Presentation on theme: "Magnetic Fields CHAPTER OUTLINE 29.1 Magnetic Fields and Forces"— Presentation transcript:
1 Magnetic Fields CHAPTER OUTLINE 29.1 Magnetic Fields and Forces 29.2 Magnetic Force Acting on aCurrent-Carrying Conductor29.4 Motion of a Charged Particlein a Uniform Magnetic Field29.5 Applications InvolvingCharged Particles Moving in aMagnetic Field
2 29.1 Magnetic Fields and Forces Experiments on various charged particles moving in a magnetic field give the following results:
8 The SI unit of magnetic field is the newton per coulomb-meter per second, which is called the tesla (T):
9 Example 29.1 An Electron Moving in a Magnetic Field
10 (B) Find a vector expression for the magnetic force on the electron
11 29.2 Magnetic Force Acting on a Current-Carrying Conductor The direction of magnetic fieldMagnetic field lines coming out of the paper are indicated by dots, representing the tips of arrows coming outward.Magnetic field lines going intothe paper are indicated by crosses,representing the feathers of arrowsgoing inward.
12 Magnetic Force Acting on a Current-Carrying Conductor considering a straight segment of wire of length L and cross-sectional area A, carrying a current I in a uniform magnetic field B, as shown in Figure. The magnetic force exerted on a charge q moving with a drift velocity vd isnAL is the number of charges in the segment.We can write this expression in a more convenient form by noting that, from Equation 27.4, the current in the wire is I = nqvdA. Therefore,
13 where L is a vector that points in the direction of the current I and has a magnitude equal to the length L of the segment. Note that this expression applies only to a straight segment of wire in a uniform magnetic field.
14 Example 29.2 Force on a Semicircular Conductor A wire bent into a semicircle of radius R forms a closed circuit and carries a current I. The wire lies in the xy plane, and a uniform magnetic field is directed along the positive y axis, as shown in Figure. Find the magnitude and direction of the magnetic force actingon the straight portion of the wire and on the curvedportion.The magnetic force F acting on the straight portion has a magnitude F = ILB = 2IRB because L = 2R.
15 29.4 Motion of a Charged Particle in a Uniform Magnetic Fieldconsider a positively charged particle moving in a uniform magnetic field with the initial velocity vector of the particle perpendicular to thefield. Let us assume that the direction of the magnetic field is into the page, as in Figure. As the particle changes the direction of its velocity in response to the magnetic force, the magnetic force remains perpendicular to the velocity. If the force is always perpendicular to the velocity, the path of the particle is a circle.
17 the radius of the path is proportional to the linear momentum mv of the particle and inversely proportional to the magnitude of the charge on the particle and to the magnitude of the magnetic field. The angular speed of the particle isThe period of the motion (the time interval the particle requires to complete one revolution) is equal to the circumference of the circle divided by the linear speed of the particle:
18 Example 29.6 A Proton Moving Perpendicular to a Uniform Magnetic Field
21 29.5 Applications Involving Charged Particles Moving in a Magnetic FieldA charge moving with a velocity v in the presence of both an electric field E and a magnetic field B experiences both an electric force qE and a magnetic force q vx B. The total force (called the Lorentz force) acting on the charge is