Chapter 27 Magnetic Field and Magnetic Forces

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1 Chapter 27 Magnetic Field and Magnetic Forces Lecture by Dr. Hebin Li

2 Goals for Chapter 27 To study magnets and the forces they exert on each other To calculate the force that a magnetic field exerts on a moving charge To contrast magnetic field lines with electric field lines To analyze the motion of a charged particle in a magnetic field To see applications of magnetism in physics and chemistry To analyze magnetic forces on current-carrying conductors To study the behavior of current loops in a magnetic field

3 Introduction How does magnetic resonance imaging (MRI) allow us to see details in soft nonmagnetic tissue? How can magnetic forces, which act only on moving charges, explain the behavior of a compass needle? In this chapter, we will look at how magnetic fields affect charges.

Magnetic monopole has not been observed.

4 Permanent magnets Permanent magnet has two poles: north and south. North and south poles always appear in pairs. Magnetic monopole has not been observed. Opposite poles attract and like poles repel. Magnets also attract certain metals (the metal can be magnetized and act like a magnet) The earth is a big magnet!

5 Magnetic field of the earth The earth itself is a magnet. Figure below shows its magnetic field.

6 Magnetic monopoles Breaking a bar magnet does not separate its poles, as shown in Figure at the right. There is no experimental evidence for magnetic monopoles.

7 Electric current and magnets In 1820, Hans Oersted discovered that a currentcarrying wire causes a compass to deflect. This discovery revealed a connection between moving charge and magnetism.

8 The magnetic field A moving charge (or current) creates a magnetic field in the surrounding space. The magnetic field exerts a force on any other moving charge (or current) that is present in the field.

9 The magnetic force on a moving charge The magnetic force on q is perpendicular to both the velocity of q and the magnetic field. The magnitude of the magnetic force is F = q vb sin. Note: The force depends on the velocity of the charge. The SI unit of magnetic field is tesla (T) 1 T = 1 N/A m Another unit is gauss (G), 1 G = 10-4 T.

10 Magnetic force as a vector product We can write the magnetic force as a vector product. The right-hand rule gives the direction of the force on a positive charge.

11 Example The figure shows a uniform magnetic field directed into the plane of the paper. A particle with a positive charge moves in the plane. Which of the three paths does the particle follow? +

12 Equal velocities but opposite signs Two charges of equal magnitude but opposite signs moving in the same direction in the same field will experience magnetic forces in opposite directions. (See Figure 27.8 below.)

13 Example If the particle has negative charge, which of the three paths does the particle follow?

Example: Magnetic force on a proton. A beam of protons (q = 1.6 10 19 C) moves at 3.0 10 5 m/s through a uniform 2.

14 Example: Magnetic force on a proton. A beam of protons (q = C) moves at m/s through a uniform 2.0-T magnetic field directed along the positive z-axis. The velocity of each proton lies in the xz-plane and is directed at 30 to the +z-axis. Find the force on a proton. (Example 27.1 on page 888).

15 Determining the direction of a magnetic field A cathode-ray tube can be used to determine the direction of a magnetic field, as shown in the figure below.

16 A moving charge in both electric and magnetic fields When a charged particle moves through a region where both electric and magnetic fields are present, both fields exert forces on the particle. The total force is F = q(e + v B)

17 Example: A particle with a charge of C is moving with instantaneous velocity v = m i + ( m ) j. s s What is the force exerted on this particle by a magnetic field B = (1.5 T) i

18 Magnetic field lines The figure below shows the magnetic field lines of a permanent magnet. Important properties of magnetic field lines: At each point a magnetic field line is tangent to the direction of B at that point. The more densely the field lines are packed, the stronger the field, and vice versa. Magnetic flux through an area is defined in analogous way to electric flux. The net magnetic flux through any closed surface is zero. Magnetic field lines do not end and they always close on themselves.

19 Magnetic field lines are not lines of force It is important to remember that magnetic field lines are not lines of magnetic force.

20 Magnetic flux We define the magnetic flux through a surface just as we defined electric flux. See the figure below. Gauss s law for magnetism. The magnetic flux through any closed surface is zero. The SI unit is the weber (Wb): 1 Wb = 1 T m 2 Gauss s law for magnetism.

21 Example: A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field B = T (a) in the +z-direction; (b) at an angle of 53.1 from the +z-direction.

