Transcript Physics 21
Physics Chapter 21: Magnetism Magnetism ☺Magnets ☺Caused by the Polarization of Iron Molecules ☺Material Containing Iron (Fe) Magnetism ☺Polarization ☺The Alignment of Charged Molecules Within the Material N S N S N S N N S S N S N S Magnetism ☺Magnetic Field (B) ☺Surrounds Magnet ☺Possess Magnitude and Direction ☺“The Direction of the Magnetic Field at Any Point in Space is the Direction Indicated by the North Pole of a Small Compass Needle Placed at that Point” Magnetism ☺Magnetic Field (B) Magnetism ☺Magnetic Field ☺Magnetic Field Lines Currently 11.50 Between Geographic North Pole and the Magnetic North Pole Magnetism ☺Electromagnetism ☺Current Passing through a Wire Creates a Magnetic Field Magnetism ☺Electromagnetism Magnetism ☺Electromagnetism ☺Right Hand Rule Magnetism ☺Electromagnetism ☺Single Loop Magnetism ☺Electromagnetism ☺Multiple Loops Magnetism ☺Forces in Magnetic Fields ☺Magnetic Forces on a Charge ☺The Charge Must Be Moving ☺Magnetic Force Does Not Act On a Stationary Charge ☺The Velocity of the Moving Charge Must Have a Component that is Perpendicular to the Direction of the Magnetic Field (Right Hand Rule) Magnetism ☺Forces in Magnetic Fields ☺Right Hand Rule (1) ☺Extend Your Right Hand so Your Fingers Represent the Direction of the Magnetic Field and Your Thumb Represents the Velocity of the Charge ☺Your Palm Then Indicates the Direction of the Maximum Magnetic Force Acting on a Positive Charge ☺If the Charge is Negative, the Back of Your Hand Indicates the Direction of the Maximum Magnetic Force Acting on the Charge Magnetism ☺Forces in Magnetic Fields Magnetism ☺Forces in Magnetic Fields ☺Magnitude ☺F = Magnitude of Magnetic Force on the Test Charge ☺q = Test Charge ☺v = Velocity of the Test Charge ☺q = Angle of the Test Charge Velocity in Relation to the Magnetic Force F B q(v sin q ) Magnetism ☺Forces in Magnetic Fields ☺Magnitude (Units) F (newton )(second ) Tesla (T ) q(v sin q ) (coulomb )( meter ) F B q(v sin q ) Magnetism ☺Forces in Magnetic Fields ☺Magnitude (Units) ☺Tesla ☺A Test Charge Perpendicular to B Experiences 1N @ 1m/s ☺For Much Smaller Quantities ☺1 Gauss = 1x10-4 Tesla F B q(v sin q ) Magnetism ☺ The Motion of a Charged Particle in a Magnetic Field ☺ Remember: In an Electric Field (like a parallel plate capacitor) a Positively Charged Particle Entering Perpendicular to the Field will Move Toward the Negatively Charged Plate ☺ In a Magnetic Field, a Positively Charged Particle Entering Perpendicular to the Field will Move Between the Sources of the Field due to the Horizontal Position of the Magnetic Force Particle Will Come Out of the Screen Toward You ++++++++++++++++ N v E F ---------------- q0 F v q 0 B S Magnetism ☺The Motion of a Charged Particle in a Magnetic Field ☺While an Electric Field will Accelerate a Charged Particle Placed Stationary Within the Field, a Magnetic Field Cannot Change the Kinetic Energy of the Particle ☺In A Magnetic Field, Particle Speed is Constant Magnetism ☺ The Motion of a Charged Particle in a Magnetic Field ☺ Circular Trajectory ☺ The Magnetic Force is Perpendicular to the Magnetic Field ☺ RHR1 Suggests a Change in Trajectory of a Particle Due to the