21.2

The Force that a Magnetic Field Exerts on a Moving Charge

CONCEPTS AT A GLANCE When a charge is placed in an electric field, it experiences an electric force, as Section 18.6 discusses. When a charge is placed in a magnetic field, it also experiences a force, provided that certain conditions are met, as we will see. The magnetic force, like all the forces we have studied (e.g., the gravitational, elastic, and electric forces), may contribute to the net force that causes an object to accelerate. Thus, when present, the magnetic force must be included in Newton’s second law, as the Concepts-at-a-Glance chart in Figure 21-7 illustrates. This chart is an extension of the charts in Figures 4-9, 10-4, 11-5, and 18-4 and emphasizes that the second law is remarkable for taking into account the forces that arise in such a wide variety of situations.

Figure 21-7 CONCEPTS AT A GLANCE The magnetic force, like other forces we have encountered, may contribute to the net force that acts on an object. According to Newton’s second law, the acceleration of an object is directly proportional to the net force. The photograph shows a magnetically levitated train developed in Germany. This train uses magnetic forces to support itself above the track and to propel itself forward. (© argus/Hartmut Schwarzbach/Peter Arnold, Inc.)

The following two conditions must be met for a charge to experience a magnetic force when placed in a magnetic field:

1. | The charge must be moving, for no magnetic force acts on a stationary charge.
2. | The velocity of the moving charge must have a component that is perpendicular to the direction of the magnetic field.

To examine the second condition more closely, consider Figure 21-8, which shows a positive test charge +q0 moving with a velocity through a magnetic field. In the drawing, the magnetic field vector is labeled by the symbol . The field is produced by an arrangement of magnets not shown in the drawing and is assumed to be constant in both magnitude and direction. If the charge moves parallel or antiparallel to the field, as in part a of the drawing, the charge experiences no magnetic force. If, on the other hand, the charge moves perpendicular to the field, as in part b, the charge experiences the maximum possible force . In general, if a charge moves at an angle * with respect to the field (see part c of the drawing), only the velocity component v , which is perpendicular to the field, gives rise to a magnetic force. This force is smaller than the maximum possible force. The component of the velocity that is parallel to the magnetic field yields no force.

Figure 21-8 (a) No magnetic force acts on a charge moving with a velocity that is parallel or antiparallel to a magnetic field . (b) The charge experiences a maximum force when the charge moves perpendicular to the field. (c) If the charge travels at an angle with respect to , only the velocity component perpendicular to gives rise to a magnetic force , which is smaller than . This component is v .

Figure 21-8 shows that the direction of the magnetic force is perpendicular to both the velocity and the magnetic field ; in other words, is perpendicular to the plane defined by and . As an aid in remembering the direction of the force, it is convenient to use Right-Hand Rule No. 1 (RHR-1), as Figure 21-9 illustrates:

> Right-Hand Rule No. 1. Extend the right hand so the fingers point along the direction of the magnetic field and the thumb points along the velocity of the charge. The palm of the hand then faces in the direction of the magnetic force that acts on a positive charge.

It is as if the open palm of the right hand pushes on the positive charge in the direction of the magnetic force. Problem solving insight: If the moving charge is negative instead of positive, the direction of the magnetic force is opposite to that predicted by RHR-1. Thus, there is an easy method for finding the force on a moving negative charge. First, assume that the charge is positive and use RHR-1 to find the direction of the force. Then, reverse this direction to find the direction of the force acting on the negative charge.
We will now use what we know about the magnetic force to define the magnetic field, in a procedure that is analogous to that used in Section 18.6 to define the electric field. Recall that the electric field at any point in space is the force per unit charge that acts on a test charge q0 placed at that point. In other words, to determine the electric field , we divide the electrostatic force by the charge q0: = /q0. In the magnetic case, however, the test charge is moving, and the force depends not only on the charge q0, but also on the velocity component that is perpendicular to the magnetic field. Therefore, to determine the magnitude of the magnetic field, we divide the magnitude of the magnetic force by the magnitude |q0| of the charge and also by v , according to the following definition:

| Definition of the Magnetic Field
| The magnitude B of the magnetic field at any point in space is defined as

| (21.1)

where F is the magnitude of the magnetic force on a test charge, is the magnitude of the test charge, and v is the magnitude of the charge’s velocity, which makes an angle with the direction of the magnetic field. The magnetic field is a vector, and its direction can be determined by using a small compass needle.
SI Unit of Magnetic Field:

