Magnetism Permanent magnets Earth s magnetic field Magnetic force Motion of charged particles in magnetic fields Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. June 1, 2016 University Physics, Chapter 27 1
Permanent Magnets Examples of permanent magnets include refrigerator magnets and magnetic door latches They are all made of compounds of iron, nickel, or cobalt If you touch an iron needle to a piece of magnetic lodestone, the iron needle will be magnetized If you then float this iron needle in water, the needle will point toward the north pole of the Earth (approximately!) We call the end of the magnet that points north the north pole of the magnet and the other end the south pole of the magnet June 1, 2016 University Physics, Chapter 27 2
Permanent Magnets - Poles Magnets exert forces on one another --- attractive or repulsive depending on orientation. If we bring together two permanent magnets such that the two north poles are together or two south poles are together, the magnets will repel each other If we bring together a north pole and a south pole, the magnets will attract each other June 1, 2016 University Physics, Chapter 27 3
Magnetic Field Lines (1) Permanent magnets interact with each other at a distance, without touching In analogy with the electric field, we define a magnetic field to describe the magnetic force As we did for the electric field, we may represent the magnetic field using magnetic field lines The magnetic field direction is always tangent to the magnetic field lines June 1, 2016 University Physics, Chapter 27 4
Magnetic Field Lines (2) The magnetic field lines from a permanent bar magnet are shown below Two dimensional computer calculation Three dimensional real-life June 1, 2016 University Physics, Chapter 27 5
Broken Permanent Magnet If we break a permanent magnet in half, we do not get a separate north pole and south pole When we break a bar magnet in half, we always get two new magnets, each with its own north and south pole Unlike electric charge that exists as positive (proton) and negative (electron) separately, there are no separate magnetic monopoles (an isolated north pole or an isolated south pole) Scientists have carried out extensive searches for magnetic monopoles; all results are negative Magnetism is not caused by magnetic particles! Magnetism is caused by electric currents June 1, 2016 University Physics, Chapter 27 6
Magnetic Field Lines For the electric field, the electric force points in the same direction as the electric field and the electric force was defined in terms of a positive test particle However, because there is no magnetic monopole, we must employ other means to define the magnetic force We can define the direction of the magnetic field in terms of the direction a compass needle would point A compass needle, with a north pole and a south pole, will orient itself in equilibrium such that its north pole points in the direction of the magnetic field Thus the direction of the field can be measured at any point by moving a compass needle around in a magnetic field and noting the direction that the compass needle points June 1, 2016 University Physics, Chapter 27 7
Magnetic Force We define the magnetic field in terms of its effect on an electrically charged particle (q). Recall that an electric field exerts a force on a particle with charge q given by A magnetic field exerts no force on a stationary charge But a magnetic field does exert a force on a charge that moves across the field The direction of the force is perpendicular to both the velocity of the moving charged particle and the magnetic field; the magnetic force is sideways The Lorentz force F qe( x) F qv B( x) June 1, 2016 University Physics, Chapter 27 8
Right Hand Rule (1) The direction of the cross product (v x B ) is given by the right hand rule To apply the right hand rule Use your right hand! Align thumb in the direction of v Align your index finger with the magnetic field Your middle finger will point in the direction of the cross product v x B June 1, 2016 University Physics, Chapter 27 9
Right Hand Rule (2) What about the sign of the charge? If q is positive, then F is in the same direction as v x B If q is negative, F is in the opposite direction June 1, 2016 University Physics, Chapter 27 10
Magnitude of Magnetic Force The magnitude of the magnetic force on a moving charge is F qvb sin B where is the angle between the velocity of the charged particle and the magnetic field. Do you see that there is no magnetic force on a charged particle moving parallel to the magnetic field? (Because is zero and sin(0)=0) Do you see when the magnetic force is most strong? (Max. force is for = 90 degrees; then F = qvb) June 1, 2016 University Physics, Chapter 27 11
Units of Magnetic Field Strength The magnetic field strength has received its own named unit, the tesla (T), named in honor of the physicist and inventor Nikola Tesla (1856-1943) 1 T = 1 Ns Cm 1 N Am Check unit consistency: F = q v B N = C (m/s) T A tesla is a rather large unit of the magnetic field strength Sometimes you will find magnetic field strength stated in units of gauss (G), (not an official SI unit) -4 1 G = 10 T 10 kg = 1 T June 1, 2016 University Physics, Chapter 27 12
