22.2

Motional emf

| The emf Induced in a Moving Conductor

When a conducting rod moves through a constant magnetic field, an emf is induced in the rod. This special case of electromagnetic induction arises as a result of the magnetic force (see Section 21.2) that acts on a moving charge. Consider the metal rod of length L moving to the right in Figure 22-4a. The velocity of the rod is constant and is perpendicular to a uniform magnetic field . Each charge q within the rod also moves with a velocity and experiences a magnetic force of magnitude F = |q|vB, according to Equation 21.1. By using RHR-1, it can be seen that the mobile, free electrons are driven to the bottom of the rod, leaving behind an equal amount of positive charge at the top. (Remember to reverse the direction of the force that RHR-1 predicts, since the electrons have a negative charge. See Section 21.2.) The positive and negative charges accumulate until the attractive electric force that they exert on each other becomes equal in magnitude to the magnetic force. When the two forces balance, equilibrium is reached and no further charge separation occurs.

Figure 22-4 (a) When a conducting rod moves at right angles to a constant magnetic field, the magnetic force causes opposite charges to appear at the ends of the rod, giving rise to an induced emf. (b) The induced emf causes an induced current I to appear in the circuit.

The separated charges on the ends of the moving conductor give rise to an induced emf, called a motional emf because it originates from the motion of charges through a magnetic field. The emf exists as long as the rod moves. If the rod is brought to a halt, the magnetic force vanishes, with the result that the attractive electric force reunites the positive and negative charges and the emf disappears. The emf of the moving rod is analogous to that between the terminals of a battery. However, the emf of a battery is produced by chemical reactions, whereas the motional emf is created by the agent that moves the rod through the magnetic field (like the hand in Figure 22-4b.)
The fact that the electric and magnetic forces balance at equilibrium in Figure 22-4a can be used to determine the magnitude of the motional emf E. Acccording to Equation 18.2, the magnitude of the electric force acting on the positive charge q at the top of the rod is Eq, where E is the magnitude of the electric field due to the separated charges. And according to Equation 19.7a (without the minus sign), the electric field magnitude is given by the voltage between the ends of the rod (the emf E) divided by the length L of the rod. Thus, the electric force is . Since we are dealing now with a positive charge, the magnetic force is , according to Equation 21.1, because the charge q moves perpendicular to the magnetic field. Since these two forces balance, it follows that . The emf, then, is

| (22.1)

As expected, when v = 0 m/s, because no motional emf is developed in a stationary rod. Greater speeds and stronger magnetic fields lead to greater emfs for a given length L. As with batteries, E is expressed in volts. In Figure 22-4b the rod is sliding on conducting rails that form part of a closed circuit, and L is the length of the rod between the rails. Due to the emf, electrons flow in a clockwise direction around the circuit. Positive charge would flow in the direction opposite to the electron flow, so the conventional current I is drawn counterclockwise in the picture. Example 1 illustrates how to. determine the electrical energy that the motional emf delivers to a device such as the light bulb in the drawing.

| Example 1 | | | Operating a Light Bulb with Motional Emf

Suppose the rod in Figure 22-4b is moving at a speed of 5.0 m/s in a direction perpendicular to a 0.80-T magnetic field. The rod has a length of 1.6 m and a negligible electrical resistance. The rails also have negligible resistance. The light bulb, however, has a resistance of . Find (a) the emf produced by the rod, (b) the current induced in the circuit, (c) the electric power delivered to the bulb, and (d) the energy used by the bulb in 60.0 s.
Reasoning The moving rod acts like an imaginary battery and supplies a motional emf of vBL to the circuit. The induced current can be determined from Ohm’s law as the motional emf divided by the resistance of the bulb. The electric power delivered to the bulb is the product of the induced current and the potential difference across the bulb (which, in this case, is the motional emf). The energy used is the product of the power and the time.
Solution

a. | The motional emf is given by Equation 22.1 as

b. | According to Ohm’s law, the induced current is equal to the motional emf divided by the resistance of the circuit:

| (20.2)

c. | The electric power P delivered to the light bulb is the product of the current I and the potential difference across the bulb:

| (20.6a)

d. | Since power is energy per unit time, the energy E used in 60.0 s is the product of the power and the time:

| (6.10b)

| Motional Emf and Electrical Energy

Motional emf arises because a magnetic force acts on the charges in a conductor that is moving through a magnetic field. Whenever this emf causes a current, a second magnetic force enters the picture. In Figure 22-4b, for instance, the second force arises because the current I in the rod is perpendicular to the magnetic field. The current, and hence the rod, experiences a magnetic force whose magnitude is given by Equation 21.3 as . Using the values of I, L, and B from Example 1 , we see that . The direction of is specified by RHR-1 and is opposite to the velocity of the rod, and thus points to the left (see Figure 22-5). By itself, would slow down the rod, and here lies the crux of the matter. To keep the rod moving to the right with a constant velocity, a counterbalancing force must be applied to the rod by an external agent, such as the hand in the picture. This force is labeled in the drawing. The counterbalancing force must have a magnitude of 0.086 N and must be directed opposite to the magnetic force . If the counterbalancing force were removed, the rod would decelerate under the influence of and eventually come to rest. During the deceleration, the motional emf would decrease and the light bulb would eventually go out.