Perpendicular Force ☺ This Causes the Force to Always be Directed to the Center of a Circular Path of the Particle ☺ Centripetal Force is then Exerted on the Particle (Fc) Magnetism ☺ The Motion of a Charged Particle in a Magnetic Field ☺ Circular Trajectory x x x x x x x x x x F x x x +q0 x x +q0 x Magnetism ☺ The Motion of a Charged Particle in a Magnetic Field ☺ Circular Trajectory mv Fc r 2 x x x x x x x x x x F x x x +q0 x x +q0 x Magnetism ☺ The Motion of a Charged Particle in a Magnetic Field ☺ Circular Trajectory mv Fc r 2 mv r qB x x x x x x x x x x F x x x +q0 x x +q0 x Magnetism ☺The Force on a Current in a Magnetic Field ☺Current Traveling through a Wire Within a Magnetic Field is Affected by the Field Magnetism ☺The Force on a Current in a Magnetic Field ☺Force Magnitude ☺I = Current ☺L = Length of Wire in Meters ☺B = Magnetic Field Magnitude F BIL sin q Magnetism ☺The Force on a Charged Particle in a Magnetic Field F BIL sin q qv I L qv F B L sin q L F Bqv sin q Magnetism ☺Problem ☺Conventional current is running through a wire. Sketch the magnetic field that the current generates. Magnetism ☺Solution Magnetism ☺Problem ☺This figure shows the end view of an electromagnet with the current as shown. What is the direction of the magnetic field inside the loop and outside the loop? Magnetism ☺Solution Magnetism ☺Problem ☺A current-carrying wire is placed between the poles of a magnet. What is the direction of the force on the wire? Magnetism ☺Solution Magnetism ☺Problem ☺The current through a wire 0.80m long is 5.0A. The wire is perpendicular to a 0.60T magnetic field. What is the magnitude of the force on the wire? Magnetism ☺Solution F BIL sin q (0.60T )(5.0 A)(0.80m) 2.4 N Magnetism ☺Problem ☺A wire 35cm long is parallel to a 0.53T uniform magnetic field. The current through the wire is 4.5A. What force acts on the wire? Magnetism ☺Solution ☺The wire is parallel to the magnetic field. No force acts on the wire. ☺q 0 F BIL sin q (0.53T )( 4.5 A)(0.35m)(sin 0) 0 Magnetism ☺Problem ☺A wire 625m long is in a 0.40T magnetic field. A 1.8N force acts on the wire. What current is in the wire? Magnetism ☺Solution ☺L = 625m ☺B = 0.40T ☺F = 1.8N ☺q 90 (assume perpendicular if angle is not given) F 1.8 N I 7.2mA BL sin q (0.40T )(625m)(sin 90) Magnetism ☺Problem ☺The force on a 0.80m wire that is perpendicular to Earth’s magnetic field is 0.12N. What is the current in the wire? Magnetism ☺Solution ☺L = 0.80m ☺B = 5x10-5T ☺F = 0.12N ☺q 90 F 0.12 N I 3kA 5 BL sin q (5 x10 T )(0.80m)(sin 90) Magnetism ☺Problem ☺A power line carries a 225A current from east to west parallel to the surface of Earth. What is the magnitude and direction of the force resulting from Earth’s magnetic field acting on each meter of the wire? Magnetism ☺Solution ☺L = 1m ☺B = 5x10-5T ☺I = 225A ☺q 90 F 5 IB (225 A)(5 x10 T ) 0.011N / m L Magnetism ☺Problem ☺A beta particle (high-speed electron) is traveling at right angles to a 0.60T magnetic field. It has a speed of 2.5x107 m/s. What size force acts on the particle? Magnetism ☺Solution ☺B = 0.60T ☺q = 1.6x10-19C ☺v = 2.5x107 m/s F Bqv (0.60T )(1.6 x1019 C )(2.5x107 m / s) 2.4 x1012 N Magnetism ☺Problem ☺A magnetic field of 16T acts in