Lesson Explainer: Motion of Straight Conductors in Uniform Magnetic Fields Physics • 9th Grade

In this explainer, we will learn how to relate the potential difference induced across straight conductors to their motion in uniform magnetic fields.

Consider a straight wire moving through a uniform magnetic field, as shown below.

Because this wire is a conductor, it contains electrons that are free to move throughout the wire.

As the wire moves through the field, a magnetic force acts on these electrons.

The direction of the magnetic force on the electrons can be determined by recalling the right-hand rule for the direction of the magnetic force on charges moving in a magnetic field.

Using our right hand, we first identify the direction of π‘žπ‘£, where π‘ž is the charge of the particle being considered and 𝑣 is the particle’s velocity.

Since electrons have a negative charge, π‘žπ‘£ in this instance points opposite the direction of 𝑣. That is, in our wire that moves to the right, π‘žπ‘£ for electrons points to the left.

We point the fingers of our right hand in the direction of π‘žπ‘£, as shown in the following figure.

The next step in using the right-hand rule is to curl our fingers in the direction of the magnetic field. In this instance, that field points out of the screen.

With our right hand arranged this way, the thumb points in the direction of the magnetic force on the charged particle, as shown below.

Electrons in the wire will therefore be pushed toward the wire’s top.

Overall, as the wire moves, negative charge will concentrate at its top and positive charge will concentrate at its bottom, as shown below.

The electric potential near the negative-charge concentration is more negative than the potential near the positive-charge concentration. Therefore, a potential difference is established across the wire.

Note that in this scenario, the magnetic field points out of the screen, while the wire’s velocity points to the right. Therefore, the angle between these two vectors is 90∘.

Next, consider a situation where the wire instead moves out of the screen, as shown below.

The angle between the wire’s velocity and the external field is now 0∘. No potential difference is induced across the wire.

Potential difference is measured in units of volts, and so is a quantity called electromotive force. This β€œforce” is an amount of energy imparted to charges. It is symbolized as πœ– and described mathematically in the following way.

Formula: Electromotive Force in a Straight Conductor Moving in a Uniform Magnetic Field

If a straight conductor of length 𝑙 is moving with velocity 𝑣 through a uniform magnetic field 𝐡 such that the angle between 𝐡 and 𝑣 is πœƒ, then the electromotive force (πœ–) induced in the conductor is πœ–=𝑙𝑣𝐡(πœƒ).sin

The induced electromotive force (often written β€œemf” for short) is equal to the potential difference induced across a wire moving through a uniform magnetic field.

In this equation, 𝑙 is the length of the dimension of the conductor along which emf is induced.

Example 1: Determining Field Strength and Motion Direction of a Straight Conductor Moving in a Uniform Magnetic Field

A 15 cm long conducting rod has a potential difference across it, as shown in the diagram. The rod moves through a uniform magnetic field at 0.32 m/s. The magnitude of the induced potential difference is 9.6 mV. The rod moves in the plane of the screen.

  1. What is the strength of the magnetic field?
  2. Which side of the field-containing region is the rod moving toward?

Answer

Part 1

The potential difference induced across the rod is equal in magnitude to the electromotive force (πœ–) across the rod. This β€œforce” is given by the equation πœ–=𝑙𝑣𝐡(πœƒ).sin

Here, 𝑙 is the length of the rod, 𝑣 is its velocity, 𝐡 is the strength of the magnetic field through which the rod moves, and πœƒ is the angle between 𝑣 and 𝐡.

We seek the strength of the magnetic field in this scenario. The field strength 𝐡 can be isolated on one side of the equation above by dividing both sides by 𝑙, 𝑣, and sin(πœƒ): 𝐡=πœ–π‘™π‘£(πœƒ).sin

The magnitude of the potential difference, which is equal to the electromotive force, is given as 9.6 mV. The rod length is 15 cm, and its speed is 0.32 m/s. The rod moves perpendicularly to the magnetic field, meaning πœƒ equals 90∘.

Before substituting in these values and computing 𝐡, we must make their units consistent with one another. We can do this by converting the emf from units of millivolts to units of volts and the length from units of centimetres to units of metres.

We know that 1β€Žβ€‰β€Ž000 mV equals 1 V, so 9.6 mV is equal to 0.0096 V.

Similarly, 100 cm equals one metre, so 15 cm equals 0.15 m.

We may now substitute our converted values into the equation for 𝐡: 𝐡=0.00960.15Γ—0.32/Γ—(90)=0.00960.15Γ—0.32/=0.2.VmmssinVmmsT∘

This result corresponds to answer D.

Part 2

The direction of the rod’s motion affects the direction of the magnetic force acting on the charges in the rod.

