# Magnetic Field Lines. Uniform Magnetic Field. Earth s Magnetic Field 6/3/2013

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1 Chapter 33: 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 0 o to 25 o (magnetic declination/angle of dip) Flips geologically Uniform Magnetic Field Fields more uniform in middle Vary at the edges Like an electric field 1

2 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 = m o qv sin q 4p r 2 m o = X 10-6 Tm/A (permeability constant) q = Angle between point and moving charge r q direction of current test charge 2

3 A proton (1.60 x C) moves along the x-axis with a velocity of 1.0 X 10 7 m/s. a. Calculate the magnetic field at point (1,0) [0] b. Calculate the magnetic field at point (0,1) [1.6 X T] c. Calculate the magnetic field at point (1,1) [0.57 X T] Perform all calculations assuming the proton is at the origin. B = m o I 2pd Magnetic Field in Wires m o = 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 o I 2pd B = (4p X 10-7 T m/a)(25a) (2p)(0.10 m) B = 5.0 X 10-5 T 3

4 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. B = m o NI 2R Coils Derivation N = number of turns I = Current R = Radius of loop Derive the formula for the magnetic field at the center. z = 0 If there are multiple coils: 4

5 A 5 turn, 10.0 cm diameter wire coil has 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 segments (line integrals) Smooth the curve out and.. B = m o I 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? 5

6 Solenoids Long coil of wire Doorbells, car starters, switches, electromagnets Magnetic field is parallel to the coil B is fairly uniform inside the coil I = NI (N = # of turns) Along the bottom (sides and top are 0) 6

7 B = m o ni m o = 4p X 10-7 Tm/A n = N/l = number of loops/length Solenoids: Ex 1 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 = m o ni B = (4p X 10-7 T m/a)(4000m-1)(2.0 A) B = 0.01 T A 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. 7

8 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. Uses of Toroids Magnetic tape heads Fusion reactors (Tokamak) Transformers 8

9 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) 9

10 A proton has a speed of 5.0 X 10 6 m/s and feels a force of 8.0 X N toward the west as it moves vertically upward. a. Calculate the magnitude of the magnetic field. b. Predict its direction v proton F west Earth Using right-hand rule: B must be towards geographic north v proton F = qvbsinq B = F/qvsinq = 8.0 X N (1.6X10-19 C)(5.0X10 6 m/s)(sin90 o ) = 0.10 T F west Earth A long horizontal wire carries a 10.0 A current. An electron is travelling to the right at a speed of 1.00 X 10 7 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 10 7 m/s in a plane perpendicular to a T magnetic field. Describe its path. 10

11 Path is circular (right-hand rule, palm positive) F = mv 2 r qvb = mv 2 r r = mv qb r = (9.1 X kg)(2.0 X 10 7 m/s ) (1.6 X C)(0.010 T) r = m r = mv qb Mass Spectrometers Used to separate different isotopes and to identify compounds (e.i. CH 3 COOH) 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) Mass 1 = Radius 1 Mass 2 Radius 2 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 At what radius would you expect to see the isotope Carbon-13? Force on Current Carrying Wires Experimentally, current carrying wires experience a force 11

12 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 60 o. F = IlBsinq F = (30A)(0.12 m)(0.90 T)sin60 o F = 2.8 N into the page A loop of wire carries 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 = IlBsin90 o F = IlB(1) F = IlB B = F Il B = 3.48 X 10-2 N = 1.42 T (0.245 A)(0.100m) 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 I 1 I2 F F 12

13 Current in opposite directions Force is repulsive South to South orientation I 1 I 2 Force per unit length F = m o l I 1 I 2 2p d F F d = separation l = length of wire Wires: Ex 1 Wires: Ex 2 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 = m o I 1 I 2 l 2p d 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 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 = m o l I 1 I 2 2p d mg = m o l I 1 I 2 2p d Solve for I 2 I 2 = 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 13

14 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 = pr 2 = (p)(0.100 m) 2 = m 2 Maximum Torque q = 90 o = NIABsin q = (10)(3.00 A)(2.00 T)(sin 90 o ) = 1.88 Nm Minimum Torque q = 0 o sin 0 o = 0 = 0 A circular loop of wire 50.0 cm in radius is oriented at 30 o to a magnetic field (0.50 T). If the current in the loop is 2.0 A, what is the torque? = 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 14

15 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 ( windings ) produces a much steadier torque AC Motors AC Motor Can work without commutators since current already switches Often use electromagnets rather than permanent magnets 15

16 Loudspeaker Variation of current in the coil Varies the force caused by the permanent magnet Speaker cone (cardboard) moves in and out mt 4. a) -3.2 X j T b) 0 c) X10-15 i T X k T A, 250 A, 5000 A, 50,000 A, 500,000 A R 14.B a = 6.7 X 10-5 T, out of the page B b = 2.0 X 10-4 T, into the page B c = 6.7 X 10-5 T, out of the page 16. a) 3.1 X 10-4 A m 2 b) 5.0 X 10-7 T 20. Line integral is zero A, out of the page A 26. a) 5.7 X j N b) 0 N 26. a) 5.7 X j N b) mt T X 10-4 N, up, 0 N, 2.5 X 10-4 N, down 38. a) Both in eq b) Loop 2, unstable 42. a) 0.50 cm b) 8.0 cm 44. (2.0 X 10-4 )e -t/10 ms cm mt, into the page T 16

17 17

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