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Chapter 24 - Magnetism Only a few substances are magnetic ( ferromagnetic ) Iron (Fe), Cobalt (Co), and Nickel (Ni) + a couple rare earth metals Some other.

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Presentation on theme: "Chapter 24 - Magnetism Only a few substances are magnetic ( ferromagnetic ) Iron (Fe), Cobalt (Co), and Nickel (Ni) + a couple rare earth metals Some other."— Presentation transcript:

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 - 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. 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. 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 N S N S N S N S

10 5. 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
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 – B = 1,000 T

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

15 x x x x x x x x x x x x x x x x Ch24 HW#2 7 – 10
7) 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
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
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 B 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
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 B 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 . x . . x x x . x . 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 3x106 m/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? = ( 2.5 T )( +.5 C )( 500 m/s ) = 625 N Ex2) An electron travels at 3x106 m/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 ) = ( 4x10-2 T )( -1.6x10-19 C )( 3x106 m/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 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 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 ) = T q∙v ( .06 C )( 250 m/s ) Ch24 HW#3 11 – 14

26 B F = B∙q∙v = (9x10-2T)(+1x10-6C)(3x104m/s) vel = x = .
Ch24 HW#3 11 – 14 11) Given +1 μC at 3x104 m/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 9x106 m/s along (-y) axis and enters a B-field of 4x10-2 T (+z). Find force. vel B F = B∙q∙v = x .

27 FB = (9x10-2T)(+1x10-6C)(3x104m/s) vel = 0.0027 N
11) Given +1 μC at 3x104 m/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 = N 12) Given +2.2 nC at 9x106 m/s along (-y) axis and enters a B-field of 4x10-2 T (+z). Find force. vel B F = B∙q∙v = x .

28 FB = (9x10-2T)(+1x10-6C)(3x104m/s) vel = 0.0027 N
11) Given +1 μC at 3x104 m/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 = N 12) Given +2.2 nC at 9x106 m/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 vel = (1.0x10-2T)(+0.5x10-3C)(2.5x102 m/s)
13) Given an electron at 4x106 m/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 vel = (1.0x10-2T)(+0.5x10-3C)(2.5x102 m/s) = x .
13) Given an electron at 4x106 m/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 vel = (1.0x10-2T)(+0.5x10-3C)(2.5x102 m/s) = 0.00125 N x .
13) Given an electron at 4x106 m/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) = N FB - x . .

32 B I . 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? B I 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? B I .

33 Ex1) A 5. 0 A current passes through a section of wire that is 0
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? B I FB = 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? B I FB = B∙I∙L FB = (.5T)(2.5A)(1.13m) = 1.6 N x .

34 x Ex3) A +2 μC charge is travelling South at 3x103 m/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 4x105 m/s through a 8.2 μT B-field oriented East. Find the force. B vel x

35 x x Ex3) A +2 μC charge is travelling South at 3x103 m/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 4x105 m/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
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
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 FB vel Fg FB = FG B.q∙v = m∙g B.(+0.5x10-6C)(1x105m/s) = ( .01 kg )( 9.8 m/s2 ) B = 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 I FB = (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 I FB = (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 vel = (2x10-3 T)(+5.2x10-6C)(2x106 m/s) =
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 1.3x104 m/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 vel = (2x10-3 T)(+5.2x10-6C)(2x106 m/s) =
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 1.3x104 m/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∙v B 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∙v B 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 B F B B F F F F F F (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 B F B B F F F F F F (V or I) t What value do you use for I for a generator? . x

49 PAC = Ieff2∙R or PAC = Ieff ∙ Veff
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? 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

51 Ch25 HW#1 1 – 6 A straight wire 0.5 m long is moved straight up at a speed of 20 m/s through a 0.4 T B-field as shown. a) Find EMF b) R = 6.0 Ω, Find I S N 2) A straight wire 25 m long moves at 125 m/s perpendicular to Earth’s B-field of 5x10-5 T. Find EMF.

52 Ch25 HW#1 1 – 6 A straight wire 0.5 m long is moved straight up at a speed of 20 m/s through a 0.4 T B-field as shown. a) Find EMF EMF = B∙L∙V = (.4T)(.5m)(20m/s) = 4V b) R = 6.0 Ω, Find I I = V = 4V = A S N R Ω 2) A straight wire 25 m long moves at 125 m/s perpendicular to Earth’s B-field of 5x10-5 T. Find EMF.