From the diagram, we see that the top of the rod has a net positive charge, while the bottom of the rod possesses a net negative charge.

We know, therefore, that the rod is moving in such a way that the force on the free electrons is toward the bottom side of the region containing the magnetic field.

This force direction is given by the right-hand rule. The rule states that if the fingers on our right hand point first in the direction of π‘žπ‘£, where 𝑣 is the velocity of the charge π‘ž, and then curl into the direction of the external magnetic field 𝐡, the thumb of that hand points in the direction of the magnetic force on the charge π‘ž.

Given that in this situation the force on positive charges acts toward the top side of the region containing the magnetic field and the magnetic field points out of the screen, we can use the right-hand rule to determine the direction of the rod’s motion.

With our right thumb pointing up, while our curled fingers point out of the screen, as shown in the first image below, we see that the only direction our fingers could point when they are straightened is to the left, as shown in the second image.

This finger direction shows that for positive charges to experience an upward force, the velocity of the rod must be to the left.

Therefore, the rod is moving toward the left side of the field-containing region.

A straight conductor moving through a uniform magnetic field may be part of a closed conducting loop, as shown in the following figure.

In this scenario, the potential difference (or equivalently, emf) induced across the moving conductor creates current through the circuit.

We have seen that the induced potential difference is given by πœ–=𝑙𝑣𝐡(πœƒ).sin

When this potential difference generates a current 𝐼 in a circuit of resistance 𝑅, Ohm’s law indicates that πœ–=𝐼𝑅.

Note that while the induced current does generate a magnetic field, this induced field is typically negligibly small compared to the uniform external field 𝐡. We therefore do not account for it in calculations of πœ–, 𝐼, or 𝑅.

Example 2: Solving for the Resistance of a Straight Conductor Moving through a Uniform Magnetic Field

A conducting rod moves on conducting rails that form a circuit that contains a resistor, as shown in the diagram. The rod travels the full distance across the rails in a time of 36 s, moving at a constant speed. The magnetic field around the circuit has a strength of 275 mT. The current in the circuit is 32 ΞΌA. Find the resistance of the rod to one decimal place.

Answer

As the rod moves through the magnetic field, a potential difference given by the equation πœ–=𝑙𝑣𝐡(πœƒ)sin will be induced across it. The moving rod essentially functions as a cell for the electric circuit shown. We can relate the potential difference πœ–, the circuit current 𝐼, and the total circuit resistance 𝑅 using Ohm’s law as follows: πœ–=𝐼𝑅.

The total circuit resistance 𝑅 consists of the resistance of the resistor (24 Ξ©) plus the rod’s resistance. Labeling the resistance of the rod 𝑅r, we can write 𝑅=24+𝑅.Ξ©r

Therefore, πœ–=𝐼×(24+𝑅),Ξ©r or equivalently 𝑙𝑣𝐡(πœƒ)=𝐼×(24+𝑅).sinΞ©r

We begin solving for 𝑅r by rearranging the equation above so 𝑅r is the subject. Multiplying the current 𝐼 across the right side of the equation gives us 𝑙𝑣𝐡(πœƒ)=𝐼×24+𝐼𝑅.sinΞ©r

Subtracting 𝐼×24 Ξ© from both sides and then dividing both sides by 𝐼, we find 𝑅=𝑙𝑣𝐡(πœƒ)βˆ’πΌΓ—24𝐼.rsinΞ©

The rod length 𝑙 is given to be 9.5 cm. This is equal to 0.095 m.

The velocity, 𝑣, of the rod is given by the expression 𝑣=𝑑𝑑, where 𝑑 is the distance traveled by the rod and 𝑑 is the time taken to travel it. The rod moves 125 cm in 36 s, meaning that it has a velocity of 125 cm per 36 s or 3.472… cm/s. In units of metres per second (m/s), the rod’s velocity is 0.0342… m/s.

The magnetic field strength 𝐡 is 275 mT. Since there are 1β€Žβ€‰β€Ž000 mT in 1 T, this equals 0.275 T.

Regarding the angle πœƒ, the rod moves perpendicularly to the field, so πœƒ is 90∘. Note that the sine of 90∘ is 1.

The current 𝐼 in the circuit is 32 ΞΌA, or equivalently 3.2Γ—10 A.

Substituting all these values into our equation for 𝑅r, we find 𝑅=(0.095)Γ—(0.03472…/)Γ—(0.275)Γ—(1)βˆ’ο€Ή3.2Γ—10×243.2Γ—10=4.3474….rmmsTAΞ©AΩ

Rounded to one decimal place, the resistance of the rod is 4.3 ohms.

Even though potential difference is a scalar quantity, it is nonetheless induced along a certain direction. That direction is determined using the right-hand-rule as we have seen.