Figure 22-5 A magnetic force is exerted on the current I in the moving rod and is directed opposite to the rod’s velocity . Since the force counterbalances the magnetic force , the rod moves to the right at a constant velocity.

We can now answer an important question—Who or what provides the 26 J of electrical energy that the light bulb in Example 1 uses in 60.0 seconds? The provider is the external agent that applies the 0.086-N counterbalancing force needed to keep the rod moving. This agent does work, and Example 2 shows that the work done is equal to the electrical energy used by the bulb.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 2 | | | The Work Needed to Keep the Light Bulb Burning

As we saw in Example 1, an induced current of 0.067 A exists in the circuit due to the moving rod. As Figure 22-5 shows, the hand provides a force that keeps the rod moving to the right. Determine the work done by this force in a time of 60.0 s. Assume, as in Example 1 , that the magnetic field has a magnitude of 0.80 T and that the rod has a length of 1.6 m and moves at a constant speed of 5.0 m/s.
Reasoning According to the discussion in Section 6.1, the work done by the hand is equal to the product of (1) the magnitude Fhand of the force exerted by the hand, (2) the magnitude x of the rod’s displacement, and (3) the cosine of the angle between the force and the displacement. Since the rod moves to the right at a constant speed, it has no acceleration and is, therefore, in equilibrium (see Section 4.11). Thus, the force exerted by the hand must be equal in magnitude, but opposite in direction, to the magnetic force exerted on the rod, since they are the only two forces acting on the rod along the direction of the motion. We will determine the magnitude F of the magnetic force by using Equation 21.3, and this will enable us to find Fhand. Since the rod moves at a constant speed, the magnitude x of its displacement is the product of its speed and the time of travel.
Knowns and Unknowns The data for this problem are:

Description | Symbol | Value
Current | I | 0.067 A
Length of rod | L | 1.6 m
Speed of rod | v | 5.0 m/s
Time during which rod moves | t | 60.0 s
Magnitude of magnetic field | B | 0.80 T
Unknown Variable |  |
Work done by hand | W | ?

Modeling the Problem

| Step 1 Work
| The work done by the hand in Figure 22-5 is given by (Equation 6.1). In this equation, Fhand is the magnitude of the force that the hand exerts on the rod, x is the magnitude of the rod’s displacement, and is the angle between the force and the displacement. Since the force and displacement point in the same direction, , so

Two forces act on the rod; the force , which points to the right, and the magnetic force , which points to the left. Since the rod moves at a constant speed, the magnitudes of these two forces are equal, so that Fhand = F. Substituting this relation into the expression for the work gives Equation 1 at the right. At this point, neither F nor x is known, and they will be evaluated in Steps 2 and 3, respectively.

| (1)

| Step 2 Magnetic Force Exerted on a Current-Carrying Rod
| We have seen in Section 21.5 that a rod of length L that carries a current I in a magnetic field of magnitude B experiences a magnetic force of magnitude F. The magnitude of the force is given by (Equation 21.3), where is the angle between the direction of the current and the magnetic field. In this case, the current and magnetic field are perpendicular to each other (see Figure 22-5), so . Thus, the magnitude of the magnetic force is

The quantities I, L, and B are known, and we substitute this expression into Equation 22.10 at the right.

| Step 3 Kinematics
| Since the rod is moving at a constant speed, its acceleration is zero. Therefore, the distance x it travels is the product of its speed v and the time t:

The variables v and t are known. We can also substitute this relation into Equation 22.10, as shown in the right column.

