2Magnetic Poles and Magnetic Fields There are two magnetic poles, north and south.The north pole of a compass magnet is defined as the end that points towards the Earth’s north pole.Like poles repel, unlike poles attract.Magnetic poles come in pairs.
3Magnetic Poles and Fields Magnetic fields are drawn from north to south. Field lines that are closer together denote a stronger magnetic field.The direction of a magnetic field, B, at any location is given by the direction of the north pole of a compass needle.Magnetism is caused by moving electric charges!
4Magnetic FieldsMagnetic field lines around a magnet – drawn from north to south.Magnetic field, B, is measured in N/Am or Tesla [T]
6Magnetic Field Strength and Magnetic Force When a charged object moves in a magnetic field, B, a magnetic force acts on the charged object.When the velocity of the charge, v, is perpendicular to B,F = qvB andB = F/qv
7Magnetic Force F = qvB ; B = F/qv B is measured in N/(Cm/s) or N/Am [N/(Cm/s)] = [Tesla] = [T]
9Magnetic Force Fm When v and B are at right angles, Fm= qvB When v and B are not at right angles,Fm = qvBsinθThe magnetic force is perpendicular to both the velocity and to the magnetic field.Note: q is assumed positive
10ExamplesIn a linear accelerator, a beam of protons travels horizontally northward. To deflect the protons eastward with a uniform magnetic field, the field should point in which direction?A particle with charge of –5.0 X 10-4 C and mass of 2.0 X 10-9 kg moves at a speed of 1.0 X 103 m/s in the + x direction. It enters a uniform magnetic field of 0.20 T that points in the +y direction. Which way will the particle deflect? What is the magnitude of the force on the particle? What is the radius of curvature?
11Homework Read Section 19.1 and 19.2 Do # 1 – 3, 6 – 8, 13, 15, 17, 18, on page 649
12Summary of Magnetic Force on Moving Charges When a charge moves at an angle with respect to a magnetic field, the charge feels a force ofF = qvBsinθThe force is always perpendicular to both the velocity and the magnetic field.Magnetic force does not change speed, only directionWhen θ = 0, the charge feels no force.
13Summary of Magnetic Force on Moving Charges Charged particles traveling in a uniform magnetic field always have circular arcs as their trajectories.
14ApplicationsAn electron beam can be deflected by magnetic fields and can cause a fluorescent material to glow…Cathode-Ray Tubes are vacuum tubes that accelerate beams of electrons which are deflected by magnetic fields and cause a fluorescent screen to glow.Oscilloscopes, TV’s and computer monitors are made of cathode ray tubes.
15ApplicationsA mass spectrometer uses magnetic fields to measure the mass of an ion (charged particle)It selects charged particles with a particular (known) velocity, then sends them into a uniform magnetic field where the radius of curvature can be measured. This allows for a calculation of the ion’s mass.
16Mass SpectrometerParallel plates are set up perpendicular to magnetic fields. The only charges to go straight through without curving will be those for which Fe = FmqE = qvBv = E/B
18ExampleOne electron is removed from a methane molecule before it enters the mass spectrometer. After passing through the velocity selector, the ioni has a speed of 1.00 X 103 m/s. It then enters the main magnetic field region in which T = 6.70 X T. It follows a circular path in which it is detected 5.00 cm from the field entrance. Determine the mass of this molecule.