Example: Proton in B Field Consider a region of uniform magnetic field (green dots) coming out of the page with magnitude B = 1.2 mt. A proton with kinetic energy K = 8.4810-13 J enters the field, moving vertically from the bottom to the top. Question: Calculate the magnetic deflecting force on the proton. Answer: Use F = qvbsin. First, figure out the velocity v. K 1 2 mv2 v 2K m v 2 8.481013 J 1.67 10 27 kg 3.2107 m/s June 1, 2016 University Physics, Chapter 27 13
F F F Example: Proton in B Field (2) Angle between B-field direction and velocity of the proton: deg qvb sin 19 7 3 1.6010 C 3.210 m/s 1.210 T sin 90 15 6.110 N F a 3.710 m/s m 15 6.110 N 27 1.6710 kg 12 2 Small force but large acceleration for light particle Direction of F: Use Right Hand Rule June 1, 2016 University Physics, Chapter 27 14
Example: Cathode Ray Tube (1) Consider a cathode-ray tube. In this tube electrons form an electron beam when accelerated horizontally by a voltage of 136 V in an electron gun. The mass of an electron is 9.109410-31 kg while the elementary charge is 1.602210-19 C. Question: Calculate the velocity of the electrons in the beam after leaving the electron gun. Answer: K U qv 1 2 6 2 mv ev implies v 6.9210 m/s June 1, 2016 University Physics, Chapter 27 15
Example: Cathode Ray Tube (2) Question: If the tube is placed in a uniform magnetic field, in what direction is the electron beam deflected? Answer: downward Question: Calculate the magnitude of acceleration of an electron if the field strength is 3.65 10-4 T. Answer: F ma qvb qvb a m 1.60221019 C 6.9210 6 m/s 9.109410 31 kg 3.6510 4 T 4.4410 14 m/s 2 June 1, 2016 University Physics, Chapter 27 16
Particle Orbits in a Uniform B (1) Tie a string to a rock and twirl it at constant speed in a circle over your head. The tension of the string provides the centripetal force that keeps the rock moving in a circle. The tension on the string always points to the center of the circle and creates a centripetal acceleration. A particle with charge q and mass m moves with velocity v perpendicular to a uniform magnetic field B. The particle will move in a circle with a constant speed v and the magnetic force F = qvb will keep the particle moving in a circle. June 1, 2016 University Physics, Chapter 27 17
Particle Orbits in Uniform B (2) Recall centripetal acceleration a v2 r Newton s second law F ma So, for a charged particle q in circular motion in a magnetic field B m v2 r qvb Or m v r qb p r qb June 1, 2016 University Physics, Chapter 27 18
Example: Moving Electrons In this photo, an electron beam is initially accelerated by an electric field. Then the electrons move in a circle perpendicular to the constant magnetic field created by a pair of Helmholtz coils. Question: Is the magnetic field into F the page or out of the page? (Remember that the v magnetic force on an electron is opposite that on a proton.) Answer: The magnetic field is out of the page. June 1, 2016 University Physics, Chapter 27 19
Orbits in a Constant Magnetic Field If a particle performs a complete circular orbit inside a constant magnetic field, then the period of revolution of the particle is just the circumference of the circle divided by the speed T 2r 2m v qb From the period we can get the frequency and angular frequency f 1 qb 2 T 2 m f qb m The frequency of the rotation is independent of the speed of the particle. Isochronous orbits Basis for the cyclotron June 1, 2016 University Physics, Chapter 27 20
Cyclotrons A cyclotron is a particle accelerator The D-shaped pieces (descriptively called dees ) have alternating electric potentials applied to them such that a positively charged particle always sees a negatively charged dee ahead when it emerges from under the previous dee, which is now positively charged The resulting electric field accelerates the particle Because the cyclotron sits in a strong magnetic field, the trajectory is curved The radius of the trajectory is proportional to the momentum, so the accelerated particle spirals outward June 1, 2016 University Physics, Chapter 27 21
Force on a Current Carrying Wire (1) Consider a long, straight wire carrying a current i in a constant magnetic field B The magnetic field will exert a force on the moving charges in the wire The charge q flowing in the wire in a given time t in a length L of wire is given by L q ti i v where v is the drift velocity of the electrons June 1, 2016 University Physics, Chapter 27 22
Force on a Current Carrying Wire (2) The magnitude of the magnetic force is then F qvbsin L v i vbsin ilbsin is the angle between the current and the magnetic field The direction of the force is perpendicular to both the current and the magnetic field and is given by the right hand rule This equation can be expressed as a vector cross product F il B F ilb for L B il represents the current in a length L of wire June 1, 2016 University Physics, Chapter 27 23
Reminder: Force on a Current Carrying Wire The force on a current carrying wire is F ilb The magnitude of the magnetic force is F ilbsin il represents the current in a length L of wire is the angle between the current and the magnetic field The direction of the force is perpendicular to both the current and the magnetic field and is given by the right hand rule June 1, 2016 University Physics, Chapter 28 24
