Chapter 30: Magnetism
• Ferromagnetism – Iron, cobalt, gadolinium
strongly magnetic
• Can cut a magnet to produce more magnets (no
magnetic monopole)
• Electric fields can magnetize nonmagnetic
metals
• Heat and shock can demagnetize metals
• Curie Temperature: Temperature above which a
magnet cannot form (1043 K for iron)
• Magnetism caused by the spin of an electron
Magnetic Field Lines
• Arrows from ALWAYS POINT SOUTH
• Compass used to find field lines
• North on compass points to south on magnet
Earth’s Magnetic Field
• North pole is really a south magnetic pole
• North geographic pole 0o to 25o (magnetic
declination/angle of dip)
• Flips geologically
Uniform Magnetic Field
• Fields more
uniform in middle
• Vary at the edges
• Like an electric
field
Magnetic Fields
1. Magnetic field present at all points surrounding
1. Permanent magnet
2. Moving charge
2. Vector quantity
3. Exerts a force on charged particles(inverse
square)
Electric Currents Produce Magnetic
Fields
• Right-hand rule
N
S
Biot-Savart Law
B = magnetic field
–
–
–
–
–
Vector (North to south)
Tesla (SI unit, N/Am)
Gauss 1 G = 10-4 T
Earth’s magnetic field ~ ½ G
Strong magnet ~ 2 – 10T
Biot-Savart Law
Magnetic field felt by a point charge
B = mo qv sin q
4p r2
mo = 1.257 X 10-6 Tm/A (permeability constant)
q = Angle between point and moving charge
r
test charge
q
direction of current
A proton (1.60 x 10-19 C) moves along the x-axis
with a velocity of 1.0 X 107 m/s.
a. Calculate the magnetic field at point (1,0) [0]
b. Calculate the magnetic field at point (0,1) [1.6
X 10-13 T]
c. Calculate the magnetic field at point (1,1) [0.57
X 10-13 T]
Perform all calculations assuming the proton is at
the origin.
Magnetic Field in Wires
B = moI
2pd
mo = 4p X 10-7 T m/A (permeability of free space)
I = Current
d = distance from wire
(Assumes wire is infinitely long)
Derivation
Converting from Charge to Current
Consider a long straight wire with current I. Find
the magnetic field at point “d” from the wire.
A wire carries a current of 25 A. What is the
magnetic field 10 cm from this wire?
B = m oI
2pd
B = (4p X 10-7 T m/A)(25A)
(2p)(0.10 m)
B = 5.0 X 10-5 T
A 1.00 m long, 1.00 mm diameter nichrome
heater wire is connected to a 12 V battery.
The resistivity of nichrome is 1.5 X 10-6 W m.
a. Calculate the resistance in the wire (Remember
R = rL/A, and that this is a circular wire)
b.Calculate the current flowing through the wire
c. Calculate the magnetic field strength 1.0 cm
from the wire.
Two parallel wires carry currents in the
opposite directions. The wires are 10.0 cm
apart and carry currents of 5.0 A and 7.0 A.
a. Calculate the magnetic field of each wire at a
point halfway between the two. (2.0 X 10-5 T,
2.8 X 10-5 T)
b. Calculate the net magnetic field at that point.
(4.8 X 10-5 T)
c. Use the right hand rule to verify that these two
fields add together rather than subtract.
Coils
B = moNI
2R
N = number of turns
I = Current
R = Radius of loop
Derivation
Derive the formula for the magnetic field at the
center.
z=0
If there are multiple coils:
A 5 turn, 10.0 cm diameter wire coil has 0.0500
A of current passing through it.
a. Calculate the magnetic field it produces at the
center.
b. Calculate the current that would be needed to
cancel the earth’s magnetic field, 5.0 X 10-5 T.
(0.80 A)
Derive the formula for the magnetic field at the
center of the quarter circular loop shown below.
Assume the end segments do not matter.
Ampere’s Law
• Formulas so far only valid for straight wires
• Ampere’s Law – Valid for all shapes
• Cuts any shape into many small, straight wires
(line integrals)
Smooth the curve out and..
