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Chapter 24 - Magnetism

Chapter 24 - Magnetism. Only a few substances are magnetic ( ferromagnetic ) Iron (Fe), Cobalt (Co), and Nickel (Ni) + a couple rare earth metals . Some other metals, like Aluminum, can be magnetic if in the presence

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Chapter 24 - Magnetism

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  1. Chapter 24 - Magnetism • Only a few substances are magnetic ( ferromagnetic ) • Iron (Fe), Cobalt (Co), and Nickel (Ni) + a couple rare earth metals • Some other metals, like Aluminum, can be magnetic if in the presence • of a very strong magnet. ( Paramagnetic ) • All elements have electrons – moving and spinning • - spin creates magnetic field. • - most elements have spins cancelling out • - some only have a small amount and do not cancel • - Fe, Co, Ni have larger amounts that don’t cancel.

  2. Magnetic poles - always creates a North and South pole N S N S N S => Magnetic domains - groups of atoms with their magnetic fields aligned ~1020 atoms aligned in one domain. Unmagnetized Weak magnet Strong magnet

  3. Magnetic Field Lines - always from North to South N S N S N S N S

  4. Magnetic Field Lines - always from North to South N S Electromagnetism - When current runs through a wire, sets up a magnetic field (Moving E-field creates a B-field ) I  N S N S N S

  5. Electromagnetism - When current runs through a wire, sets up a magnetic field (Moving E-field creates a B-field ) I  Right-Hand Rule #1 - in a coil, a B-field overlaps itself as many times as there are coils. I Iin Iout I - place an iron core in the coil, its atoms (domains) align, gets an even stronger electromagnet - more current = stronger - more coils = stronger Chapter 24 HW# 1 1 – 6

  6. Lab24.1 – Magnetic Fields - due next day - Ch24 HW#1 due at beginning of period

  7. Ch24 HW#1 1 – 6 1. a) Two Norths: b) Two Souths: c) North & South: 2. 3. 4. N S N S N S N S

  8. Ch24 HW#1 1 – 6 1. a) Two Norths: Repel b) Two Souths: Repel c) North & South: Attract 2. 3. 4. N S N S N S N S

  9. Ch24 HW#1 1 – 6 1. a) Two Norths: Repel b) Two Souths: Repel c) North & South: Attract 2. S N N S Atoms in nail re-align 3. 4. N S N S N S N S

  10. 5. I 6. I I Increase I Increase # of coils Put iron core inside

  11. Ch24.2 – Magnetic Force Right – Hand Rule #2 All 3 fingers must be perpendicular to each other Magnetic Force on a wire FB = B ∙ I ∙ L FB – magnetic force B – magnetic field Units of B-field:Teslas I – current T: N/A∙m L – length of wire

  12. FB = B ∙ I ∙ L FB – magnetic force B – magnetic field I – current L – length of wire Ex1) A straight wire 0.10 m long carries 5.0 A from east to west, in a uniform magnetic field of 0.4 T, oriented south, what is the magnitude and direction of the force on the wire? Ex2) A 0.25 m wire carries 2.1 A out of the page, while in a 0.15 T uniform B-field oriented along the (-x) axis. What is the magnitude and direction of the resulting force? Ex3) A 0.3 m wire carries 0.01 A along the (-y) axis, while in an unknown B-field. A force of 3 N pushes the wire towards the (+x) axis. What is the strength and direction of the B-field?

  13. Ex1) A straight wire 0.10 m long carries 5.0 A from east to west, in a uniform magnetic field of 0.4 T, oriented south, what is the magnitude and direction of the force on the wire? FB = B ∙ I ∙ L = ( .4 T )( 5 A )( .1 m ) = .2 N Ex2) A 0.25 m wire carries 2.1 A out of the page, while in a 0.15 T uniform B-field oriented along the (-x) axis. What is the magnitude and direction of the resulting force? FB = B ∙ I ∙ L = ( .15 T )( 2.1 A )( .25 m ) = ( .08 N ) Ex3) A 0.3 m wire carries 0.01 A along the (-y) axis, while in an unknown B-field. A force of 3 N pushes the wire towards the (+x) axis. What is the strength and direction of the B-field? FB = B ∙ I ∙ L ( 3 N ) = B ∙ ( .01 A )( .3 m ) Ch24 HW#2 7 – 10 B = 1,000 T