Parallel Current Carrying Wires (1) Consider the case in which two parallel wires are carrying currents The two wires will exert a magnetic force on each other because the magnetic field of one wire will exert a force on the moving charges in the second wire The magnitude of the magnetic field created by a current carrying wire is given by 0i Br () 2 r This magnetic field is always perpendicular to the wire with a direction given by the right hand rule June 1, 2016 University Physics, Chapter 28 25
Parallel Current Carrying Wires (2) Let s start with wire one carrying a current i 1 to the right The magnitude of the magnetic field a distance d from wire one is 01 i B1 2 d Now consider wire two carrying a current i 2 in the same direction as i 1 placed a distance d from wire one The magnetic field due to wire one will exert a magnetic force on the moving charges in the current flowing in wire two June 1, 2016 University Physics, Chapter 28 26
Parallel Current Carrying Wires (3) The charge q 2 flowing in wire two in a given time t in a length L of wire is given by L q2 ti2 i2 v where v is the drift speed of the charge carriers The magnetic force is then L F qvb i vb i LB v 2 1 2 1 Putting in our expression for B 1 we get F i L 12 2 0i1 0i1i 2L 2d 2d June 1, 2016 University Physics, Chapter 28 27
Example: Straight Parallel Wires The figure shows three long, straight, parallel, equally spaced wires with identical currents either into or out of the page. Question: What is the force on wire (a) due to the other two? Answer: Step 1: Find the net magnetic field from wires (b) and (c): B bc Step 2: Find the force on wire (a) B b at (a) is down B c at (a) is up The net field is F i LB abc a bc Force on (a) by (b) and (c) B b 0i b 2d B 0i c c 2 2d B bc 0i 2d 0i 4d 0i 4d F abc i a LB bc 0 i2 L 4d June 1, 2016 University Physics, Chapter 28 28
Torque on a Current-Carrying Loop (1) Electric motors rely on the magnetic force exerted on a current carrying wire This force is used to create a torque that turns a shaft A simple electric motor is depicted below consisting of a single loop carrying current i in a constant magnetic field B The two magnetic forces, F and -F, shown in the figure are of equal magnitude and opposite direction These forces create a torque that tends to rotate the loop around its axis June 1, 2016 University Physics, Chapter 27 29
Torque on a Current-Carrying Loop (2) As the coil turns in the field, the forces on the sides of the loop perpendicular to the magnetic field will change The forces act on the sides of the square loop below, where is the angle between a normal vector, n, and the magnetic field B The normal vector is perpendicular to the plane of the wire loop and points in a direction given by the right hand rule based on the current flowing in the loop June 1, 2016 University Physics, Chapter 27 30
Torque on a Current-Carrying Loop (3) Here the current is flowing upward in the top segment and downward in the lower segment as illustrated by the arrow feathers and arrowhead The force of each of the vertical segments is F iab (remember F ilb) The force on the other two sides is parallel or anti-parallel to the axis of rotation and cannot cause a torque June 1, 2016 University Physics, Chapter 27 31
Torque on a Current-Carrying Loop (4) The sum of the torque on the upper side plus the torque on the lower side gives the torque exerted on the coil about the center of the loop rf a 2 1 iab where A = a 2 sin iab a 2 sin ia2 Bsin iabsin June 1, 2016 University Physics, Chapter 27 32
Example: Current Carrying Loop (1) A single-turn current loop, carrying a current of 4 A, is in the shape of a triangle with sides 50, 120, and 130 cm. The loop is in a uniform magnetic field of strength B = 75 mt (direction parallel to the 130 cm side of the loop). Question: What is the magnitude of the magnetic force on the 130 cm side? Answer: For the 130 cm side: F il B il B F 0 June 1, 2016 University Physics, Chapter 27 33
Example: Current Carrying Loop (2) Question: What is the magnitude of the magnetic force on the 120 cm side? Answer: Look at the x and y components of the magnetic field! B B x y Bcos Bsin 1 tan 50 cm /120 cm 22.6 120 There is no contribution from B on il since B il Force due to x y B is il B y y x 120 x x F il B 4 A 1.20 m 0.075 T sin 22.6 0.138 N June 1, 2016 University Physics, Chapter 27 34
Reminder - Cross Product Force F exerted on charge q with velocity v moving at angle relative to magnetic field B. F qvb F ilb June 1, 2016 University Physics, Chapter 27 35
Magnetic Flux To quantify the amount of magnetic field lines we define the magnetic flux in analogy to the electric flux When we introduced Gauss s Law for the electric field, we defined the electric flux as E E da For the magnetic field, we can define magnetic flux in analogy as B The unit of magnetic flux is the weber (Wb) B da 2 1 Wb 1 Tm June 1, 2016 University Physics, Chapter 29 36
Magnetic Flux - Special Case Consider the special case of a flat loop of area A in a constant magnetic field B In this case we can re-write the magnetic flux as B BAcos is the angle between the surface normal vector of the plane of the loop and the magnetic field If the magnetic field is perpendicular to the plane of the loop = 0, B = BA If the magnetic field is parallel to the plane of the loop = 90, B = 0 June 1, 2016 University Physics, Chapter 29 37