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Charges in Magnetic Fields

Original source retained for reference; this draft does not establish AP coverage.

Original author draft. Original wording and recorded answer states are preserved without claiming reviewed scientific correctness or AP alignment.

Physics 2 original textbook · Chapter 5

Section Overview

  1. Magnetic Force
  2. Example 5.2.1
  3. Charges in Magnetic Fields
  4. Cyclotron Motion
  5. Example 5.2.2
  6. The Lorentz Force
  7. Section Review

For motion perpendicular to a uniform magnetic field, the force magnitude is F=∣q∣v⊥BF=|q|v_\perp B and the nonrelativistic circular radius is R=mv⊥/(∣q∣B)R=mv_\perp/(|q|B). A parallel velocity component produces helical motion.

Charges in Magnetic Fields

Suppose we somehow induced a uniform magnetic field and then we shot a charged object through it. What would you expect to happen? It is perfectly conceivable that the object would move straight through the field without any interference, and it does in certain circumstances, but what is truly interesting is that in other circumstances the path of the charged object will actually curve.

Since this object is changing it’s path we know that there must be some force acting on it and after careful experimentation we can figure out exactly what equation describes this force. Unlike gravitation this force does not depend upon the mass of the object but instead on the objects charge, the objects velocity, and the orientation of the magnetic field which the object is placed in. If we use B to denote the magnetic field, let v be the velocity of the object, and use Q as the quantity of charge that the object has then the force F on that object is given by

Interactive 1.1 What the Cross Product means

Interactive 1.1 What the Cross Product means
Original diagram
Parallel

Once the magnetic field and the particles velocity are parallel (or anti-parallel) there ceases to be a force.

Closing the Gap

Once the velocity of the particle and the magnetic field cease to be perpendicular the magnitude of force on the particle begins to decrease.

Perpendicular Vectors

When the magnetic field and velocity are perpendicular the force on a moving charged particle is the strongest.

Example 5.2.1

Optional arbitrary-angle particle-force extension. For positive v and b, the stated velocity makes a 45° angle with the field. Prerequisites: vector components and a cross product; AP quantitative particle-force tasks are limited to 0°, 90° and 180°. This preserved source example makes no AP coverage claim.

A particle of charge Q has v=(0,v,v) in B=(0,b,0). Find instantaneous force.

Authored clarification: No electric force is specified.

This problem requires nothing more from us than to use our magnetic force law to calculate the force on the particle using the given parameters and our understanding of the mathematics involved.

The first thing we must do is setup the necessary matrix. For our problem we get that

Using an augmented matrix to calculate the determinant we get the required result

As a result of the magnetic force law it is important to realize that the sign of the charge which is moving through a magnetic field will effect the path it takes. In fact since the charge scales the entire equation if two charges have the same magnitude of charge, the same velocity, and are in the same magnetic field if the signs of the charge are opposite the forces acting on those charged particles will be opposite.

Gallery 1.3 Charges in a Magnetic Field

Gallery 1.3 Charges in a Magnetic Field
Original frame 1

A rectangular tube is permeated by a magnetic field pointing upwards.

Original image
Gallery 1.3 Charges in a Magnetic Field
Original frame 2

As charges particles move into box depending upon their charge they are either pulled to the left of right by the magnetic field. How far forward they get before hitting the side depends on their mass and velocity.

Original image

Using this idea it is possible to create a charge selector using only a magnetic field. We start by creating some type of rectangular tube and inside induce a uniform magnetic field which points in a single direction and is perpendicular to one side of the tube. When charged particles are shot through the tube they will be deflected one of two ways depending upon the orientation of the magnetic field and the charge of the particle.

Interactive 1.2 Cyclotron Motion

Interactive 1.2 Cyclotron Motion
Original diagram
The Magnetic Field

The magnetic field here points perpendicular to the plane of the circle.

The Magnetic Force

As the particle moves along the circular path the magnetic force remains pointing toward the center of the circle. As a result it is a centripetal acceleration which forces the particle into circular motion.

The Velocity

THe magnitude of the particles velocity stays fixed. THe direction of this velocity does move though as a result of the magnetic field.

