Some History of Magnetism
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1 Magnetism
2 Some History of Magnetism The ancient Greeks were the first to observe magnetism. They studied the mineral magnetite. The poles of a magnet were observed to be south or north seeking. These properties were found useful in navigation. They were used by the Chinese, Arabs, and the Italians,in the first 10 centuries A.D.
3 In 1820 Hans Oersted noticed a relationship between electricity and magnetism. A current carrying wire produced a magnetic field. *There is some connection between moving charge (current) and magnetic field. This idea is one of the major themes in this unit. There were many contributors to electromagnetism, but the greatest was James Clerk Maxwell who devised the most complete theory. In this unit we are examining pre-maxwell electromagnetic theory.
4 Magnets and Magnetic Fields Magnets have two ends poles called north and south. Like poles repel; unlike poles attract.
5 However, if you cut a magnet in half, you don t get a north pole and a south pole you get two smaller magnets.
6 Magnetic fields can be visualized using magnetic field lines, which are always closed loops. These are the field lines produced by a typical bar magnet (magnetic dipole has 2 magnetic poles).
7 A magnetic dipole will align itself in any external magnetic field. The north pole of the dipole points in the direction of the magnetic field. One can see the compass below aligned in the magnetic field of the bar magnet. S N
8 The magnetic field lines of two bar magnets, with like poles opposing. Magnetic field lines always leave a north pole and enter a south pole
9 The magnetic field lines of two bar magnets with unlike poles opposing. Magnetic field lines inside the bar magnet go from south to north.
10 Domain Theory of Magnetism Ferromagnetic materials are those that can become strongly magnetized, such as iron and nickel. These materials are made up of tiny regions called domains; the magnetic field in each domain is in a single direction.
11 When the material is unmagnetized, the domains are randomly oriented. They can be partially or fully aligned by placing the material in an external magnetic field. It should be noted that magnetism is connected with (moving charge) but any typical bar magnet is neutrally charged.
12 The Earth s magnetic field is similar to that of a bar magnet. Note that the Earth s North Pole is really a south magnetic pole, as the north ends of magnets are attracted to it. From evidence in the geologic record it appears that the Earth s magnetic field has flipped periodically
13 A uniform magnetic field is constant in magnitude and direction. The field between these two wide poles is nearly uniform. The evenly spaced field lines indicate a uniform field.
14 Field produced by current carrying wire Experiment shows that an electric current produces a magnetic field. In the case of a long straight wire the field lines are concentric circles around the wire.
15 This leads us to our first right hand rule. Right hand rule (RHR) #1: Point thumb in direction of conventional current (direction + charge moves) Fingers wrap in direction of magnetic field lines
16 The magnetic field due to a long straight wire can be calculated by the following formula: r B = u 0I 2πr B= the magnetic field (T) I= the current (A) r= perpendicular distance from from the wire ( m) u 0 =permeability of free space 4πx10-7 Tm/A The magnetic field has units called the Tesla (T) u 0 is an important constant in magnetism. It is a measure of the resistance in forming a magnetic field
17 Unit of B: the tesla, T. 1 T = 1 N/A m. Another unit sometimes used: the gauss (G). 1 G = 10-4 T.
18 Example: What is the magnitude and direction of conventional current in the wire below, if the field at point p (10 cm from the wire) has a value of 1.00x10-5 T into the page? 10 cm p B= 1.00x10-5 T Often to indicate a vector direction into the page and x is used, and to indicate a vector out of the page a. is used. Imagine an arrow or dart where the flights (back of arrow) represent the x and the dot. represents the point of the arrow head.
19 Solution: B = u 0 I 2πr I = 2πrB u 0 I = 2π (0.1)(1x10 5 ) 4πx10 7 = 5.00 A According to the right hand rule #1, the conventional current must be upward in the wire.