53 Ch25 HW#1 1 – 6 A straight wire 0.5 m long is moved straight up at a speed of 20 m/s through a 0.4 T B-field as shown. a) Find EMF EMF = B∙L∙V = (.4T)(.5m)(20m/s) = 4V b) R = 6.0 Ω, Find I I = V = 4V = A S N R Ω I 2) A straight wire 25 m long moves at 125 m/s perpendicular to Earth’s B-field of 5x10-5 T. Find EMF. EMF = B∙L∙V = (5x10-5T)(25m)(125m/s) = .16 V

54 3) A straight wire 30 m long moves at 2. 0 m/s to the right in a 1
3) A straight wire 30 m long moves at 2.0 m/s to the right in a 1.0 T B-field as shown. a) Find EMF b) R = 15Ω, Find I 4) A horseshoe magnet is mounted so that the field lines are vertical. A wire between the poles is pulled out, making a current flow from left to right. Which is the North pole? I FB .

55 3) A straight wire 30 m long moves at 2. 0 m/s to the right in a 1
3) A straight wire 30 m long moves at 2.0 m/s to the right in a 1.0 T B-field as shown. a) Find EMF b) R = 15Ω, Find I EMF = B∙L∙V = (1.0T)(30m)(2.0m/s) = 60V I = V = 60V = 4A into the page R Ω 4) A horseshoe magnet is mounted so that the field lines are vertical. A wire between the poles is pulled out, making a current flow from left to right. Which is the North pole? I FB .

56 3) A straight wire 30 m long moves at 2. 0 m/s to the right in a 1
3) A straight wire 30 m long moves at 2.0 m/s to the right in a 1.0 T B-field as shown. a) Find EMF b) R = 15Ω, Find I EMF = B∙L∙V = (1.0T)(30m)(2.0m/s) = 60V I = V = 60V = 4A into the page R Ω 4) A horseshoe magnet is mounted so that the field lines are vertical. A wire between the poles is pulled out, making a current flow from left to right. Which is the North pole? S I N FB .

57 Veff = .707( Vmax ) Ieff = .707 ( Imax )
5) A generator develops a max voltage of 170 V Find Veff Veff = .707(170V) = A 60 W light bulb placed across the generator with Imax = .70 A. Find Ieff Ieff = .707(0.70A) = c) Resistance of light bulb: R = Veff Ieff 6) The effective voltage is 117 V. a) Vmax = ? V = .707(Vmax) b) Ieff = 5.5 A, find 5.5A = .707(Imax)

58 Veff = .707( Vmax ) Ieff = .707 ( Imax )
5) A generator develops a max voltage of 170 V Find Veff Veff = .707(170V) = 120 V A 60 W light bulb placed across the generator with Imax = .70 A. Find Ieff Ieff = .707(0.70A) = 0.49A c) Resistance of light bulb: R = Veff = 120V = 245Ω Ieff A 6) The effective voltage is 117 V. a) Vmax = ? V = .707(Vmax) b) Ieff = 5.5 A, find 5.5A = .707(Imax)

59 Veff = .707( Vmax ) Ieff = .707 ( Imax )
5) A generator develops a max voltage of 170 V Find Veff Veff = .707(170V) = 120 V A 60 W light bulb placed across the generator with Imax = .70 A. Find Ieff Ieff = .707(0.70A) = 0.49A c) Resistance of light bulb: R = Veff = 120V = 245Ω Ieff A 6) The effective voltage is 117 V. a) Vmax = ? V = .707(Vmax) Vmax = 165V b) Ieff = 5.5 A, find 5.5A = .707(Imax) Imax = 7.8A

60 Ch25.2 Transformers Lenz’s Law – When a changing B-field creates a current, that current then creates a B-field that opposes the original B-field. ( Law of Inertia for charges ) ( Explains back EMF’s in motors ) N S

61 Transformers – device used to increase or decrease voltage.
– must be AC Ex: Step up transformer A step up transformer increases voltage and decreases current. A step down transformer decreases voltage and increases current.

62 Transformers – device used to increase or decrease voltage.
– must be AC Ex: Step up transformer Pin = Pout A step up transformer increases voltage and decreases current. A step down transformer decreases voltage and increases current. primary voltage # of turns of wire on primary Notice the N’s are reversed! secondary voltage # of turns of wire on 2ndary

63 Ex: Hoover dam Across desert Victorville Power pole Ur House P = I2.R Turbines generate Step up to Step down Step down energy at ~ 22,000V 100,000+ V to 10,000V to 120V

64 Ex: A step up transformer has 200 turns on the primary and
3,000 turns on the secondary coil. The voltage to the primary coil is 90V. a) What is the induced secondary voltage? b) If the primary current is 30A, what is the secondary current? c) What is the power input and output? Ch25 HW#2 7 – 9

