D2.5 Investigate particle behaviour in a field

Ontario Grade 12 Physics

Gravitational, Electric, and Magnetic Fields

Electric and magnetic fields, forces, and particle paths

A field describes how an object can experience a force in a region of space. In SPH3U, you studied forces and motion: a net force can change an object’s velocity. Here, apply that idea to charged particles in electric and magnetic fields. A particle’s path depends on its charge, its velocity, and the field’s direction. A vector has both magnitude and direction; a scalar has magnitude only. Force, velocity, and acceleration are vectors. Charge, mass, and speed are scalars.

For each situation, define the system as the particle being studied. Use a reference frame fixed to the laboratory or diagram, and state the positive directions before assigning signs. A reference frame is the viewpoint and coordinate system used to describe motion. In the examples, right is positive

x

, up is positive

y

, and out of the page is positive

z

. Field arrows and particle paths are described from that frame.

What you will learn

Describe how an electric or magnetic field can change a charged particle’s motion.
Use force and motion relationships to predict a particle’s acceleration or path.
Plan an investigation that distinguishes predictions from measured or simulated evidence.

1. Connect fields to forces and motion

An electric field is a region where a charged particle experiences an electric force. Its direction is defined as the direction of the force on a positive test charge. A negative particle feels a force opposite the field. Charge is measured in coulombs (

\mathrm{C}

), and electric field strength in newtons per coulomb (

\mathrm{N/C}

).

A magnetic field is a region where a moving charged particle can experience a magnetic force. Magnetic field strength is measured in teslas (

\mathrm{T}

). A particle at rest has no magnetic force from its motion through a magnetic field. A particle moving parallel or anti-parallel to the field also has no magnetic force. The force is greatest when the particle moves perpendicular to the field.

In either case, use Newton’s second law: the net force determines acceleration. Acceleration is the change in velocity per time and is measured in metres per second squared (

\mathrm{m/s^2}

). A force can change speed, direction, or both. In a uniform electric field, the force on a particle is constant if its charge and the field do not change. In a uniform magnetic field, the force may continually change direction as the particle moves.

The electric force is proportional to charge and field strength. The magnetic force depends on charge, speed, field strength, and the angle between velocity and field. The sign of charge determines the force direction; use the right-hand rule for a positive charge in a magnetic field, then reverse the direction for a negative charge.

\vec F_E=q\vec E\qquad F_B=|q|vB\sin\theta\qquad \vec F_{\mathrm{net}}=m\vec a

Electric force acts on a charged particle, including one at rest.
Magnetic force requires motion with a component perpendicular to the field.
Always include direction when reporting a vector quantity.

2. Read field and motion diagrams

A field diagram uses arrows to show field direction. In a uniform field, the arrows are parallel and equally spaced. A particle diagram should also show the particle’s velocity and the force acting on it. Do not treat a field arrow as the particle’s path: the field describes the force direction for a positive test charge, not the direction the particle must travel.

For an electric field pointing right, a positive particle feels force right and a negative particle feels force left. If the force is along the particle’s motion, its speed changes. If the force is perpendicular to its motion, its velocity direction begins to change.

For a magnetic field, determine the force direction from the particle’s velocity and the field. For a positive particle, point the fingers of your right hand along velocity and curl them toward the field; your thumb gives the force direction for a perpendicular case. For a negative particle, reverse that result. When the velocity is perpendicular to a uniform magnetic field, the magnetic force stays perpendicular to the velocity. It changes direction of motion without changing speed, so the path is circular. If the velocity has both parallel and perpendicular components, the particle follows a helical path; this lesson’s calculations focus on the perpendicular case.

In an investigation, keep the frame, positive directions, particle properties, and field conditions clear. A prediction from a model is not a measurement. If you use a simulation, identify its output as simulated evidence. If you collect data with equipment, describe what was actually measured and report uncertainty where appropriate.

r=\frac{mv}{|q|B}

Draw separate arrows for the field, velocity, and force.
A negative charge reverses the electric or magnetic force direction relative to a positive charge.
A circular path in a magnetic field requires velocity perpendicular to the field.