20 Field produced by loop of wire The direction of the field is given by another right-hand rule.
21 Field produced by loop of wire Notice that this single loop of wire carrying current produces field lines similar to a magnetic dipole (bar magnet. The right hand rule (RHR#2): 1) Wrap fingers in the direction of conventional current 2) The thumb points in the direction of field in the centre of loop
22 Field produced by loop of wire You can see that on top of the loop there would be a north pole and a south pole underneath the loop. N A single circular loop carrying current creates a field like a magnetic dipole. S
23 There is another useful mnemonic with loops of wire carrying current. When looking a loop in the plane of the page, if the current were clockwise the near pole (pole on this side of page) would be a south, and if the current were counter-clockwise the near pole would be a north (see diagram). clockwise counter-! clockwise
24 The Solenoid A solenoid is a long coil of wire. It may be wrapped several times. If it is tightly wrapped, the magnetic field in its interior is almost uniform. A solenoid can be used as an electromagnet. On can see its field lines are similar to a bar magnet (dipole)
25 The magnetic field inside the solenoid can be calculated by the following formulae: B = µ 0 ni = µ 0 N l I l = the length of solenoid (m) I = current (A) N= total number of turns(loops) n = N l = #turns per unit length
26 RHR#2 can be used for the solenoid also: The right hand rule (RHR#2): 1) Wrap fingers in the direction of conventional current 2) The thumb points in the direction of the field in the centre of the solenoid
27 If a piece of iron is inserted in the solenoid, the magnetic field greatly increases. Such electromagnets have many practical applications.
28 Example: A solenoid of length 25 cm is connected to the power sources as shown. It has a magnetic field in its centre of T and a current of 4.50A. a) Label the magnetic poles, and sketch the field lines around the solenoid. b) How many total turns does the solenoid have?
29 Solution: a) Conventional current leaves the positive terminal. Using the RHR#2 the thumb points right. The field inside the solenoid points right. This means the pole on the right side is a north. S N I
30 Solution: b) B = µ 0 N l I, Solve for N: N = Bl µ 0 I N = (0.045)(0.25) π (4x10 7 )(4.50) N= 1.99x10 3 turns
31 Charged Particle in Magnetic Field The force on a moving charge in a magnetic field is given by the relationship: F B = Magnetic force (N) q = charge (C) υ = velocity (m/s) B = Magnetic field (T) F B = qυbsinθ θ = angle between υ and B vectors Once again, the direction is given by a right-hand rule.
32 There is a right hand rule (RHR#3) that relates force, velocity, and magnetic field. RHR#3(flat hand) for (+) charge 1. Thumb points in direction of v 2. Fingers point in direction of B 3. Palm pushes in direction of force (F B ) or Force perpendicular to palm. The force on a charged particle is the vector cross product of its velocity with the magnetic field.
33 If it were a negative charge travelling in a magnetic field the force would be in the opposite direction to that of a positive charge. Example: If there were a positive charge travelling to the right, and there was a magnetic field directly into the page, in what direction is the force exerted on the charge? q+ v
34 Solution: The force on the charge must be upwards, according to the RHR#3. F B q+ v
35 Example: A positively charged alpha particle is travelling at an angle of 35 0 to a magnetic field (B= T) with a speed of 284 m/s. What is the force exerted on the alpha particle? v B q 35 0
36 Solution: F B = qυbsinθ = (2)(1.6x10-19 )(284)(0.36)sin(35 0 ) = 1.88x10-17 N into the page
37 Example: A positive charge has a velocity (as shown) next to a straight wire carrying current. In what direction is the magnetic force on the charge? + v I
38 Solution: The charge is traveling in the magnetic field created by the wire. One can determine the direction of the field created by the wire at the charge s position (RHR#1): This would be into the page. Then using RHR one can determine the direction of force. The force would be upwards. F B B + v I
39 Question: When would a charged particle in a magnetic field experience no force? One can use the formula to deduce this. The answer is on the next slide.
40 Answer: 1) One possibility is when v=0, or in other words the charge is not moving. If v=0, then F B =0 2) Another possibility occurs when sinθ=0, This occurs when θ=0 0 or In other words if the charge particle is moving parallel to the magnetic field lines it experiences no force.
41 Force on Electric Charge Moving in a Magnetic Field If a charged particle is moving perpendicular to a uniform magnetic field, its path will Usually be a circle. It is possible that its path may be a spiral. Do you know why? Split the initial velocity of the particle into components parallel and perpendicular to the field and it becomes more obvious.
