D2.4 Solve magnetic-force problems for moving charges and currents

Ontario Grade 12 Physics

Gravitational, Electric, and Magnetic Fields

SPH4U D2.4 | Calculate magnetic-force magnitude and direction

In SPH3U, you learned that a force can change an object's motion and that magnetic fields can exert forces. A magnetic force is a vector: it has both magnitude and direction. In this lesson, the physical system will be either one moving charged particle or a straight section of current-carrying wire. We use the reference frame of the lab, with the usual right-handed axes:

+x

to the right,

+y

upward, and

+z

out of the page. The opposite direction, into the page, is

-z

. These axes let us describe force directions consistently.

What you will learn

Use the magnetic-force relationship for a moving charged particle.
Use the magnetic-force relationship for a current-carrying wire.
Determine how the angle between motion or current and the magnetic field affects force.
Find the force direction using a right-hand rule and account for negative charge.

1. The model for a moving charge

A charged particle is an object with electric charge, measured in coulombs (C). Its velocity is its speed and direction of motion, measured in metres per second (m/s). A magnetic field is described by the vector

\vec B

and measured in teslas (T).

For a charge moving through a magnetic field, the force magnitude depends on the charge, speed, field strength, and angle between the velocity and the field. The angle is measured from the direction of

\vec v

to the direction of

\vec B

. If the particle moves along the field, the magnetic force is zero. If it moves at right angles to the field, the force is greatest for those values of charge, speed, and field.

The force direction is perpendicular to both the particle's velocity and the magnetic field. For a positive charge, point your right-hand fingers along velocity and turn them toward the field through the smaller angle. Your thumb gives the force direction. For a negative charge, reverse that direction. The right-hand rule gives direction; the equation gives magnitude.

A dot symbol,

\odot

, represents a vector coming out of the page toward you. A cross symbol,

\otimes

, represents a vector going into the page away from you. These symbols are useful when the three directions cannot all be drawn on a flat page.

F_B=|q|vB\sin\theta

The moving-charge relationship is $F_B=|q|vB\sin\theta$ .
Charge is measured in C, speed in m/s, field in T, and force in N.
The force is zero for motion parallel or opposite to the field, and greatest for perpendicular motion.
Reverse the right-hand-rule direction for a negative charge.

2. The model for a current-carrying wire

Electric current is the rate at which charge passes a point. Current is measured in amperes (A). In a wire, conventional current is the direction positive charge would move. Use this direction in the magnetic-force rule, even though electrons in a metal move in the opposite direction.

For a straight wire segment in a uniform magnetic field, the force depends on current, the length of wire in the field, field strength, and the angle between the current direction and the field. The wire force is perpendicular to both current direction and field. Use your right hand with your fingers along conventional current, turning them toward the field; your thumb points along the force.

The length

L

is the length of the wire segment within the magnetic field, in metres. If the wire is parallel to the field, the force is zero. If it is perpendicular, the force is greatest for the stated current, length, and field. The relationship applies to the straight segment described in the problem.

For either type of problem, first identify the system, choose directions, and record known quantities and the unknown. Find the angle before substituting. Then use the direction rule and report both force magnitude and direction.

F_B=ILB\sin\theta

For a straight wire segment, use $F_B=ILB\sin\theta$ .
Use conventional current to determine wire-force direction.
The wire segment length must be in metres; current is in amperes.
A force answer needs a magnitude and a direction.

3. A reliable problem-solving method

Draw or describe the directions of motion or current and the magnetic field. State the reference frame and the positive directions being used. For example, specify whether each vector points right, upward, into the page, or out of the page. This prevents a correct magnitude from being paired with an incorrect direction.

Use the angle between the two relevant vectors, not an angle measured from an unrelated axis. If a diagram gives the complementary angle, convert it to the angle between velocity or current and field before using the sine. A component parallel to the field does not contribute to magnetic force; the sine factor accounts for this.

Check units after substituting. For a charge, the tesla unit is equivalent to a newton per ampere-metre, and an ampere is a coulomb per second. For a wire, amperes multiplied by metres and teslas also give newtons. Round to a sensible number of significant figures based on the given values.

Finally, check the limiting cases and the direction. A result should be zero when the relevant motion or current is parallel to the field. A perpendicular arrangement should give the largest force for the same other values. If reversing a positive charge's velocity or the current reverses the force direction, the answer is consistent with the direction rule.

Set the system, frame, directions, known values, and unknown before calculating.
Use the angle between the motion or current and the field.
Check units, significant figures, direction, and whether the result fits the arrangement.

Worked example

Force on a positive moving charge

A positive particle with charge

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

moves at

3.0\times10^4\,\mathrm{m/s}

to the right through a

0.40\,\mathrm{T}

magnetic field directed upward. Find the magnetic force.

Set the system and directions
The system is the charged particle in the lab frame. Take right as $+x$ , upward as $+y$ , and out of the page as $+z$ . The velocity is $+x$ and the field is $+y$ , so the angle is $90^\circ$ . The unknown is the force vector.
Calculate the magnitude
Use the moving-charge relationship. The sine of $90^\circ$ is $1$ , so the force magnitude is the product of the charge magnitude, speed, and field strength.
$F_B=(2.0\times10^{-6}\,\mathrm{C})(3.0\times10^4\,\mathrm{m/s})(0.40\,\mathrm{T})\sin 90^\circ=2.4\times10^{-2}\,\mathrm{N}$
Determine direction and check
For a positive charge, the right-hand rule gives $+z$ : out of the page. The units reduce to newtons. The result is reasonable because the motion is perpendicular to the field, so the force is at its maximum for these values.