Motion of charged particles in a magnetic field If the velocity of the particle is perpendicular to the magnetic field, the particle moves in a circle of radius

22 Motion of charged particles in a magnetic field If the velocity of the particle is perpendicular to the magnetic field, the particle moves in a circle of radius R. F = q vb; mv 2 R So, R = mv q B = q vb The number of revolutions of the particle per unit time is the cyclotron frequency. T = 2πR v = 2πm q B f = 1 T = B q 2πm

23 Example: An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of kg) traveling horizontally at 35.6 km/s enters a uniform, vertical, 1.10-T magnetic field. (a)what is the diameter of the path followed by this alpha particle? (b)what effect does the magnetic field have on the speed of the particle? (c)what are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d)explain why the speed of the particle does not change even though an unbalanced external force acts on it?

24 Helical motion If the particle has velocity components parallel to and perpendicular to the field, its path is a helix. (See Figure at the right.) The speed and kinetic energy of the particle remain constant. F = q(e + v B)

25 Velocity selector A velocity selector uses perpendicular electric and magnetic fields to select particles of a specific speed from a beam. A positive charge experiences two forces: F E = qe; F B = qvb qe = qvb v = E/B Only particles having speed v = E/B pass through undeflected.

26 Thomson s e/m experiment Thomson s experiment measured the ratio e/m for the electron. His apparatus is shown in the figure below. The particle gains kinetic energy during the acceleration Therefore,

27 Mass spectrometer A mass spectrometer measures the masses of ions. The Bainbridge mass spectrometer (see the figure at the right) first uses a velocity selector. Then the magnetic field separates the particles by mass. The radius can be measured So, R = mv/qb m = RqB v where v = E/B.

Example: In the mass spectrometer shown above, singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and

28 Example: In the mass spectrometer shown above, singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V/m and the magnetic field is T. The ions next enter a uniform magnetic field of magnitude T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) The particles enter the second magnetic field and complete a half circle. The distance from the point where they land on the detector to the point where they enter the second magnetic field is 34.0 cm. What is their mass?

The number of moving charges is nal, so F = nal qv d B = (nqv d A)(lB) Since the current I = nqv d A, we have F = IlB The

29 The magnetic force on a current-carrying conductor Consider a conductor with current I in a uninform magnetic field, the magnetic force is the total force on all moving charges in the conductor. The number of moving charges is nal, so F = nal qv d B = (nqv d A)(lB) Since the current I = nqv d A, we have F = IlB The direction of the force is always perpendicular to both the conductor and the field, determined by the right-hand rule. If the field is not perpendicular to the current, F = IlB = IlB sin φ

The magnetic force on a current-carrying conductor We represent a segment of wire with a vector l along the wire in direction of the current, then the force on this

30 The magnetic force on a current-carrying conductor We represent a segment of wire with a vector l along the wire in direction of the current, then the force on this segment can be written in the vector form: F = I l B If the conductor is not straight, we can divide it into infinitesimal segments d l, the force on each segment is df = Id l B

31 Example: how the battery should be connected How the battery should be connected to have the force show in the figure?

32 Example: Magnetic force on a straight conductor Find the direction and magnitude of the magnetic force. or The direction is upward.

33 Force and torque on a current loop The net force on a current loop in a uniform magnetic field is zero. But the net torque is not, in general, equal to zero. The torque is maximal when the magnetic field B is in the plane of the loop. The torque is zero when the magnetic field B is perpendicular to the loop.

34 Force and torque on a current loop The forces on the sides with length b have zero torque. The torque due to the forces on the sides with length a is given by τ = 2F b 2 sin φ = (IBa)(b sin φ) Define the magnetic dipole moment of the loop as μ = IA. Then, τ = μb sin φ The potential energy for a magnetic dipole in a magnetic field:

35 Example: A rectangular coil of wire, 22.0 cm by 35.0 cm and carrying a current of 1.50 A, is oriented with the plane of its loop perpendicular to a uniform 2.50-T magnetic field, as shown in the figure. (a) Calculate the net force and torque that the magnetic field exerts on the coil. (b) The coil is rotated through a 60.0 angle about the axis shown, with the left side coming out of the plane of the figure and the right side going into the plane. Calculate the net force and torque that the magnetic field now exerts on the coil. (c) What is the change in potential energy of the coil when it is rotated from 0 to 60.0?

Chapter 29. Magnetic Fields

Chapter 29. Magnetic Fields Chapter 29 Magnetic Fields Outline 29.1 Magnetic Fields and Forces 29.2 Magnetic Force Acting on a Current-Carrying Conductor 29.4 Motion of a Charged Particle in a Uniform Magnetic Field 29.5 Applications