  14. Lab24.2 Electric Motors - due tomorrow - Go over Ch24 HW#2 first

  15. Ch24 HW#2 7 – 10 7) 0.50 m wire, 8 A along (+x) axis, 0.4 T B-field along (+y) axis. What is the magnitude and direction of force? B F = B ∙ I ∙ L = (.4T)(8A)(.5m) I→ 8) A wire 75 cm long with 6.0 A pointing North in a 1.2 T B-field into the page. What is the magnitude and direction of force? x x x x B I FB x x x x x x x x x x x x

  16. 7) 0.50 m wire, 8 A along (+x) axis, 0.4 T B-field along (+y) axis. What is the magnitude and direction of force? B F = B ∙ I ∙ L = (.4T)(8A)(.5m) = 1.6 N . I→ FB (out of the page) 8) A wire 75 cm long with 6.0 A pointing North in a 1.2 T B-field into the page. What is the magnitude and direction of force? x x x x B I FB x x x x x x x x x x x x

  17. 7) 0.50 m wire, 8 A along (+x) axis, 0.4 T B-field along (+y) axis. What is the magnitude and direction of force? B F = B ∙ I ∙ L = (.4T)(8A)(.5m) = 1.6 N . I→ FB (out of the page) 8) A wire 75 cm long with 6.0 A pointing North in a 1.2 T B-field into the page. What is the magnitude and direction of force? F = B ∙ I ∙ L = ( 1.2 T )( 6 A )( .75 m ) = 5.4 N x x x x B I FB x x x x x x x x x x x x

  18. 9) A wire 150 cm long carries 20A along (-x) axis. The force on the wire is 0.6 N in (+y) direction. Find magnitude and direction of B-field. F ←I 10) A 40 cm long copper wire has a current of 6 A and weighs 0.35 N. The B-field produces a force that balances the wire. Find the B-field. FB I Fg

  19. 9) A wire 150 cm long carries 20A along (-x) axis. The force on the wire is 0.6 N in (+y) direction. Find magnitude and direction of B-field. F ←IB F = B ∙ I ∙ L 0.6N = B ∙ (20A)(1.50m) B = 0.02T 10) A 40 cm long copper wire has a current of 6 A and weighs 0.35 N. The B-field produces a force that balances the wire. Find the B-field. I .

  20. 9) A wire 150 cm long carries 20A along (-x) axis. The force on the wire is 0.6 N in (+y) direction. Find magnitude and direction of B-field. F ←IB F = B ∙ I ∙ L 0.6N = B ∙ (20A)(1.50m) B = 0.02T 10) A 40 cm long copper wire has a current of 6 A and weighs 0.35 N. The B-field produces a force that balances the wire. Find the B-field. FB B I Fg F = B ∙ I ∙ L 35N = B ∙ (6A)(0.4m) B = 0.15T . x

  21. Lab24.2 – Electric Motors I I N S As the wire spins around: . x . . x x N S N S x . x . N S N S

  22. Ch24.3 – Force on Charges Magnetic force on a charge particle FB = B ∙ q ∙ v B – magnetic field q – charge v – velocity Ex1) A +0.5 C charged particle travelling at 500 m/s into the page passes through a 2.5 T B-field oriented North. What is the resulting force? Ex2) An electron travels at 3x106m/s to the right through a B-field of 4x10-2 T oriented along the (-y) axis. What is the magnitude and direction of the force? ( qe = -1.6x10-19 C )

