
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19.10 Magnetic Field of a solenoid
19.9 Magnetic Field of a current loop
Solenoid magnet consists of a wire coil with multiple
loops.
It is often called an electromagnet.
Magnetic field produced by a wire can be enhanced
by having the wire in a loop.
Δx1
B
I
Δx2
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Solenoid Magnet
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Solenoid Magnet
B = μo nI
Field lines inside a solenoid magnet are parallel, uniformly spaced
spaced
and close together.
The field inside is uniform and strong.
The field outside is non uniform and much weaker.
One end of the solenoid acts as a north pole, the other as a south
south
pole.
For a long and tightly looped solenoid, the field inside has a value:
value:
n = N/l : number of (loop) turns per unit length.
I : current in the solenoid.
μo = 4π ×10−7 Tm / A
B = μo nI
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Comparison:
Electric Field vs. Magnetic Field
Example: Magnetic Field inside a Solenoid.
Consider a solenoid consisting of 100 turns of wire and length of
of
10.0 cm. Find the magnetic field inside when it carries a current
current of
0.500 A.
N = 100
l = 0.100 m
I = 0.500 A
μo = 4π ×10−7 Tm/A
Source
Acts on
Force
Direction
N 100 turns
=
= 1000 turns/m
l
0.10 m
B = μo nI
n=
(
)
= 4π ×10−7 Tm/A (1000 turns/m )( 0.500 A)
B = 6.28 ×10−4 T
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Magnetic
Charges
Charges
F = Eq
Parallel E
Moving Charges
Moving Charges
F = q v B sin(θ)
Perpendicular to v,B
Field Lines
Opposites
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Electric
Charges Attract
Currents Repel
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Introduction
Previous chapter: electric currents produce magnetic
fields (Oersted
(Oersted’’s experiments)
Is the opposite true: can magnetic fields create electric
currents?
Chapter 20
Induced EMF and Induction
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20.1 Induced EMF and magnetic flux
Problem: determining a flux
Definition of Magnetic Flux
A square loop 2.00m on a side is placed in a magnetic field of
strength 0.300T. If the field makes an angle of 50.0°
50.0° with the
normal to the plane of the loop, determine the magnetic flux
through the loop.
Just like in the case of electric flux,
consider a situation where the magnetic
field is uniform in magnitude and
direction. Place a loop in the BB-field.
The flux, Φ, is defined as the product of
the field magnitude by the area crossed
by the field lines.
Φ = B⊥ A = BA cos θ
where B⊥ is the component of B
perpendicular to the loop, θ is the angle
between B and the normal to the loop.
Units: T·
T·m2 or Webers (Wb)
Wb)
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The value of magnetic flux is proportional to the total number of
magnetic field lines passing through the loop.
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A square loop 2.00m on a side is placed in a magnetic field of strength 0.300T. If the
field makes an angle of 50.0° with the normal to the plane of the loop, determine the
magnetic flux through the loop.
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20.1 Induced EMF and magnetic flux
Faraday’s experiment
Solution:
Given:
L = 2.00 m
B = 0.300 T
θ = 50.0˚
From what we are given, we use
Φ = BA cosθ = ( 0.300T )( 2.00m ) cos(50.0 )
2
= 0.386 Tm2
Find:
Φ=?
Picture © Molecular Expressions
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Two circuits are not connected:
no current?
However, closing the switch
we see that the compass’
compass’
needle moves and then goes
back to its previous position
Nothing happens when the
current in the primary coil is
steady
But same thing happens when
the switch is opened,
opened, except
for the needle going in the
opposite direction…
direction…
What is going on?
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