65 Ex: A step up transformer has 200 turns on the primary and
3,000 turns on the secondary coil. The voltage to the primary coil is 90V. a) What is the induced secondary voltage? VS = NS VS_ = VS = 1,350V VP NP V b) If the primary current is 30A, what is the secondary current? IP = VS A = V IS = 2A IS VP IS V c) What is the power input and output? Pin = Pout IP ∙ VP = IS ∙ VS ( 30 )( 90 ) = ( 2 )( 1350 ) 2,700 W = 2,700 W Ch25 HW#2 7 – 9

66 Lab25.1 – EM Induction and Transformers
- due tomorrow - Ch25 HW#2 beginning of period

67 Ch25 HW#2 7 – 9 7. A step down transformer has 7500 turns on the primary and 125 turns on the secondary coil. The voltage to the primary coil is 7.2kV. a) What is the induced secondary voltage? b) If the secondary current is 36A, what is the primary current?

68 Ch25 HW#2 7 – 9 7. A step down transformer has 7500 turns on the primary and 125 turns on the secondary coil. The voltage to the primary coil is 7.2kV. a) What is the induced secondary voltage? b) If the secondary current is 36A, what is the primary current? 8. A step up transformer has 500 turns on the primary and 15,000 turns on the secondary coil. The EMF to the primary coil is 120V. a) What is the induced secondary EMF? b) If the secondary current is 3.0A, what is the primary current? c) What is the power input and output?

69 8. A step up transformer has 500 turns on the primary and 15,000 turns on the secondary coil. The EMF to the primary coil is 120V. a) What is the induced secondary EMF? b) If the secondary current is 3.0A, what is the primary current? c) What is the power input and output? Pin = Pout IP ∙ VP = IS ∙ VS ( 90A )( 120V ) = ( 3A )( 3600V ) 10,800 W = 10,800 W 9. A step up transformer has 300 turns on the primary and 90,000 turns on the secondary coil. The voltage to the primary coil is 60 V. a) What is the induced secondary voltage? b) If the secondary current is 0.50A, what is the primary current?

70 9. A step up transformer has 300 turns on the primary and 90,000 turns
on the secondary coil. The voltage to the primary coil is 60 V. a) What is the induced secondary voltage? b) If the secondary current is 0.50A, what is the primary current?

71 Ch24,25 Rev 1) A 1.50 m wire carries a current of 10.0 A to the right in a uniform magnetic field that creates a force on the wire of 0.60 N upwards. What is the magnitude and direction of the magnetic field? FB I 2) Draw the magnetic field associated with the following currents in wires: Straight wire, current out of page b) Circular wire, current clockwise I . 3) What is the direction of the force on this wire? I S N x S N

72 4) A beam of electrons travel out of the page with a velocity of 2
4) A beam of electrons travel out of the page with a velocity of 2.5x107 m/s, in a B-field of 6.0x10-2 T to the right. What is the magnitude and direction of the force acting on each electron? ( qe = -1.6x10-19 C ) 5) A wire 0.5 m long is yanked out of a 0.25 T B-field, with a speed of 10 m/s as shown. What voltage is induced in the wire? What direction will the induced current travel through the wire? F If the wire is part of a circuit with 2Ω of resistance, what current passes through? . . . . . . . . . . . . . . . . . .

73 6) An AC generator develops a max voltage of 150 V.
What is the effective voltage? 7) A step up transformer has 80 turns on its primary coil and 1200 turns on its secondary coil. 120V AC is applied to its primary. What is the secondary voltage? b) If the current to the primary is 2.0 A, what current is induced in the secondary? c) Find the input and output power.

74 6) An AC generator develops a max voltage of 150 V.
What is the effective voltage? Veff = ( .707 )∙Vmax = ( .707 )( 150V ) = 106V 7) A step up transformer has 80 turns on its primary coil and 1200 turns on its secondary coil. 120V AC is applied to its primary. What is the secondary voltage? b) If the current to the primary is 2.0 A, what current is induced in the secondary? c) Find the input and output power. Pin = IP ∙ VP Pout = IS ∙ VS = ( 2 )( 120 ) = ( .13 )( 1800 ) = 240 W = 240 W

75 Chapter Extra Credit A coin lies submerged at the bottom of a pan of water. Does refraction of light from the coin make it appear deeper, or make it appear shallower than it really is? Explain: Kent stands 1 meter in front of the dresser mirror and looks at the flower on the top of his head in a small mirror held .5 meter behind his head. How far in back of the dresser mirror does he see the image of the flower? Explain:

76 Chapter 22 + 23 Extra Credit - Physics
Which is more dangerous, touching a faulty 110-volt light bulb or a Van de Graff generator charged to 100,000 volts? Why? Compared to the huge force that attracts an iron tack to a strong magnet, the force that the tack exerts on the magnet is Relatively small Equally huge


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