  23. Magnetic force on a charge particle FB = B ∙ q ∙ v B – magnetic field q – charge v – velocity Ex1) A +0.5 C charged particle travelling at 500 m/s into the page passes through a 2.5 T B-field oriented North. What is the resulting force? FB = B ∙ q ∙ v = ( 2.5 T )( +.5 C )( 500 m/s ) = 625 N Ex2) An electron travels at 3x106m/s to the right through a B-field of 4x10-2 T oriented along the (-y) axis. What is the magnitude and direction of the force? ( qe = -1.6x10-19 C ) FB = B ∙ q ∙ v = ( 4x10-2 T )( -1.6x10-19 C )( 3x106m/s ) = -1.92x10-14 N

  24. Magnetic force on a charge particle FB = B ∙ q ∙ v B – magnetic field q – charge v – velocity Ex3) A +0.06 C charge moves along the (+x) axis at 250 m/s parallel to the earth’s surface. It has a mass of 0.01 kg. What minimum B-field will keep it traveling straight?

  25. Magnetic force on a charge particle FB = B ∙ q ∙ v B – magnetic field q – charge v – velocity Ex3) A +0.06 C charge moves along the (+x) axis at 250 m/s parallel to the earth’s surface. It has a mass of 0.01 kg. What minimum B-field will keep it traveling straight? FG = FB m∙g = Bqv B = m∙g = ( .01 kg )( 9.8 m/s2 ) = .006 T q∙v ( .06 C )( 250 m/s ) Ch24 HW#3 11 – 14

  26. Ch24 HW#3 11 – 14 11) Given +1 μC at 3x104m/s into the page, and B-field is 9x10-2 T East. What is the force? B F = B∙q∙v = (9x10-2T)(+1x10-6C)(3x104m/s) vel = 12) Given +2.2 nC at 9x106m/s along (-y) axis and enters a B-field of 4x10-2 T (+z). Find force. vel B F = B∙q∙v = x .

  27. 11) Given +1 μC at 3x104m/s into the page, and B-field is 9x10-2 T East. What is the force? B F = B∙q∙v FB = (9x10-2T)(+1x10-6C)(3x104m/s) vel = 0.0027 N 12) Given +2.2 nC at 9x106m/s along (-y) axis and enters a B-field of 4x10-2 T (+z). Find force. vel B F = B∙q∙v = x .

  28. 11) Given +1 μC at 3x104m/s into the page, and B-field is 9x10-2 T East. What is the force? B F = B∙q∙v FB = (9x10-2T)(+1x10-6C)(3x104m/s) vel = 0.0027 N 12) Given +2.2 nC at 9x106m/s along (-y) axis and enters a B-field of 4x10-2 T (+z). Find force. vel B F = B∙q∙v FB = (4x10-2T)(2.2x10-9C)(9x106 m/s) = 7.9x10-4 N x .

  29. 13) Given an electron at 4x106m/s West in a B-field of 0.5T North, • what is the force? • B FB = B∙q∙v • vel = (.5T)(-1.6x10-19C)(4x106 m/s) • = • 14) A stream of +0.5mC ionized particles have a speed of 2.5x102 m/s out • of the page, and perpendicular to a 1.0x10-2T B-field oriented South. Force? • B FB = B∙q∙v • vel = (1.0x10-2T)(+0.5x10-3C)(2.5x102 m/s) • = - .

  30. 13) Given an electron at 4x106m/s West in a B-field of 0.5T North, • what is the force? • B FB = B∙q∙v • vel = (.5T)(-1.6x10-19C)(4x106 m/s) • = -3.2x10-13 N • FB(full credit) • FB(extra credit) • 14) A stream of +0.5mC ionized particles have a speed of 2.5x102 m/s out • of the page, and perpendicular to a 1.0x10-2T B-field oriented South. Force? • B FB = B∙q∙v • vel = (1.0x10-2T)(+0.5x10-3C)(2.5x102 m/s) • = - x . .

  31. 13) Given an electron at 4x106m/s West in a B-field of 0.5T North, • what is the force? • B FB = B∙q∙v • vel = (.5T)(-1.6x10-19C)(4x106 m/s) • = -3.2x10-13 N • FB(full credit) • FB(extra credit) • 14) A stream of +0.5mC ionized particles have a speed of 2.5x102 m/s out • of the page, and perpendicular to a 1.0x10-2T B-field oriented South. Force? • B FB = B∙q∙v • vel = (1.0x10-2T)(+0.5x10-3C)(2.5x102 m/s) • = 0.00125 N • FB - x . .

