21.10

Concepts & Calculations

Both magnetic fields and electric fields can apply forces to an electric charge. However, there are distinct differences in the way the two types of fields apply their forces. This chapter discusses the magnetic field, and Chapter 18 deals with the electric field. The following example serves to review how the two types of fields behave.

| Concepts & Calculations Example 12 | | | Magnetic and Electric Fields

Figure 21-45 shows a particle that carries a charge of . It is moving along the +y axis at a speed of . A magnetic field of magnitude is directed along the +z axis, and an electric field of magnitude 123 N/C points along the −x axis. Determine the magnitude and direction of the net force that acts on the particle.

Figure 21-45 The net force acting on the moving charge in this drawing is calculated in Example 12.

Concept Questions and Answers What is the net force?

Answer The net force is the vector sum of all the forces acting on the particle, which in this case are the magnetic force and the electric force.

How do you determine the direction of the magnetic force acting on the negative charge?

Answer First, remember that the direction for either a positive or negative charge is perpendicular to the plane formed by the magnetic field vector and the velocity vector . Then, apply Right-Hand Rule No. 1 (RHR-1) to find the direction of the force as if the charge were positive. Finally, reverse the direction indicated by RHR-1 to determine the direction of the force on the negative charge.

How do you determine the direction of the electric force acting on the negative charge?

Answer The direction for either a positive or negative charge is along the line of the electric field. The electric force on a positive charge is in the same direction as the electric field, while the force on a negative charge is in the direction opposite to the electric field.

Does the fact that the charge is moving affect the values of the magnetic and electric forces?

Answer The fact that the charge is moving affects the value of the magnetic force, for without motion, there is no magnetic force. The electric force, in contrast, has the same value whether or not the charge is moving.

Solution An application of RHR-1 (see Figure 21-9) reveals that if the charge were positive, the magnetic force would point along the +x axis. But the charge is negative, so the magnetic force points along the −x axis. According to Equation 21.1, the magnitude of the magnetic force is , where is the magnitude of the charge. Since the velocity and the magnetic field are perpendicular, , and we have .

Problem solving insight The direction of the magnetic force exerted on a negative charge is opposite to that exerted on a positive charge, assuming both charges are moving in the same direction in the same magnetic field.

The electric field points along the −x axis and would apply a force in that same direction if the charge were positive. But the charge is negative, so the electric force points in the opposite direction, or along the +x axis. According to Equation 18.2, the magnitude of the electric force is .
Using plus and minus signs to take into account the different directions of the magnetic and electric forces, we find that the net force is

The fact that the answer is negative means that the net force points along the −x axis.

A number of forces can act simultaneously on a rigid object, some of them producing torques, as Chapter 9 discusses. If, as a result of the forces and torques, the object has no acceleration of any kind, it is in equilibrium. Section 9.2 discusses the equilibrium of rigid objects. The next example illustrates that the magnetic force can be one of the forces keeping an object in equilibrium.

| Concepts & Calculations Example 13 | | | Equilibrium

A 125-turn rectangular coil of wire is hung from one arm of a balance, as Figure 21-46 shows. With the magnetic field turned off, a mass M is added to the pan on the other arm to balance the weight of the coil. When a constant 0.200-T magnetic field is turned on and there is a current of 8.50 A in the coil, how much additional mass m must be added to regain the balance?

Figure 21-46 This setup is the focus of Concepts & Calculations Example 13.

Concept Questions and Answers In a balanced, or equilibrium, condition the device has no angular acceleration. What does this imply about the net torque acting on the device?

Answer According to Newton’s second law as applied to rotational motion (Equation 9.7), when the angular acceleration is zero, the net torque is zero. Counterclockwise torques must balance clockwise torques for an object that is in equilibrium.

What is a torque?

Answer According to Equation 9.1, the torque produced by a force is the magnitude of the force times the lever arm of the force. The lever arm is the perpendicular distance between the line of action of the force and the axis of rotation. Counterclockwise torques are positive, and clockwise torques are negative.

In calculating the torques acting on an object in equilibrium, where do you locate the axis of rotation?

Answer The location of the axis is arbitrary because, if there is no angular acceleration, the torques must add up to zero with respect to any axis whatsoever. Choose an axis that simplifies your calculations.

Solution In a condition of equilibrium the sum of the torques acting on the device is zero. For the purpose of calculating torques, we use an axis that is perpendicular to the page at the knife-edge supporting the balance arms (see the drawing). The forces that can produce torques are the weights of the pans, the masses in the left pan, the coil hanging from the right pan, and the magnetic force on the coil. The weight of each pan is Wpan, and the weight of the coil is Wcoil. The weights of the masses M and m are Mg and mg, where g is the magnitude of the acceleration due to gravity. The lever arm for each of these forces is (see the drawing). When the magnetic field is turned off and the mass m is not present, the sum of the torques is

When the field is turned on, a magnetic force is applied to the 0.0150-m segment of the coil that is in the field. Since the magnetic field points into the page and the current is from right to left, an application of RHR-1 reveals that the magnetic force points downward. According to Equation 21.3, the magnitude of the magnetic force is , where I is the current and L is the length of the wire in the field. Since the current and the field are perpendicular, , and we have F = ILB. The sum of the torques with the field turned on and the mass m added to the left pan is

From the field-off equation, it follows that . With this substitution, the field-on equation simplifies to the following:

Solving this expression for the additional mass m gives

In this calculation we have used the fact that the coil has 125 turns, so the length of the wire in the field is 125 (0.0150 m).

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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