The unit of magnetic field strength that follows from Equation 21.1 is the N·s/(C·m). This unit is called the tesla (T), a tribute to the Croatian-born American engineer Nikola Tesla (1856–1943). Thus, one tesla is the strength of the magnetic field in which a unit test charge, traveling perpendicular to the magnetic field at a speed of one meter per second, experiences a force of one newton. Because a coulomb per second is an ampere (1 C/s = 1 A, see Section 20.1), the tesla is often written as 1 T = 1 N/(A·m).
In many situations the magnetic field has a value that is considerably less than one tesla. For example, the strength of the magnetic field near the earth’s surface is approximately 10−4 T. In such circumstances, a magnetic field unit called the gauss (G) is sometimes used. Although not an SI unit, the gauss is a convenient size for many applications involving magnetic fields. The relation between the gauss and the tesla is

Example 1 deals with the magnetic force exerted on a moving proton and on a moving electron.

| Example 1 | | | Magnetic Forces on Charged Particles

A proton in a particle accelerator has a speed of 5.0 × 106 m/s. The proton encounters a magnetic field whose magnitude is 0.40 T and whose direction makes an angle of with respect to the proton’s velocity (see Figure 21-8c). Find (a) the magnitude and direction of the magnetic force on the proton and (b) the acceleration of the proton. (c) What would be the force and acceleration if the particle were an electron instead of a proton?
Reasoning For both the proton and the electron, the magnitude of the magnetic force is given by Equation 21.1. The magnetic forces that act on these particles have opposite directions, however, because the charges have opposite signs. In either case, the acceleration is given by Newton’s second law, which applies to the magnetic force just as it does to any force. In using the second law, we must take into account the fact that the masses of the proton and the electron are different.
Solution

a. | The positive charge on a proton is , and according to Equation 21.1, the magnitude of the magnetic force is . Therefore,

The direction of the magnetic force is given by RHR-1 and is directed upward in Figure 21-8c, with the magnetic field pointing to the right.
b. | The magnitude a of the proton’s acceleration follows directly from Newton’s second law as the magnitude of the net force divided by the mass mp of the proton. Since the only force acting on the proton is the magnetic force F, it is the net force. Thus,

| (4.1)

The direction of the acceleration is the same as the direction of the net force (the magnetic force).
c. | The magnitude of the magnetic force on the electron is the same as that on the proton, since both have the same velocity and charge magnitude. However, the direction of the force on the electron is opposite to that on the proton, or downward in Figure 21-8c, since the electron charge is negative. Furthermore, the electron has a smaller mass me and, therefore, experiences a significantly greater acceleration:

The direction of this acceleration is downward in Figure 21-8c.

Check Your Understanding 1

Two particles, having the same charge but different velocities, are moving in a constant magnetic field (see the drawing, where the velocity vectors are drawn to scale). Which particle, if either, experiences the greater magnetic force? (a) Particle 1 experiences the greater force, because it is moving perpendicular to the magnetic field. (b) Particle 2 experiences the greater force, because it has the greater speed. (c) Particle 2 experiences the greater force, because a component of its velocity is parallel to the magnetic field. (d) Both particles experience the same magnetic force, because the component of each velocity that is perpendicular to the magnetic field is the same. (The answer is given at the end of the book.)

Background: The magnetic force depends on the electric charge on the object, the velocity of the object, and the magnetic field. In particular, the force depends on how the velocity is oriented relative to the field.
For similar questions (including calculational counterparts), consult Self-Assessment Test 21.1, which is described at the end of Section 21.4.

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