19Magnetic Force on Current Carrying Wire Since current is defined as moving charges, a wire carrying current in a magnetic field will feel a magnetic force. Consider a wire of length LFm = qvBsinθ and v = L/tFm =q(L/t)BsinθFm = (q/t)LBsinθFm = ILBsinθ where θ give the angle between the current and the magnetic field
20Magnetic Force on Current Carrying Wire A wire of length L carrying a current, I, in a uniform magnetic field feels a total forceFm = ILB sinθThe direction of the force can be found using the right hand rule
21ExampleBecause a current carrying wire is acted on by a magnetic force, it would seem possible to suspend such a wire at rest above the ground using the Earth’s magnetic field. If a long straight wire is located at the equator, in what direction would the current have to be (up, down, east or west?)Calculate the current required to suspend the wire, assuming the Earth’s magnetic field is 0.40 G (gauss) at the equator and the wire is 1.0 meter long with mass of 30 grams. 1 G = T
22Homework Read 19.3, 19.4 Pay attention to Example 19.5 on page 635 Do # 22, 23, 30, 32, 36, 38, 42, 44, 45 – 48, 50 on page 651
23Magnetic Forces on Moving Charges Summary:Fm = qvBsinθFm = ILBsinθ
24Applications of Force on Current Carrying Wire Galvanometer!A loop of wire placed inside a magnetic field will feel a torque when a current passes through…Recall T = rFsinθ
26Galvanometer T = r Fsinθ Two forces create torque on a rectangular loop of length L and width wTnet = 2r Fsinθ= 2(½w)(ILB)sinθ=wILBsinθ= AIBsinθ (A = area)For a coil of N turns,T net= NIABsinθ(m = NIA = magnetic moment)
27GalvanometerWhen a current is detected, the needle swings due the torque created by the magnetic force on the current carrying wire.A galvanometer with high resistance can be used as a voltmeterA galvanometer with low resistance can be used as an ammeter
28Other Applications of Current Carrying Wires in Magnetic Fields DC Motors - current passing through coil creates torque (spinning motion). The spinning is maintained by continually switching the direction of current. See page 636 diagramElectronic Balance – A downward force of gravity (the weight applied) is offset by an upward force created by coils in a magnetic field. The current needed to balance the weight is measured and converted into a ‘mass’ reading.
29Ampere’s Law Recall that magnetism is created by moving charges. Ampere’s Law – relates the current in a wire to the resulting magnetic field created by the moving charges in the wireAmpere’s Law can be used to find B near current carrying wires.
30Ampere’s LawAmpere’s Law allows us to find the magnetic field at some distance from a current carrying wire.Direction of B is found using right hand rule…
31Find Magnetic FieldAt a perpendicular distance, d, from a long straight wire,B(2πd) = µ0Iwhere µ0 is permeability of free space = 4π X Tm/AB = µ0I/2πd
32Find Magnetic FieldAt the center of a current carrying loop of radius r,B = µ0I/2rFor a coil of N loops,B = µ0NI/2r
33SolenoidA solenoid is a coil of wire, length L, in which the interior magnetic field caused by current is constant and uniform.B = µ0NI/L
34ExampleThe maximum household current in a wire is about 15 A. What are the magnitude and direction of the magnetic field the current produces 1.0 cm below the wire?
35ExampleA solenoid of 300 turns and length 0.30 m carries a current of 15.0 A. What is the magnitude and direction of the magnetic field at the center of the solenoid? Compare your result with the field near the single wire carrying the same current.
36Example: Two long parallel wires carry currents in the same direction. a) Is the magnetic force between the wires attractive or repulsive? Make a sketch to show your result.b) If both wires carry the same 5.0 A current, have length of 50 cm, and are separated by 30.0 mm, determine the force on each wire.
37Magnetic Materials Moving charges create magnetism One moving charge sets up a “magnetic domain” in the space around it.When a material has its magnetic domains aligned, the material acts like a magnet.Magnetic domains can be aligned using electric currents.
38ElectromagnetsElectromagnets are magnets which can be switched on or off.They set up a magnetic field inside a material using a current carrying coil.
39Electromagnets Recall B = µ0NI/L for solenoid where µ0 = 4π X 10-7 Now B = µNI/L for an electromagnetwhere μ = μ0κm gives magnetic permeabilityand κm gives the relative permeability of core
40Chapter 19 SummaryMoving charges experience forces when placed in magnetic fieldsFm = qvBsinθFm = ILBsinθMoving charges set up magnetic fields!B = µ0I/2πd near long straight wireB = µ0NI/2r at center of circular loopB = µ0NI/L at center of a solenoid