Chapter 30: Magnetism

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• 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

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Magnetic Field Lines
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• Arrows from ALWAYS POINT SOUTH
• Compass used to find field lines
• North on compass points to south on magnet

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Earths Magnetic Field

• North pole is really a south magnetic pole
• North geographic pole 0o to 25o (magnetic
  declination/angle of dip)
• Flips geologically

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Uniform Magnetic Field

• Fields more uniform in middle
• Vary at the edges
• Like an electric field

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Magnetic Fields

1. Magnetic field present at all points surrounding
2. Permanent magnet
3. Moving charge
4. Vector quantity
5. Exerts a force on charged particles(inverse
   square)

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Electric Currents Produce Magnetic Fields

• Right-hand rule

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• N S

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Biot-Savart Law

• B magnetic field
• Vector (North to south)
• Tesla (SI unit, N/Am)
• Gauss 1 G 10-4 T
• Earths magnetic field ½ G
• Strong magnet 2 10T

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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

test charge
r
q
direction of current
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• A proton (1.60 x 10-19 C) moves along the x-axis
  with a velocity of 1.0 X 107 m/s.
• Calculate the magnetic field at point (1,0) 0
• Calculate the magnetic field at point (0,1) 1.6
  X 10-13 T
• Calculate the magnetic field at point (1,1) 0.57
  X 10-13 T
• Perform all calculations assuming the proton is
  at the origin.

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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)

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Derivation

• Converting from Charge to Current

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• Consider a long straight wire with current I.
  Find the magnetic field at point d from the
  wire.

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• A wire carries a current of 25 A. What is the
  magnetic field 10 cm from this wire?
• B moI
• 2pd
• B (4p X 10-7 T m/A)(25A)
• (2p)(0.10 m)
• B 5.0 X 10-5 T

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• 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.
• Calculate the resistance in the wire (Remember R
  rL/A, and that this is a circular wire)
• Calculate the current flowing through the wire
• Calculate the magnetic field strength 1.0 cm from
  the wire.

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• 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.
• 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)
• Calculate the net magnetic field at that point.
  (4.8 X 10-5 T)
• Use the right hand rule to verify that these two
  fields add together rather than subtract.

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Coils

• B moNI
• 2R
• N number of turns
• I Current
• R Radius of loop

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• Derivation

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• Derive the formula for the magnetic field at the
  center.
• z 0
• If there are multiple coils

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• A 5 turn, 10.0 cm diameter wire coil has 0.0500 A
  of current passing through it.
• Calculate the magnetic field it produces at the
  center.
• Calculate the current that would be needed to
  cancel the earths magnetic field, 5.0 X 10-5 T.
  (0.80 A)

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• Derive the formula for the magnetic field at the
  center of the quarter circular loop shown below.
  Assume the end segments do not matter.

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Amperes Law

• Formulas so far only valid for straight wires
• Amperes Law Valid for all shapes
• Cuts any shape into many small, straight wires
• (line integrals)

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• Smooth the curve out and..
• B moI
• 2pR

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• 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.

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• 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?

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Solenoids

• Long coil of wire
• Doorbells, car starters, switches, electromagnets
• Magnetic field is parallel to the coil
• B is fairly uniform inside the coil

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• B monI
• mo 4p X 10-7 Tm/A
• n N/l number of loops/length

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• 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

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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.

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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.

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• There is a magnetic field due to the inner wire
  in the insulating sleeve
• The field outside the cable is zero

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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

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Doorbell
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Toroid

• Use Amperes law to derive the magnetic field
  strength inside and outside a toroid.

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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)
• Earths magnetic field funnels dangerous
  particles to the poles

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I
B
F

• Force out for positive charge and conventional
  current(Palm positive)
• Force in for negative charges

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• F qvBsinq

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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)

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• 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.
• Calculate the magnitude of the magnetic field.
• Predict its direction

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vproton
Fwest
Earth
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• Using right-hand rule
• B must be towards geographic north

vproton
Fwest
Earth
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• F qvBsinq
• B F/qvsinq
• 8.0 X 10-14N
• (1.6X10-19C)(5.0X106m/s)(sin90o)
• B 0.10 T

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• 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.
• Calculate the magnetic field at the 1.00 cm mark.
• Calculate the force experienced by the electron.
• Sketch the wire and path of the electron

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• An electron travels at 2.0 X 107 m/s in a plane
  perpendicular to a 0.010-T magnetic field.
  Describe its path.

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• Path is circular (right-hand rule, palm positive)
• F mv2
• r
• qvB mv2
• r
• r mv
• qB

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• 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

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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)

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• Mass1 Radius1
• Mass2 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 14
• x 26.4 cm

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• At what radius would you expect to see the
  isotope Carbon-13?

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Force on Current Carrying Wires

• Experimentally, current carrying wires experience
  a force

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• F IlBsinq
• I current
• l length
• B magnetic field

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• 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

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• 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?

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• F IlBsinq
• F IlBsin90o
• F IlB(1)
• F IlB
• B F
• Il
• B 3.48 X 10-2 N 1.42 T
• (0.245 A)(0.100m)

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Magnetic Field in Wires

• Moving current carries a magnetic field
• Wires in your house generate a magnetic field

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Forces between Parallel Wires

• Current in same direction
• Force is attractive
• North to South orientation

I1
I2
F
F
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• Current in opposite directions
• Force is repulsive
• South to South orientation

I1
I2
F
F
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• 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

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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

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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?

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• F mo l I1I2
• 2p L
• mg mo l I1I2
• 2p L
• Solve for I2
• I2 15 A

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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

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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

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• Fr
• F IaB
• t IaBb IaBb
• 2 2
• NIABsin q
• loops
• I Current
• A area of loop
• B magnetic field

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• 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

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• Maximum Torque
• q 90o
• NIABsin q
• (10)(3.00 A)(2.00 T)(sin 90o)
• 1.88 Nm
• Minimum Torque
• 0o
• sin 0o 0
• t 0

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• 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)2sin 30o
• 0.39 Nm

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Galvanometers

• I is from the device we are testing
• Spring keeps loop from rotating full around
• Deflection indicates current
• Also used in EKG machines

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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

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commutator
brushes
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• Increasing the number of coils (wingdings)
  produces a much steadier torque

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AC Motors

• Can work without commutators since current
  already switches
• Often use electromagnets rather than permanent
  magnets

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AC Motor
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Loudspeaker

• Variation of current in the coil
• Varies the force caused by the permanent magnet
• Speaker cone (cardboard) moves in and out