Answer: The magnetic force is

2.4\times10^{-2}\,\mathrm{N}

out of the page.

Check: The force is perpendicular to both rightward motion and the upward field. Its units are newtons, and the three significant-figure inputs support a two-significant-figure result.

Worked example

Force on a negative moving charge

An electron has charge magnitude

1.60\times10^{-19}\,\mathrm{C}

. It moves at

2.5\times10^6\,\mathrm{m/s}

to the right in a

0.30\,\mathrm{T}

field directed out of the page. Find the force magnitude and direction.

Set the system and directions
The system is the electron in the lab frame. Use right as $+x$ , upward as $+y$ , and out of the page as $+z$ . The angle between velocity and field is $90^\circ$ . The unknown is the force vector.
Calculate the magnitude
The force-magnitude equation uses the absolute value of charge, so use the stated charge magnitude. Since the angle is $90^\circ$ , the sine factor is $1$ .
$F_B=(1.60\times10^{-19}\,\mathrm{C})(2.5\times10^6\,\mathrm{m/s})(0.30\,\mathrm{T})\sin 90^\circ=1.2\times10^{-13}\,\mathrm{N}$
Reverse the direction for negative charge
For a positive charge moving right with field out of the page, the right-hand rule points downward, or $-y$ . An electron has negative charge, so reverse this direction. Its force is upward, or $+y$ .

Answer: The electron experiences a force of

1.2\times10^{-13}\,\mathrm{N}

upward.

Check: The magnitude has units of newtons. Reversing the charge sign reverses the force direction, not its magnitude.

Worked example

Force on a current-carrying wire

A straight wire segment of length

0.25\,\mathrm{m}

carries a conventional current of

4.0\,\mathrm{A}

to the right. It lies in a

0.60\,\mathrm{T}

field directed into the page. Find the magnetic force on the segment.

Set the system and directions
The system is the wire segment in the lab frame. Take right as $+x$ , upward as $+y$ , and out of the page as $+z$ . Current is along $+x$ , and the field is along $-z$ , so they are perpendicular. The unknown is the force on the segment.
Calculate the magnitude
Use the wire relationship with the segment length in metres. The angle between current and field is $90^\circ$ , making the sine factor equal to $1$ .
$F_B=(4.0\,\mathrm{A})(0.25\,\mathrm{m})(0.60\,\mathrm{T})\sin 90^\circ=0.60\,\mathrm{N}$
Determine direction and check
Point your right-hand fingers along the conventional current to the right and turn them toward the field into the page. Your thumb points upward. The units give newtons, and the perpendicular arrangement supports a maximum force for these values.

Answer: The force on the wire segment is

0.60\,\mathrm{N}

upward.

Check: The force is perpendicular to both current and field. The answer has two significant figures, matching the input values.

Common mistakes and how to avoid them

Using the particle's charge sign to make the force magnitude negative.

Correction: Use |q| in the magnitude equation. Use the charge sign only to decide whether to keep or reverse the right-hand-rule direction.

Using the angle from the page or from an axis instead of the angle between motion or current and the field.

Correction: Identify the two vectors in the equation and find the angle between them. Use that angle in the sine.

Using electron-flow direction to find the force on a wire.

Correction: Use conventional current direction in the wire relationship and its right-hand rule.

Reporting only a force magnitude.

Correction: A force is a vector. Report its magnitude and direction, including whether it points into or out of the page when needed.

Lesson summary

A moving charge experiences magnetic force with magnitude $F_B=|q|vB\sin\theta$ .
A straight current-carrying wire segment experiences magnetic force with magnitude $F_B=ILB\sin\theta$ .
For positive charge or conventional current, use the right-hand rule for direction. Reverse the result for negative charge.
Check the vector angle, SI units, significant figures, and physical reasonableness.

Check your understanding

Question 1

A positive charge moves parallel to a magnetic field. What is the magnetic-force magnitude?

Zero
The product of charge, speed, and field strength
The charge multiplied by the field strength, regardless of speed
correctIndex

Show answer and explanation

Zero

Parallel motion gives an angle of

0^\circ

, and

\sin 0^\circ=0

, so the magnetic force is zero.

Question 2

A wire carries conventional current to the right while the magnetic field points out of the page. What is the force direction?

Upward
Downward
Into the page
correctIndex

Show answer and explanation

Downward

The right-hand rule for current to the right and field out of the page gives a force downward.

Question 3

A negative particle moves right while the field points upward. Which direction is its magnetic force?

Out of the page
Into the page
Upward
correctIndex

Show answer and explanation

Into the page

For a positive charge, right crossed with upward gives out of the page. A negative charge reverses that direction, so its force is into the page.

Key terms

Magnetic field: A vector description of the magnetic influence in a region. Its SI unit is the tesla.
Conventional current: The direction positive charge would move in a circuit or wire.
Vector: A quantity with both magnitude and direction, such as velocity or force.
Scalar: A quantity with magnitude but no direction, such as speed or charge magnitude.
Right-hand rule: A hand method used to find the direction of magnetic force from the directions of motion or current and magnetic field.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation D2.4. 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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D2.4 · Solve magnetic-force problems for moving charges and currents

Ontario Grade 12 Physics

SPH4U D2.4 | Calculate magnetic-force magnitude and direction

What you will learn

1. The model for a moving charge

2. The model for a current-carrying wire

3. A reliable problem-solving method

Force on a positive moving charge

Force on a negative moving charge

Force on a current-carrying wire

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