  32. Ch24 – Mid Chapter Review Ex1) A 5.0 A current passes through a section of wire that is 0.6 m long oriented with the current heading West. If the section of wire is in a 1.2 T uniform magnetic field oriented North, what is the magnitude and direction of the force? BI Ex2) A 1.13 m long wire oriented out of the page has a current of 2.5 A running through it. It is in a 0.5 T B-field oriented North. What is the magnitude and direction of FB? BI .

  33. Ex1) A 5.0 A current passes through a section of wire that is 0.6 m long oriented with the current heading West. If the section of wire is in a 1.2 T uniform magnetic field oriented North, what is the magnitude and direction of the force? BIFB = B∙I∙L FB = (1.2T)(5A)(.6m) = 3.6 N Ex2) A 1.13 m long wire oriented out of the page has a current of 2.5 A running through it. It is in a 0.5 T B-field oriented North. What is the magnitude and direction of FB? BIFB = B∙I∙L FB= (.5T)(2.5A)(1.13m) = 1.6 N x .

  34. Ex3) A +2 μC charge is travelling South at 3x103m/s through a 5.8 mT B-field oriented West. Find the force. B vel Ex4)A -1.5 mC charge is traveling into the page at 4x105m/s through a 8.2 μT B-field oriented East. Find the force. B vel x

  35. Ex3) A +2 μC charge is travelling South at 3x103m/s through a 5.8 mT B-field oriented West. Find the force. FB = B∙q∙v B vel = (5.8x10-3T)(+2x10-6C)(3x103m/s) FB= 3.5x10-5 N Ex4)A -1.5 mC charge is traveling into the page at 4x105m/s through a 8.2 μT B-field oriented East. Find the force. B FB (f.c.)FB = B∙q∙v vel= (8.2x10-6T)(-1.5x10-3C)(4x105m/s) FB (X.C.)= -4.9x10-3 N x x

  36. Ex5) A +0.5 μC charge traveling at 1x105 m/s has a mass of 10 g. If it is traveling East along the surface of the Earth. What kind of B-field will keep it straight? B vel x x x + x x x

  37. Ex5) A +0.5 μC charge traveling at 1x105 m/s has a mass of 10 g. If it is traveling East along the surface of the Earth. What kind of B-field will keep it straight? BFB vel Fg FB = FG B.q∙v= m∙g B.(+0.5x10-6C)(1x105m/s) = ( .01 kg )( 9.8 m/s2 ) B = 1.96 T Ch24 HW#4 15 – 18 x x x + x x x

  38. Ch24 HW#4 15 – 18 15. A wire 1.25m long carrying a current of 2A along the (+)y axis is in a 0.95T B-field oriented out of the page. Find force. FB = B ∙ I ∙ L IFB = (0.95T)∙(2A)(1.25m) B = 16. A wire 25cm long carrying a current of 1.4A to the west is placed in a 3.1T B-field pointing North. Find force. B FB = B ∙ I ∙ L = (3.1T)∙(1.4A)(0.25m) I .

  39. 15. A wire 1.25m long carrying a current of 2A along the (+)y axis is in a 0.95T B-field oriented out of the page. Find force. FB = B ∙ I ∙ L IFB = (0.95T)∙(2A)(1.25m) B = 16. A wire 25cm long carrying a current of 1.4A to the west is placed in a 3.1T B-field pointing North. Find force. B FB = B ∙ I ∙ L = (3.1T)∙(1.4A)(0.25m) I .

