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General Physics II By Dr. Cherdsak Bootjomchai (Dr.Per) Chapter 7 Magnetic Field and Magnetic Force Objectives: After completing this module, you should be able to: • Define the magnetic field, discussing magnetic poles and flux lines. • Solve problems involving the magnitude and direction of forces on charges moving in a magnetic field. • Solve problems involving the magnitude and direction of forces on current carrying conductors in a B-field. Magnetism Since ancient times, certain materials, called magnets, have been known to have the property of attracting tiny pieces of metal. This attractive property is called magnetism. S Bar Magnet S N N Magnetic Poles Iron filings N The strength of a magnet is concentrated at the ends, called north and south “poles” of the magnet. S A suspended magnet: N-seeking end and S-seeking end are N and S poles. W N S N Bar magnet S N E Compass Magnetic Attraction-Repulsion S S N N N Magnetic Forces: Like Poles Repel S S N N S Unlike Poles Attract Magnetic Field Lines We can describe magnetic field lines by imagining a tiny compass placed at nearby points. The direction of the magnetic field B at any point is the same as the direction indicated by this compass. N S Field B is strong where lines are dense and weak where lines are sparse. Field Lines Between Magnets Unlike poles Attraction N S Leave N and enter S N Like poles N Repulsion The Density of Field Lines Electric field DN Magnetic field flux lines D B DA DA S Line density DN E DA DA Df N Line density Magnetic Field B is sometimes called the flux density in Webers per square meter (Wb/m2). f Magnetic Flux Density • Magnetic flux lines are continuous and closed. B A DA Df • Direction is that of the B vector at any point. • Flux lines are NOT in direction of force but ^. When area A is perpendicular to flux: Magnetic Flux density: B ; = BA A The unit of flux density is the Weber per square meter. Calculating Flux Density When Area is Not Perpendicular The flux penetrating the area A when the normal vector n makes an angle of q with the B-field is: BA cosq n A q a B The angle q is the complement of the angle a that the plane of the area makes with the B field. (Cos q = Sin a) Origin of Magnetic Fields Recall that the strength of an electric field E was defined as the electric force per unit charge. Since no isolated magnetic pole has ever been found, we can’t define the magnetic field B in terms of the magnetic force per unit north pole. We will see instead that magnetic fields result from charges in motion—not from stationary charge or poles. This fact will be covered later. E + + v B^v Magnetic Force on Moving Charge Imagine a tube that projects charge +q with velocity v into perpendicular B field. F B v N S Experiment shows: F qvB Upward magnetic force F on charge moving in B field. Each of the following results in a greater magnetic force F: an increase in velocity v, an increase in charge q, and a larger magnetic field B. Direction of Magnetic Force The right hand rule: With a flat right hand, point thumb in direction of velocity v, fingers in direction of B field. The flat hand pushes in the direction of force F. F B v N F B v S The force is greatest when the velocity v is perpendicular to the B field. The deflection decreases to zero for parallel motion. Force and Angle of Path N N N S S S Deflection force greatest when path perpendicular to field. Least at parallel. F v sinq F v sin q q v B v Definition of B-field Experimental observations show the following: F qv sin q or F constant qv sin q By choosing appropriate units for the constant of proportionality, we can now define the B-field as: Magnetic Field Intensity B: F B qv sin q or F qvB sin q A magnetic field intensity of one tesla (T) exists in a region of space where a charge of one coulomb (C) moving at 1 m/s perpendicular to the B-field will experience a force of one newton (N). Example 1. A 2-nC charge is projected with velocity 5 x 104 m/s at an angle of 300 with a 3 mT magnetic field as shown. What are the magnitude and direction of the resulting force? Draw a rough sketch. q = 2 x 10-9 C v = 5 x 104 m/s v sin f B = 3 x 10-3 T q = 300 F B 300 v B v Using right-hand rule, the force is seen to be upward. F qvB sin q (2 x 10-9C)(5 x 104 m/s)(3 x 10-3T)sin 300 Resultant Magnetic Force: F = 1.50 x 10-7 N, upward Forces on Negative Charges Forces on negative charges are opposite to those on positive charges. The force on the negative charge requires a left-hand rule to show downward force