3. Plan an investigation of particle behaviour

A useful investigation changes one factor at a time and observes how the path responds. For example, a proposed simulation could display a charged particle entering a uniform magnetic field perpendicular to its initial velocity. Keep mass, charge, speed, and field strength fixed while comparing the predicted path for positive and negative charge. Then vary one quantity, such as field strength, and record the simulated path radius. Do not describe such a procedure as a completed laboratory experiment unless it was actually carried out.

For a physical investigation, use only approved equipment and follow its safety instructions. A possible evidence record would include the field setting, particle conditions provided by the apparatus, and an observed path or measured radius. State whether each value came from a direct measurement, an instrument display, or a model. Compare observations with the model’s prediction, and identify limitations such as uncertainty in locating the path edge.

The magnetic-force relationship predicts that, for fixed mass, speed, and charge magnitude, increasing the field strength increases the force and reduces the radius. The electric-field relationship predicts that reversing the sign of charge reverses the electric force direction. These are model-based predictions; observations are needed to test them.

F_B=\frac{mv^2}{r}

State what you will change, what you will observe, and what you will keep fixed.
Separate measured evidence from predicted or simulated results.
Use the model to make a prediction before comparing it with evidence.

Worked example

1. Electric force and acceleration

A particle with charge

+2.0\times10^{-6}\,\mathrm{C}

and mass

4.0\times10^{-6}\,\mathrm{kg}

is in a uniform electric field of

3.0\times10^3\,\mathrm{N/C}

directed right. Find its acceleration.

Set the frame and direction
The system is the particle in the laboratory frame. Let right be positive. The particle is positively charged, so its electric force points with the field.
Find the force
Use the electric-force relationship. Its units are coulombs times newtons per coulomb, which gives newtons.
$F_E=qE=(2.0\times10^{-6}\,\mathrm{C})(3.0\times10^3\,\mathrm{N/C})=6.0\times10^{-3}\,\mathrm{N}$
Find acceleration
Apply Newton’s second law. The force and acceleration are both directed right, so the acceleration is positive.
$a=\frac{F_E}{m}=\frac{6.0\times10^{-3}\,\mathrm{N}}{4.0\times10^{-6}\,\mathrm{kg}}=1.5\times10^3\,\mathrm{m/s^2}$

Answer: The particle’s acceleration is

1.5\times10^3\,\mathrm{m/s^2}

to the right.

Check: The units reduce to

\mathrm{N/kg= m/s^2}

. A positive charge accelerates with the field, as expected.

Worked example

2. Magnetic force direction and magnitude

A particle with charge

-3.0\times10^{-6}\,\mathrm{C}

moves right at

2.0\times10^3\,\mathrm{m/s}

. A uniform magnetic field of

0.40\,\mathrm{T}

points out of the page. Find the magnetic-force magnitude and direction.

Set the frame and directions
The system is the particle in the laboratory frame. Right is positive $x$ , up is positive $y$ , and out of the page is positive $z$ . The velocity is perpendicular to the field.
Calculate the magnitude
Use the magnetic-force relationship with an angle of $90^\circ$ . The magnitude uses the absolute value of charge; the sign is used to determine direction.
$F_B=|q|vB\sin\theta=(3.0\times10^{-6}\,\mathrm{C})(2.0\times10^3\,\mathrm{m/s})(0.40\,\mathrm{T})\sin90^\circ=2.4\times10^{-3}\,\mathrm{N}$
Determine the direction
For a positive charge moving right in a field out of the page, the right-hand rule gives a force down. This particle is negative, so reverse that direction: the force is up.

Answer: The magnetic force is

2.4\times10^{-3}\,\mathrm{N}

upward.

Check: The units

\mathrm{C\,m/s\,T}

reduce to newtons. The force is perpendicular to both velocity and field, and its direction is reversed for negative charge.

Worked example

3. Radius of a magnetic path

A particle of mass

1.2\times10^{-6}\,\mathrm{kg}

and charge magnitude

2.0\times10^{-6}\,\mathrm{C}

enters a uniform

0.30\,\mathrm{T}

magnetic field at

4.0\times10^2\,\mathrm{m/s}

, perpendicular to the field. Find the path radius.