Consider a straight conductor of length 𝑙 and square cross section of width w moving through a uniform magnetic field, as shown in the following diagram.

For positive charges, π‘žπ‘£ points to the right. The magnetic field 𝐡 points toward the top of the screen. Using the right-hand rule, the magnetic force on positive charges is directed out of the screen.

Therefore, positive charge will accumulate on the front-facing side of the conductor, while negative charge will be concentrated on the back face. This is shown in the following diagram, where regions of net positive charge are colored red and regions of net negative charge are colored blue.

Though emf is induced in this conductor, its magnitude is not equal to 𝑙𝑣𝐡(πœƒ)sin. Rather, it is equal to 𝑀𝑣𝐡(πœƒ)sin, where 𝑀 is the length of the conductor dimension along which charge has separated.

Example 3: Rotating Conductors in Uniform Magnetic Fields

A conducting rod is rotated with one end of it held stationary. The rod rotates uniformly within a uniform magnetic field, with the direction of the rotation of the rod relative to the magnetic field varied, as shown in the diagrams I, II, III, and IV. The rod rotates at the same rate in each diagram.

  1. In which of the diagrams does the value for the magnitude of the potential difference induced between the fixed end of the rod and the free end of the rod vary as the rod rotates?
  2. Is the magnitude of the potential difference induced between the fixed end of the rod and the free end of the rod in diagram I equal to the magnitude of the potential difference induced between the fixed end of the rod and the free end of the rod in diagram II?
  3. Is the magnitude of the potential difference induced between the fixed end of the rod and the free end of the rod in diagram III equal to the magnitude of the potential difference induced between the fixed end of the rod and the free end of the rod in diagram IV?
  4. Is the magnitude of the potential difference induced between the fixed end of the rod and the free end of the rod in diagram I equal to the magnitude of the potential difference induced between the fixed end of the rod and the free end of the rod in diagram III?

Answer

Part 1

Considering the four diagrams, we begin with diagram I.

Diagram I shows the rotating rod at four moments. We can label these positions 0∘, 90∘, 180∘, and 270∘, as shown in the figure below.

At each position, the rod has a nonzero average velocity and moves through a uniform magnetic field pointing to the right.

The following figure shows these average velocities with the corresponding magnetic field vectors.

To determine if, and in what direction, potential difference is induced in the rod at any of these positions, we use the right-hand rule.

This rule indicates that the direction of the magnetic force on a charge π‘ž moving at a velocity 𝑣 through a magnetic field 𝐡 can be determined as follows: The fingers of the right hand are pointed in the direction of π‘žπ‘£, then curled in the direction of 𝐡. The direction then indicated by the right thumb shows the direction of magnetic force acting on the charge.

Applying this rule to the four rod positions in the figure above, we find the results shown in the following figure.

Note that at all four positions, the magnetic force on a positive charge is never along the length of the conductor. Positive and negative charges do not separate along the rod’s length, and therefore the potential difference along the rod’s length is zero.

Now, consider the rod at some arbitrary rotation angle other than the angles shown above. In such an arbitrary position, the average velocity vector of the rod can be divided into a horizontal and a vertical component, as shown in the figure below.

For each component, the induced emf along the length of the rod is zero. Since the angle theta in the figure is arbitrary, the induced emf along the rod’s length in diagram I is zero at all positions.

Next, we consider diagram II. Note that the only difference between diagram II and diagram I is that diagram II is rotated 90∘ from diagram I in the counterclockwise direction.

Therefore, no potential difference is induced along the length of the rod in diagram II either.

In diagram III, the magnetic field points into the screen. Therefore, the velocity vectors and matching magnetic field vectors appear as shown in the following figure.

Again, using the right-hand rule to determine the direction of magnetic force on positive charge at each position, we find the results below, where red arrows indicate force vectors on positive charge.

In diagram III, positive charge is pushed toward the axis of rotation of the rod, meaning that negative charge is pushed toward the rod’s free end. Therefore, charge separation occurs along the rod’s length.

A potential difference is induced across the length of the rod as a result.

Recall, however, that we are asked to identify diagrams where the induced potential difference along the rod varies.

For the scenario shown in diagram III, the induced potential difference along the rod is nonzero, but it is also constantβ€”it does not change as the rod rotates.

Finally, consider the situation depicted in diagram IV.

This scenario is identical to that shown in diagram III, except that the magnetic field now points out of the screen rather than into it.

The result of this difference is that in the scenario shown in diagram IV, positive charge is forced toward the free end of the rod and negative charge is pushed toward its fixed end.

Nonzero potential difference is induced along the rod’s length.

As was true for the situation shown in diagram III, however, this potential difference does not vary as the rod rotates.