F. Right-hand rule for positive q N F B v S Left-hand rule for negative q N B F v S Indicating Direction of B-fields One way of indicating the directions of fields perpendicular to a plane is to use crosses X and dots : A field directed into the paper is denoted by a cross “X” like the tail feathers of an arrow. X X X X X X X X X X X X X X X X A field directed out of the paper is denoted by a dot “” like the front tip end of an arrow. Practice With Directions: What is the direction of the force F on the charge in each of the examples described below? X X X X F X X X Up X+ X v X X X X X X X F Up v - negative q X X Left X X v X X FX X + X X X X X X X X F Right v Crossed E and B Fields The motion of charged particles, such as electrons, can be controlled by combined electric and magnetic fields. Note: FE on electron is upward and opposite E-field. + But, FB on electron is down (left-hand rule). Zero deflection when FB = FE e- x x x x x x x x v - FE E -- e B v B FB v The Velocity Selector This device uses crossed fields to select only those velocities for which FB = FE. (Verify directions for +q) When FB = FE : qvB qE E v B Source of +q + x x x x x x x x +q v - Velocity selector By adjusting the E and/or B-fields, a person can select only those ions with the desired velocity. Example 2. A lithium ion, q = +1.6 x 10-16 C, is projected through a velocity selector where B = 20 mT. The E-field is adjusted to select a velocity of 1.5 x 106 m/s. What is the electric field E? E v B E = vB Source of +q + x x x x x x x x +q v V E = (1.5 x 106 m/s)(20 x 10-3 T); E = 3.00 x 104 V/m Circular Motion in B-field The magnetic force F on a moving charge is always perpendicular to its velocity v. Thus, a charge moving in a B-field will experience a centripetal force. mv 2 FC ; FB qvB; R FC FB Centripetal Fc = FB X mv qvB R X +X X X X RX X X X X X FX X mv R qB X X X X X The radius of path is: 2 + X + X c X X X + X X X X X X Mass Spectrometer +q - slit E v + B xx Photographic xx plate xx R xx x x x x x x x x x x x x x x x x x x x x x x x x m2 x x x x x x x x x x mv 2 qvB R m1 Ions passed through a velocity selector at known velocity emerge into a magnetic field as shown. The radius is: mv R qB The mass is found by measuring the radius R: qBR m v Example 3. A Neon ion, q = 1.6 x 10-19 C, follows a path of radius 7.28 cm. Upper and lower B = 0.5 T and E = 1000 V/m. What is its mass? +q E v + B xx Photographic - xx plate xx R xx x x x x x x x slit x x x x x x x x x x x x x x m x x x x x x x x x x E 1000 V/m v B 0.5 T v = 2000 m/s mv R qB (1.6 x 10-19C)(0.5 T)(0.0728 m) m 2000 m/s qBR m v m = 2.91 x 10-24 kg The force on a current carrying wire in a magnetic field FB ,q qv d B ) FB qv d B )nAL I vd qnA FB IL B FB BIL sin q For a straight wire in a uniform field dFB Ids B b FB I ds B a For an arbitrary wire in an arbitrary field b FB I ds B a b FB I ds B a FB IL B ) FB I ds B ds 0 FB 0 Ex. From the figure, the magnetic field (B) is uniform and perpendicular to the plane. The conductor caries a current I. Find the total magnetic force on these wire. y I I R I L I (in ) x Torque on a Current Loop in a Uniform Magnetic Field If the field is parallel to the plane of the loop For 1 and 3, LB 0 FB 0 For 2 and 4, F2 F4 IaB max max b b b b F2 F4 IaB ) IaB ) 2 2 2 2 IaB )b max IAB If the field makes an angle with a line perpendicular to the plane of the loop: a a F1 sin q F3 sin q 2 2 a a IbB sin q IbB sin q IabB sin q 2 2 IAB sin q IA B Magnetic Dipole Moment μ IA (Amperes.m2) μB If the wire makes N loops around A, Nμ loop B μ coil B The potential energy of the loop is: U μ B Hall Effect qvd B qEH E H vd B DVH vd Bd I vd nqA Hall constant or Hall coefficient IB R H IB ΔVH nqt t Summary The direction of forces on a charge moving in an electric field can be determined by the right-hand rule for positive charges and by the left-hand rule for negative charges. Right-hand rule for positive q N F B v S Left-hand rule for negative q N B F v S Summary (Continued) F v sin q q For a charge moving in a B-field, the magnitude of the force is given by: v B v F = qvB sin q Summary (Continued) The velocity selector: E v B The mass spectrometer: mv R qB qBR m v + + vq x x x x x x x x V +q - xx - xx + xx xx x x x slit x x x x x x x x E v B R x x x x x x x x x x x x x x x m The end of Chapter 7 Magnetic Field and Magnetic Force Quiz III v0 A 10.0 cm B An electron at point A in Fig. has a speed v0 of 2.94 106 m/s. Find a) the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from A to B ; b) the time required for the electron to move from A to B.