Define the system and unknown
The system is the particle in the laboratory frame. The field is uniform, and the initial velocity is perpendicular to it. Find the radius; the path curves in the direction given by the charge sign and right-hand rule.
Relate magnetic force to circular motion
For this perpendicular case, magnetic force provides the inward force that changes the velocity direction. Equate the magnetic-force magnitude to $mv^2/r$ and solve for radius.
$r=\frac{mv}{|q|B}$
Substitute with SI units
Use the given mass, speed, charge magnitude, and field strength. The radius is positive because it is a distance; the charge sign affects the curve’s direction, not its radius.
$r=\frac{(1.2\times10^{-6}\,\mathrm{kg})(4.0\times10^2\,\mathrm{m/s})}{(2.0\times10^{-6}\,\mathrm{C})(0.30\,\mathrm{T})}=8.0\times10^2\,\mathrm{m}$

Answer: The path radius is

8.0\times10^2\,\mathrm{m}

, or

800\,\mathrm{m}

.

Check: The units reduce to metres. A larger speed would produce a larger radius, while a stronger field would produce a smaller radius, consistent with the relationship.

Common mistakes and how to avoid them

Assuming the magnetic force points in the direction of motion.

Correction: For perpendicular velocity and field, magnetic force is perpendicular to both. It changes the velocity direction, not the speed.

Using a right-hand rule result for a negative particle without changing the direction.

Correction: Find the direction for a positive charge first, then reverse it for a negative charge.

Calling a simulated path a measured result.

Correction: Label it as simulated evidence. Reserve measured evidence for quantities actually obtained using measurement equipment.

Reporting a field or force direction without stating the chosen axes.

Correction: Define the reference frame and positive directions before interpreting vector signs.

Lesson summary

Electric fields exert force on charged particles, including particles at rest.
Magnetic forces act on moving charged particles when their velocity has a component perpendicular to the field.
Use charge sign to determine force direction; use charge magnitude when calculating magnetic-force magnitude or path radius.
A uniform magnetic field bends a perpendicular particle path into a circle without changing the particle’s speed.
An investigation compares a stated model prediction with clearly identified measured or simulated evidence.

Check your understanding

Question 1

A negative particle is at rest in a uniform magnetic field. What is the magnetic force due to its motion through that field?

Zero
Directed along the field
Opposite the field
It cannot be determined without the particle’s mass

Show answer and explanation

Zero

Magnetic force depends on particle velocity. At rest, the particle has no magnetic force from its motion through the field.

Question 2

A positive particle moves right through an electric field directed left. Which way is its electric force?

Right
Left
Up
Zero

Show answer and explanation

Left

The electric force on a positive charge is in the direction of the electric field, which is left.

Question 3

A particle’s velocity is perpendicular to a uniform magnetic field. If its speed increases while the other quantities remain fixed, what happens to the circular-path radius?

It increases
It decreases
It stays the same
It becomes zero

Show answer and explanation

It increases

The radius is proportional to speed when mass, charge magnitude, and field strength are fixed.

Key terms

Electric field: A region where a charged particle experiences electric force; its direction is the force direction on a positive test charge.
Magnetic field: A region where a moving charged particle can experience magnetic force.
Uniform field: A field with constant strength and direction throughout the region considered.
Reference frame: The viewpoint and coordinate system used to describe positions and motion.
Perpendicular: At a right angle; the angle between the directions is $90^\circ$ .
Simulated evidence: Output produced by a model or computer simulation, rather than direct physical measurement.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation D2.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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D2.5 · Investigate particle behaviour in a field

Ontario Grade 12 Physics

Electric and magnetic fields, forces, and particle paths

What you will learn

1. Connect fields to forces and motion

2. Read field and motion diagrams

3. Plan an investigation of particle behaviour

1. Electric force and acceleration

2. Magnetic force direction and magnitude

3. Radius of a magnetic path

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH4U

About this lesson and its review