Solution Algebraically combining the results of the three steps, we have

The work done by the force of the hand is

The 26 J of work done on the rod by the hand is mechanical energy and is the same as the 26 J of energy consumed by the light bulb (see Example 1). Hence, the moving rod and the magnetic force convert mechanical energy into electrical energy, much as a battery converts chemical energy into electrical energy.
Related Homework: Problem 6
It is important to realize that the direction of the current in Figure 22-5 is consistent with the principle of conservation of energy. Consider what would happen if the direction of the current were reversed, as in Figure 22-6. With the direction of the current reversed, the direction of the magnetic force would also be reversed and would point in the direction of the velocity of the rod. As a result, the force would cause the rod to accelerate rather than decelerate. The rod would accelerate without the need for an external force (like that provided by the hand in Figure 22-5) and would create a motional emf that supplies energy to the light bulb. Thus, this hypothetical generator would produce energy out of nothing, since there is no external agent. Such a device cannot exist because it violates the principle of conservation of energy, which states that energy cannot be created or destroyed, but can only be converted from one form to another. Therefore, the current cannot be directed clockwise around the circuit, as in Figure 22-6. In situations such as that in Examples 1 and 2, when a motional emf leads to an induced current, a magnetic force always appears that opposes the motion, in accord with the principle of conservation of energy. Conceptual Example 3 deals further with the important issue of energy conservation.

Figure 22-6 The current cannot be directed clockwise in this circuit, because the magnetic force exerted on the rod would then be in the same direction as the velocity . The rod would accelerate to the right and create energy on its own, violating the principle of conservation of energy.

| Conceptual Example 3 | | | Conservation of Energy

Figure 22-7a illustrates a conducting rod that is free to slide down between two vertical copper tracks. There is no kinetic friction between the rod and the tracks, although the rod maintains electrical contact with the tracks during its fall. A constant magnetic field is directed perpendicular to the motion of the rod, as the drawing shows. Because there is no friction, the only force that acts on the rod is its weight , so the rod falls with an acceleration equal to the acceleration due to gravity, a = 9.8 m/s2. Suppose that a resistance R is connected between the tops of the tracks, as in part b of the drawing. (a) Does the rod now fall with the acceleration due to gravity? (b) How does the principle of conservation of energy apply to what happens in Figure 22-7b?

Figure 22-7 (a) Because there is no kinetic friction between the falling rod and the tracks, the only force acting on the rod is its weight . (b) When an induced current I exists in the circuit, a magnetic force also acts on the rod.

Reasoning and Solution (a) As the rod falls perpendicular to the magnetic field, a motional emf is induced between its ends. This emf is induced whether or not the resistance R is attached between the tracks. However, when R is present, a complete circuit is formed, and the emf produces an induced current I that is perpendicular to the field. The direction of this current is such that the rod experiences an upward magnetic force , opposite to the falling motion and the weight of the rod (see part b of the drawing and use RHR-1). The net force acting downward on the rod is , which is less than the weight, since points upward and the weight points downward. In accord with Newton’s second law of motion, the downward acceleration is proportional to the net force, so the rod falls with an acceleration that is less than the acceleration due to gravity. In fact, as the speed of the rod in Figure 22-7b increases during the descent, the magnetic force becomes larger and larger. There will be a time when the magnitude of the magnetic force becomes equal to the magnitude of the rod’s weight. When this occurs, the net force on the rod will be zero, and the rod’s acceleration will also be zero. From then on, the rod will fall at a constant velocity.
(b) For the freely falling rod in Figure 22-7a, gravitational potential energy (GPE) is converted into kinetic energy (KE) as the rod picks up speed on the way down. In Figure 22-7b, however, the rod always has a smaller speed than that of a freely falling rod at the same place. This is because the acceleration is less than the acceleration due to gravity. In terms of energy conservation, the smaller speed arises because only part of the GPE is converted into KE, with part also being dissipated as heat in the resistance R. In fact, when the rod reaches the point when it falls with a constant velocity, none of the GPE is converted into KE because all of it is dissipated as heat in the resistance.
Related Homework: Problem 9

Check Your Understanding 1

Consider the induced emf being generated in Figure 22-4. Suppose the length of the rod is reduced by a factor of four. For the induced emf to be the same, what should be done? (a) Without changing the speed of the rod, increase the strength of the magnetic field by a factor of four. (b) Without changing the magnetic field, increase the speed of the rod by a factor of four. (c) Increase both the speed of the rod and the strength of the magnetic field by a factor of two. (d) All of the previous three methods may be used. (The answer is given at the end of the book.)
Background: Motional emf is one type of induced emf. It depends on the speed and length of the moving conductor, as well as the strength of the magnetic field, according to Equation 22.1.
For similar questions (including calculational counterparts), consult Self-Assessment Test 22.1, which is described at the end of Section 22.5.

Need more practice?

Interactive LearningWare22.1
A person is twirling a 1.2-m metal chain in a horizontal circle about his head at an angular speed of 6.3 rad/s. The vertical component of the earth’s magnetic field points straight down toward the earth and has a magnitude of 1.8 × 10−5 T. (a) What is the magnitude of the motional emf induced in the rotating chain? (b) When viewed from above, the chain is twirled in a counterclockwise manner. Which end of the chain is positive?

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