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Transcript
Slide 1 Fig 29-CO, p.895 The direction of the magnetic field B at any location is the direction in which a compass needle points at that location. the magnetic field lines outside the magnet point away from north poles and toward south poles. Slide 2 We can define a magnetic field B at some point in space in terms of the magnetic force FB that the field exerts on a test object, for which we use a charged particle moving with a velocity v. assuming that no electric (E) or gravitational (g) fields are present at the location of the test object. Slide 3 • The magnitude FB of the magnetic force exerted on the particle is proportional to the charge q and to the speed v of the particle • The magnitude and direction of FB depend on the velocity of the particle V and on the magnitude and direction of the magnetic field B. • When a charged particle moves parallel to the magnetic field vector (i.e., θ = 0) , the magnetic force acting on the particle is zero. • When the particle’s velocity vector makes any angle with the magnetic field, the magnetic force acts in a direction perpendicular to both v and B; FB is perpendicular to the plane formed by v and B Slide 4 • The magnetic force exerted on a positive charge is in the direction opposite the direction of the magnetic force exerted on a negative charge moving in the same direction. • The magnitude of the magnetic force exerted on the moving particle is proportional to sin , where is the angle the particle’s velocity vector makes with the direction of B. We can summarize these observations by writing the magnetic force in the form Slide 5 Fig 29-3b, p.897 Slide 6 Slide 7 Fig 29-6, p.901 An electron in a television picture tube moves toward the front of the tube with a speed of 8.0 x 106 m/s along the x axis. Surrounding the neck of the tube are coils of wire that create a magnetic field of magnitude 0.025 T, directed at an angle of 60° to the x axis and lying in the xy plane. Calculate the magnetic force on and acceleration of the electron. Slide 8 3- An electron moving along the positive x axis perpendicular to a magnetic field experiences a magnetic deflection in the negative y direction. What is the direction of the magnetic field? Slide 9 5- A proton moves in a direction perpendicular to a uniform magnetic field B at 1.0 x107 m/s and experiences an acceleration of 2.0x1013 m/s2 in the + x direction when its velocity is in the +z direction. Determine the magnitude and direction of the field. Slide 10 If a magnetic force is exerted on a single charged particle when the particle moves through a magnetic field, it should not surprise you that a current-carrying wire also experiences a force when placed in a magnetic field. the current is a collection of many charged particles in motion; hence, the resultant force exerted by the field on the wire is the vector sum of the individual forces exerted on all the charged particles making up the current. The force exerted on the particles is transmitted to the wire when the particles collide with the atoms making up the wire. Slide 11 Fig 29-7, p.901 Slide 12 Fig 29-7a, p.901 Slide 13 Fig 29-8, p.901 13- A wire having a mass per unit length of 0.500 g/cm carries a 2.00-A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward? Slide 14 16- A conductor suspended by two flexible wires as shown in Figure P29.16 has a mass per unit length of 0.040 0 kg/m. What current must exist in the conductor for the tension in the supporting wires to be zero when the magnetic Slide 15 # The magnetic force acting on a charged particle moving in a magnetic field is perpendicular to the velocity of the particle and that consequently the work done on the particle by the magnetic force is zero. ( W = F. S = F s c o s θ ) # Because FB always points toward the center of the circle, it changes only the direction of v and not its magnitude. # the rotation is counterclockwise for a positive charge. If q were negative, the Slide 16 rotation would be clockwise. The angular speed of the particle The period of the motion T (the time that the particle takes to complete one revolution) is equal to the circumference of the circle divided by the linear speed of the particle: Slide 17 A proton is moving in a circular orbit of radius 14 cm in a uniform 0.35-T magnetic field perpendicular to the velocity of the proton. Find the linear speed of the proton. Slide 18 Slide 19 Velocity Selector: Only those particles having speed v pass un-deflected through the mutually perpendicular electric and magnetic fields. The magnetic force exerted on particles moving at speeds greater than this is stronger than the electric force, and the particles are deflected upward. Those moving at speeds less than this are deflected downward. Slide 20 Slide 21 Slide 22 Slide 23 Fig 29-20, p.909 Slide 24 Fig 29-21, p.910 Slide 25 Fig 29-22, p.910 Slide 26 Fig 29-23, p.911 Slide 27 Fig 29-23a, p.911 Slide 28 Fig 29-23b, p.911 Slide 29 Fig 29-24, p.911 Slide 30 Fig 29-25a, p.912 Slide 31 Fig 29-25b, p.912 Slide 32 Fig 29-26, p.912 Slide 33 Fig 29-27a, p.913 Slide 34 Fig 29-27b, p.913 Slide 35 Fig 29-28, p.914 Slide 36 Fig 29-29, p.915 Slide 37 Fig 29-29a, p.915 Slide 38 Fig 29-29b, p.915 Slide 39 Fig Q29-9, p.917 Slide 40 Fig P29-17, p.918 Slide 41 Fig P29-1, p.918 Slide 42 Fig P29-1a, p.918 Slide 43 Fig P29-1b, p.918 Slide 44 Fig P29-1c, p.918 Slide 45 Fig P29-1d, p.918 Slide 46 Fig P29-14, p.919 Slide 47 Fig P29-15, p.919 Slide 48 Fig P29-17, p.919 Slide 49 Fig P29-18, p.920 Slide 50 Fig P29-23, p.920 Slide 51 Fig P29-48, p.922 Slide 52 Fig P29-53, p.922 Slide 53 Fig P29-55, p.922 Slide 54 Fig P29-59, p.923 Slide 55 Fig P29-61, p.923 Slide 56 Fig P29-63, p.923 Slide 57 Fig P29-64, p.923 Slide 58 Fig P29-66, p.924 Slide 59 Fig P29-69, p.924 Slide 60 Table P29-70, p.924 Slide 61 Fig P29-71, p.924 Slide 62 Fig P29-72, p.925