a direction due west. An electron is traveling due south at 8.1x105 m/s. What are the magnitude and direction of the force acting on the electron? Magnetism ☺Solution ☺B = 16T ☺q = 1.6x10-19C ☺v = 8.1x105 m/s F Bqv (16T )(1.6 x1019 C )(8.1x105 m / s) 2.1x1012 N ☺Upward, remember that electron flow is opposite of current flow. Magnetism ☺Problem ☺A wire carrying 15A of current has a length of 25cm in a magnetic field of 0.85T. What is the force on the wire if it makes an angle with the magnetic field lines of 45°? Magnetism ☺Solution ☺B = 0.85T ☺L = 0.25m ☺I = 15A F BIL sin q (0.85T )(15 A)(0.25m) sin 45 2.3N Magnetism ☺Problem ☺A magnet attracts a nail, which, in turn, attracts many small tacks. If the N-pole of the permanent magnet is the top face, which end of the nail is the N-pole? N S Magnetism ☺Solution ☺Down N S N S S N Magnetism ☺Problem ☺A student makes a magnet by winding wire around a nail and connecting it to a battery, as shown in the figure below. Which end of the nail, the pointed end or the head, will be the north pole? Magnetism ☺Solution ☺The Pointed End S N Magnetism ☺Problem ☺A wire 0.50m long carrying a current of 8.0A is at right angles to a 0.40T magnetic field. How strong a force acts on the wire? Magnetism ☺Solution ☺B = 0.40T ☺L = 0.50m ☺I = 8.0A F BIL sin q (0.40T )(8.0 A)(0.50m) sin 90 1.6 N Magnetism ☺Problem ☺A wire 75cm long carrying a current of 6.0A is at right angles to a uniform magnetic field. The magnitude of the force acting on the wire is 0.60 N. What is the strength of the magnetic field? Magnetism ☺Solution ☺F = 0.60N ☺L = 0.75m ☺I = 6.0A F (0.60 N ) B 0.13T IL sin q (6.0 A)(0.75m) sin 90 Magnetism ☺Problem ☺A copper wire 40cm long carries a current of 6.0A and weighs 0.35N. A certain magnetic field is strong enough to balance the force of gravity on the wire. What is the strength of the magnetic field? Magnetism ☺Solution ☺F = 0.35N ☺L = 0.4m ☺I = 6.0A F (0.35 N ) B 0.145T IL sin q (6.0 A)(0.4m) sin 90 Magnetism ☺Problem ☺An electron passes through a magnetic field at right angles to the field at a velocity of 4.0x106 m/s. The strength of the magnetic field is 0.50T. What is the magnitude of the force acting on the electron? Magnetism ☺Solution ☺B = 0.50T ☺q = 1.6x10-19C ☺v = 4x106 m/s F Bqv (.50T )(1.6 x10 19 C )(4 x10 m / s) 3.2 x10 6 13 N Magnetism ☺Problem ☺Doubly ionized helium atoms (alpha particles) are traveling at right angles to a magnetic field at a speed of 4.0x10–2 m/s. The force that acts on each particle is 6.4x10-22N. What is the magnitude of the magnetic field? Magnetism ☺Solution ☺F = 6.4x10-22N ☺q = 1.6x10-19C (x2) ☺v = 4x10-2 m/s F 6.4 x1022 N 2 B 5 x 10 T 19 2 qv (2)(1.6 x10 C )( 4 x10 m / s) Magnetism ☺Problem ☺A force of 5.78x10–16 N acts on an unknown particle traveling at a 90° angle through a magnetic field. If the velocity of the particle is 5.65x104 m/s and the field is 3.20x10–2 T, how many elementary charges does the particle carry? Magnetism ☺Solution ☺F = 5.78x10-16N ☺B = 3.20x10-2T ☺v = 5.65x104 m/s F 5.78x1016 N 19 q 3.20 x10 C 2 4 Bv (3.20 x10 T )(5.65 x10 m / s) 3.2 x10 19 e C 2 19 1.6 x10 C Magnetism ☺Homework: ☺Pages 782-783 ☺31 (2.1x10-3m/s) ☺32 (0.3T) ☺38 (1.9x1014m/s2)