Our answer to part 1 of the question is that none of the diagrams show scenarios where the potential difference induced along the rod’s length varies.

Part 2

Reviewing our analysis in part 1, we recall that in diagrams I and II, the induced potential difference across the rod’s length is zero. Therefore, it is true that these values are equal.

Part 3

In the scenario shown in diagram III, positive charge collects toward the fixed end of the rod and negative charge accumulates toward its free end, as shown in the following figure.

The magnitude of the potential difference induced is given by the equation πœ–=𝑙𝑣𝐡(πœƒ),sin where 𝑙 is the conductor’s length, 𝑣 is its velocity, 𝐡 is the magnetic field strength, and πœƒ is the angle between 𝐡 and 𝑣.

We are to compare this magnitude with the magnitude of the emf induced in the scenario shown by diagram IV. In this situation, positive and negative charges accumulate as shown below.

The polarity of the induced potential difference is opposite that of the scenario in diagram III. Note, however, that none of the values influencing the induced potential difference magnitude—𝑙, 𝑣, 𝐡, and πœƒβ€”have changed.

Therefore, the magnitude of potential difference induced along the length of the rod is the same in the scenarios depicted in diagrams III and IV.

Part 4

We have seen that diagram I shows a situation in which induced potential difference between the ends of the rod is zero.

In contrast, a nonzero potential difference is induced across the rod shown in diagram III. Therefore, these magnitudes are not equal.

As a conductor moves through a uniform magnetic field, its motion may be periodic rather than constant.

Consider the conductor shown below, positioned in a uniform magnetic field.

We may change our viewing perspective, observe the conductor from one of its ends, and see that the conductor moves in circles, as shown below.

As the conductor moves, the angle between the conductor’s velocity vector and the external magnetic field changes.

At the locations marked 1, 2, 3, and 4, that angle πœƒ, measured from the magnetic field vector to the conductor’s velocity vector, has values of 90∘, 0∘, 270∘, and 180∘, as shown below.

Recalling that the electromotive force induced in a conductor is given by πœ–=𝑙𝑣𝐡(πœƒ),sin we note that this quantity depends on the sine of the angle πœƒ. Therefore, the emf induced in this conductor moving in circles follows a sinusoidal pattern, as shown in the following figure.

Example 4: Varying Potential Difference Over Time in a Straight Conductor Moving through a Uniform Magnetic Field

A conducting rod that is within a uniform magnetic field moves at a constant speed along a circular path, where the direction of the circular motion is perpendicular to the length of the rod throughout the motion. When the rod is at the positions A and C, shown in the diagram, the direction of the circular motion is along the line of the direction of the magnetic field. When the rod is at the positions B and D, shown in the diagram, the direction of the circular motion is perpendicular to the line of the direction of the magnetic field. The graph shows lines of four different colors. Each line could represent the change in the potential difference across the length of the rod as it moves from A to B to C to D and back to A. Which color line correctly represents how the potential difference varies with time?

  1. The blue line
  2. The orange line
  3. The red line
  4. The green line
  5. None of the answers are correct.

Answer

Studying the diagram of the conductor moving through the field, we see it follows a circular path, in a plane parallel to the magnetic field.

If πœƒ is the angle between the field and the velocity vector of the conductor, πœƒ changes constantly. Specifically, at the points marked A, B, C, and D, πœƒ has corresponding values of 0∘, 270∘, 180∘, and 90∘, as shown below.

In general, the potential difference induced in a straight conductor moving through a uniform magnetic field is described by the equation πœ–=𝑙𝑣𝐡(πœƒ).sin

In our scenario, πœƒ varies, and this equation indicates that it does so sinusoidally.

We can conclude that the potential difference induced in the conductor is not constant over time and also that the conductor’s induced potential difference varies according to the shape of the sine function.

Reviewing our answers, answer Cβ€”the red curveβ€”shows potential difference varying with time, as it would in our moving conductor.

Key Points

  • For a straight conductor moving through a uniform magnetic field, an electromotive force (emf), which is equivalent to potential difference, is induced according to the equation πœ–=𝑙𝑣𝐡(πœƒ)sin, where πœ– is the induced emf, 𝑙 is the length of the conductor, 𝑣 is its velocity, 𝐡 is the strength of the external magnetic field, and πœƒ is the angle between 𝑣 and 𝐡.
  • When a straight conductor moving through a uniform magnetic field is part of a closed electric circuit, the conductor causes charge to flow through the circuit. The potential difference πœ–, circuit current 𝐼, and circuit resistance 𝑅 are related by Ohm’s law: πœ–=𝐼𝑅.
  • When a conductor moves periodically through a uniform magnetic field, πœƒ may change constantly, causing the emf induced to follow a sinusoidal pattern.

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