42 When a charged particle enters a uniform magnetic field its path will generally be a circle. In this case the magnetic force on the particle will equal its centripetal force. Therefore when approaching these problems one may state: F B = F C qυb = m υ 2 r = m 4π 2 r T 2
43 Example: An electron travelling at half the speed of light entered a uniform magnetic field of 2.30 mt. What must be the radius of its circular path? Solution: F B = F C qυb = m υ 2 r = mυ qb = [(9.11x10-31 )(1.5x10 8 )/[(1.6x10-19 )(2.3x10-3 )] = m r
44 Example: Can a magnetic field ever change a charged particle s speed? Answer: If the the force is always perpendicular to the velocity vector the force would do zero work and therefore not change the kinetic energy (speed) of the particle.
45 Force on an Electric Current in a Magnetic Field A magnet exerts a force on a current-carrying wire. The direction of the force is given by a right-hand rule.
46 The force on the wire depends on the current, the length of the wire, the magnetic field, and its orientation. F B = BILsinθ This equation defines the magnetic field B.
47 Derivation: B q + v L F B = qυbsinθ, υ = L t! F B = q# L " t $ &Bsinθ = q % t L ( )Bsinθ = ILBsinθ F B = BILsinθ
48 To relate how the force on a current carrying wire is related to the field one can use another right hand rule. RHR #4 (use flat hand) equivalent to RHR #3 1)Fingers point in direction of B 2)Thumb points in direction of conventional current 3)Palm pushes in direction of force. It can also be thought of as the vector cross product between the current (moving charge is a vector) and the magnetic field.
49 Example: Find the force on each segment of the wire (ab, bc, cd, and de) below if there is a magnetic field of T pointing directly to the right in the plane of the page, and conventional current of 5.00 A flows through the wire from left to right. c 10 cm 60 0 a b d e
50 Solution: For segments ab and de the force is zero since the current is parallel to the magnetic field (sinθ=0) F B = BILsinθ The force on segment bc is: F B = BILsinθ = (0.2)(5)(0.1)sin(90) = 0.10N into page The force on segment cd= 0.10 N out of the page.
51 Force between Two Parallel Wires The magnetic field produced at the position of wire 2 due to the current in wire 1 is: The force this field exerts on a length l 2 of wire 2 is: *Physics 12 students do not need to know this formula
52 Parallel currents attract; antiparallel currents repel. Can you explain(without equations) why in a) the wires attract and in b) they repel??
53 The trick is to look at the field created by one of the wires and how the other wire would respond (according to RHR#4) in that magnetic field. See if you can do this.
54 Galvanometers, Motors, Loudspeakers A galvanometer takes advantage of the torque on a current loop to measure current.
55 D.C Motor An electric D.C motor also takes advantage of the torque on a current loop, to change electrical energy to mechanical energy. The split ring commutator reverses the current every half cycle and keeps the coils turning in one direction.
56 One way to understand why the coils of the motor (armature) rotate is to think of motor consisting of one rectangular coil. One can see that there will be a torque exerted on this coil. B F F S rotation In order to keep the motor rotating, the current must be reversed every half cycle(rotation). The commutator accomplishes this.
57 The commutator connects the power from the power source to the motor. The actual connections are two brushes that make contact with the split ring commutator.
58 Loudspeakers use the principle that a magnet exerts a force on a current-carrying wire to convert electrical signals into mechanical vibrations, producing sound.
59 Mass Spectrometer The mass spectrometer is a device use to detect and determine the mass to charge ratio of ions. First of all atoms of a sample are ionized by stripping off electrons. There may be many different kinds of ions present. The charged (+) particles are then accelerated by an accelerating voltage. All the ions pass through a velocity selector and are selected at a specific velocity. Below is a diagram of a velocity selector. The particles pass through perpendicularly crossed electric ( E ) and magnetic( 1 )fields. B 1
60 The electric field points downwards, and the magnetic field points directly into the page. If the charged particles move through undeflected, then the magnetic and electric forces must be balanced. F B = F E qυb 1 = qe also qδv = 1 2 mυ 2 If accelerated from rest υ = E B 1 The charged particles then move into a region of a uniform magnetic field ( B 2 ). It will then travel in a circular path and will eventually strike a detector where it is counted.
61 All the atoms reaching the second magnetic field will have the same speed; their radius of curvature in the second region will depend on their charge and mass. B 1 B 2
62 The charged particles then move into a region of a uniform magnetic field It will then travel in a circular path and will eventually strike a detector where it is counted. F B2 = F C qυb 2 = m υ 2 r qb 2 = m υ r and recall υ = E B 1 q m = E B 1 B 2 r
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