B = m oI
2pR
A wire carrying a current I has a radius R. Derive
the formula for the magnetic field within the
wire at distance r from the center.
Using the equation from the previous problem,
what is the formula at the full radius of the wire?
What happens to the magnetic field as you move
farther away from the wire?
Solenoids
•
•
•
•
Long coil of wire
Doorbells, car starters, switches, electromagnets
Magnetic field is parallel to the coil
B is fairly uniform inside the coil
B = monI
mo = 4p X 10-7 Tm/A
n = N/l = number of loops/length
A 10 cm long solenoid has a total of 400 turns of
wire and carries a current of 2.0 A. Calculate
the magnetic field inside the solenoid.
n = 400 turns/0.10 m = 4000 m-1
B = monI
B = (4p X 10-7 T m/A)(4000m-1)(2.0 A)
B = 0.01 T
Solenoids: Ex 1
A 0.100 T magnetic field is required. A student
makes a solenoid of length 10.0 cm. Calculate
how many turns are required if the wire is to
carry 10.0 A.
Solenoids: Ex 3
A coaxial cable carries current through the central
wire, and then the return current through the
cylindrical braid. Comment on the magnetic
field between the solid wire an the braid.
• There is a magnetic field due to the inner wire in
the insulating sleeve
• The field outside the cable is zero
Doorbell
•Uses a soleniod
•Car starters also work this way
•Completing the circuit produces
a magetic field that pulls the iron
bar against the bell
Doorbell
Toroid
Use Ampere’s law to derive the magnetic field
strength inside and outside a toroid.
Force on a charge in Magnetic Field
• Charged particles can be moved by magnetic
fields
• Used to determine composition of compounds
(mass spectrometry)
• Used to control particle beams (esp. for fusion)
• Earth’s magnetic field funnels dangerous
particles to the poles
I
B
F
•Force out for positive charge and conventional current(“Palm
positive”)
•Force in for negative charges
F = qvBsinq
Forces and particles: Ex 1
Using the right hand rule, predict the direction of
the force on a proton and an electron entering the
following magnetic field.
(v is the direction of the particle)
A proton has a speed of 5.0 X 106 m/s and feels
a force of 8.0 X 10-14N toward the west as it
moves vertically upward.
a. Calculate the magnitude of the magnetic field.
b. Predict its direction
vproton
Fwest
Earth
Using right-hand rule:
B must be towards
geographic north
vproton
Fwest
Earth
F = qvBsinq
B = F/qvsinq
 = 8.0 X 10-14N
(1.6X10-19C)(5.0X106m/s)(sin90o)
 = 0.10 T
A long horizontal wire carries a 10.0 A current.
An electron is travelling to the right at a
speed of 1.00 X 107 m/s. It is 1.00 cm above
the wire.
a. Calculate the magnetic field at the 1.00 cm
mark.
b. Calculate the force experienced by the electron.
c. Sketch the wire and path of the electron
An electron travels at 2.0 X 107 m/s in a plane
perpendicular to a 0.010-T magnetic field.
Describe its path.
Path is circular (right-hand rule, palm positive)
F = mv2
r
qvB = mv2
r
r = mv
qB
r = mv
qB
r = (9.1 X 10-31 kg)(2.0 X 107 m/s )
(1.6 X 10-19 C)(0.010 T)
r = 0.011 m
Mass Spectrometers
• Used to separate different
isotopes and to identify
compounds (e.i. CH3COOH)
• Particles must be charged
(ionized) by heating or electric
current
• Can select the speed at which
something moves through the
main chamber
• Lower the mass, lower the radius
(lighter particles deflected more)
Mass1
Mass2
=
Radius1
Radius2
Two carbon isotopes are placed in a mass
spectrometer. Carbon-12 has a radius of 22.4
cm. The other isotope has a radius of 26.2 cm.
What is the other isotope?