  40. 17) A stream of +5.2μC ionized particles have a speed of 2x106 m/s traveling along +y axis in a 2.0x10-3T B-field oriented +x axis. Force? • B FB = B∙q∙v • vel = (2x10-3 T)(+5.2x10-6C)(2x106 m/s) • = • 18) Given an electron at 1.3x104m/s east in a B-field of 1.5T North, • what is the force? • B FB = B∙q∙v • vel = (1.5T)(-1.6x10-19C)(1.3x104 m/s) • =

  41. 17) A stream of +5.2μC ionized particles have a speed of 2x106 m/s traveling along +y axis in a 2.0x10-3T B-field oriented +x axis. Force? • B FB = B∙q∙v • vel = (2x10-3 T)(+5.2x10-6C)(2x106 m/s) • = • 18) Given an electron at 1.3x104m/s east in a B-field of 1.5T North, • what is the force? • B FB = B∙q∙v • vel = (1.5T)(-1.6x10-19C)(1.3x104 m/s) • =

  42. Ch25.1 – Electromagnetic Induction Ch24: A current creates a B-field I Does a B-field create a current?

  43. Ch25.1 – Electromagnetic Induction Ch24: A current creates a B-field I Does a B-field create a current? NO! Only a changing B-field creates a current! The symmetry to the physics is not current creates B-field, B-field creates current. The symmetry is Electrical energy creates movement in a B-field, then movement in the B-field creates electrical energy.

  44. Ch25.1 – Electromagnetic Induction Ch24: A current creates a B-field I Does a B-field create a current? NO! Only a changing B-field creates a current! The symmetry to the physics is not current creates B-field, B-field creates current. The symmetry is Electrical energy creates movement in a B-field, then movement in the B-field creates electrical energy. - usually this is done by moving a coil of wires in and out of a B-field (or vice versa ). Moving in creates a (+) current. Moving out creates a (-) current. - Faraday called the voltage produced: Electromotive Force (not actually a force. It’s a potential difference (V) ) EMF = B∙L∙v (voltage) (B-field) (length of wire) (velocity) Alternating current

  45. . . . EMF = B∙L∙vB vel Ex1) A 0.20 m long wire moves to the right at 7.0 m/s , through a B-field of 8.0x10-2 T oriented out of the page as shown. What is the induced EMF? b) If the wire is part of a circuit with a total resistance of 0.50 Ω, what is the magnitude and direction of the current? R = .5 Ω . . . . . . . . . . . . . . .

  46. . . . EMF = B∙L∙vB vel Ex1) A 0.20 m long wire moves to the right at 7.0 m/s , through a B-field of 8.0x10-2 T oriented out of the page as shown. What is the induced EMF? EMF = B∙L∙V = ( 8x10-2 T )( .2 m )( 7 m/s ) = .112 V b) If the wire is part of a circuit with a total resistance of 0.50 Ω, what is the magnitude and direction of the current? R = .5 Ω . . . . . . . . . . . . . . .

  47. Generator: converts mechanical energy to electrical energy. Motor: converts electrical energy to mechanical energy. AC generator: Top view: Side view: F B BF B B F F F FFF (V or I) t . x

  48. Generator: converts mechanical energy to electrical energy. Motor: converts electrical energy to mechanical energy. AC generator: Top view: Side view: F B BF B B F F F FFF (V or I) t What value do you use for I for a generator? . x

  49. Effective current ( Ieff ) = average I Ieff = .707∙ImaxVeff = .707∙Vmax PAC = Ieff2∙R or PAC = Ieff ∙ Veff Ex2) What is the effective current for a generator that produces a max current of 15 A? If the effective voltage for this generator is 120 V AC, what is the average power produced? Ch25 HW#1 1 – 6

  50. Effective current ( Ieff ) = average I Ieff = .707∙Imax Veff = .707∙Vmax PAC = Ieff2∙R or PAC = Ieff ∙ Veff Ex2) What is the effective current for a generator that produces a max current of 15 A? Ieff = ? Ieff = ( .707 )( 15A ) = 10.6 A Imax = 15A If the effective voltage for this generator is 120 V AC, what is the average power produced? Veff = 120 V PAC = Ieff ∙ Veff = ( 10.6 A )( 120 V) = 1,272 Watts Ch25 HW#1 1 – 6

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