12 = 22.4 cm
x
26.4 cm
x = 14
At what radius would you expect to see the isotope
Carbon-13?
Force on Current Carrying Wires
• Experimentally, current carrying wires
experience a force
F = IlBsinq
– I = current
– l = length
– B = magnetic field
What is the force on a wire carrying 30 A
through a length of 12 cm? The magnetic
field is 0.90 T and the angle is 60o.
F = IlBsinq
F = (30A)(0.12 m)(0.90 T)sin60o
F = 2.8 N into the page
A loop of wire carries
0.245 A and is placed
in a magnetic field.
The loop is 10.0 cm
wide and experiences a
force of 3.48 X 10-2 N
downward (on top of
gravity). What is the
strength of the
magnetic field?
F = IlBsinq
F = IlBsin90o
F = IlB(1)
F = IlB
B=F
Il
B = 3.48 X 10-2 N
(0.245 A)(0.100m)
=
1.42 T
Magnetic Field in Wires
• Moving current carries a
magnetic field
• Wires in your house generate a
magnetic field
Forces between Parallel Wires
• Current in same direction
• Force is attractive
• North to South orientation
I1
F
I2
F
• Current in opposite directions
• Force is repulsive
• South to South orientation
I1
F
I2
F
Force per unit length
F = mo l I1I2
2p L
L = separation
l = length of wire
You also can rearrange to get the total force
Wires: Ex 1
Two wires in a 2.0 m long cord are 3.0 mm apart.
If they carry a dc current of 8.0 A, calculate the
force between the wires.
F = mo I1I2 l
2p L
F = (4p X 10-7 T m/A)(8.0A)(8.0A)(2.0m)
(2p)(3.0 X 10-3 m)
F = 8.5 X 10-3 N
Wires: Ex 2
The top wire carries a current of 80 A. How much
current must the lower wire carry in order to
leviate if it is 20 cm below the first and has a
mass of 0.12 g/m?
F = mo l I1I2
2p L
mg = mo l I1I2
2p L
Solve for I2
I2 = 15 A
Definition of the Ampere
Ampere – current flowing in each of two parallel
wires 1 m apart that results in a force of 2X10-7
N/m between them
1 Coulomb = 1 A s
Torque on a Current Loop
• Loop of wire in a
magnetic field
• Important in motors and
meters (galvanometer)
• Apply right-hand rule to
show force
 = Fr
F = IaB
 = IaBb + IaBb
2
2
 = NIABsin q
 = # loops
I = Current
A = area of loop
B = magnetic field
A circular coil of wire has a diameter of 20.0 cm
and contains 10 loops. The current is 3.00 A and
the coil is placed in a 2.00 T magnetic field.
Calculate the maximum and minimum torque on
the coil
A = pr2 = (p)(0.100 m)2 = 0.0314 m2
Maximum Torque
q = 90o
 = NIABsin q
 = (10)(3.00 A)(2.00 T)(sin 90o)
 = 1.88 Nm
Minimum Torque
q = 0o
sin 0o = 0
=0
A circular loop of wire 50.0 cm in radius is
oriented at 30o to a magnetic field (0.50 T). If
the current in the loop is 2.0 A, what is the
torque?
 = NIABsin q
 =(1)(2 A)[p(0.50 m)2]sin 30o
 = 0.39 Nm
Galvanometers
• I is from the device we are
testing
• Spring keeps loop from
rotating full around
• Deflection indicates
current
• Also used in EKG
machines
DC Motors
• Coil wrapped around an iron core
• Current must be reversed to keep center rotating
• Commutators and brushes
– Commutator – mounted on the shaft
– Brushes - stationary contacts that rub against the
commutators
– Direction of current switches each half-rotation
commutator
brushes
• Increasing the number of coils (“wingdings”)
produces a much steadier torque
AC Motors
• Can work without commutators since current
already switches
• Often use electromagnets rather than permanent
magnets
AC Motor
Loudspeaker
• Variation of current in the coil
• Varies the force caused by the
permanent magnet
• Speaker cone (cardboard)
moves in and out