Perhaps the most iconic type of motion you can get in a magnetic field is what we call cyclotron motion. This type of motion can be derived by carefully considering the fact that in a uniform magnetic field a particle with an initial velocity perpendicular to the magnetic field will experience a force of constant magnitude which always pints at a particular point. This force creates a centripetal acceleration which confines the particle to circular motion. Using newtons laws we can find an equation which describes this motion.

Using newtons laws and realizing that the force and velocity vectors of our particle at separated by 90 degrees we get that

∣q∣vB=mv2/R,v⊥B⃗|q|vB=mv^2/R,\quad v\perp\vec B

Which becomes

∣q∣vB=mv2/R,R=mv/(∣q∣B)|q|vB=mv^2/R,\quad R=mv/(|q|B)

after making the appropriate substitutions for a realizing that it is a centripetal acceleration.

Example 5.2.2

A proton and electron have equal speeds perpendicular to the same uniform B. Compare their circular radii.

Authored clarification: Nonrelativistic speeds and no other forces.

Reviewed worked solution:

  1. Equate force magnitude to centripetal requirement.

    ∣q∣vB=mv2/R|q|vB=mv^2/R

  2. Solve for radius.

    R=mv/(∣q∣B)R=mv/(|q|B)

  3. Use equal speed and equal charge magnitudes.

    Rp/Re=mp/me≈1836R_p/R_e=m_p/m_e\approx1836

Answer: The proton radius is about 1836 times the electron radius; their turn directions are opposite.

Check: Radius must be positive for both signs of charge.

To end this section we present what is commonly called the Lorentz Force Law. As you may recall from the beginning of this book a charge experiences a force when placed in an electric field just as when placed in a magnetic field. A particle in both an electric and magnetic field will actually experience force from both fields. When these forces are combined we get the Lorentz Force Law

Review 1.2 Charges in Magnetic Fields

This review preserves the author’s recorded prompts, choices, and answer states. Corrected items have separate reviewed explanations and keys; unchanged draft keys are marked unreviewed. Unfinished items are not scored.

Original question 1

For a particle with charge q and velocity v, the Lorentz force due to electric and magnetic fields is F⃗=q(E⃗+v⃗×B⃗)\vec F=q(\vec E+\vec v\times\vec B).

  1. True

  2. False

Reviewed answer and reasoning

A

Electric and magnetic contributions add vectorially. Magnetic force alone is perpendicular to v; electric force can do work.

This reviewed version states the assumptions or convention needed for one determinate answer. The archived source prompt and key are retained.

Archived original question and recorded key · contains the source error

The Lorentz Force is the combined force on a particle due to both the magnetic and electric fields in a particular area.

  1. True

  2. False

Original recorded key: A

Original question 2

Which describes nonrelativistic circular cyclotron motion in a uniform static magnetic field?

  1. A charged particle with velocity perpendicular to B and no other forces moves in a circle

  2. Every charged particle in a magnetic field moves in a circle

  3. A particle moving parallel to B moves in a circle

Reviewed answer and reasoning

A

A nonzero parallel velocity adds a constant axial component, producing a helix; fully parallel motion is straight.

Authored correction or explicit assumptions; this reviewed key is separate from the archived original recorded states.

Archived original question and recorded key · contains the source error

What is Cyclotron Motion?

  1. The waving motion a moving charge undergoes when placed inside an oscillating magnetic field

  2. The circular motion a moving charge undergoes when placed in a uniform magnetic field.

  3. The motion a moving charge undergoes when it is parallel with the magnetic field.

Original recorded key: B

Original question 3

A particle of +1 C has velocity (1 m/s,0,0) in B=(2 T,1 T,0), with no electric field. Find the magnetic force magnitude.

  1. 1 N

  2. 2 N

  3. 0 N

Reviewed answer and reasoning

A

F⃗=qv⃗×B⃗=(0,0,1) N\vec F=q\vec v\times\vec B=(0,0,1)\,\mathrm N. The parallel Bₓ component produces no magnetic force.

This reviewed version states the assumptions or convention needed for one determinate answer. The archived source prompt and key are retained.

Archived original question and recorded key · contains the source error

A charge of 1 C is moving in a magnetic field given by (2 T, 1 T, 0) with a velocity of (1 m/s, 0, 0). What magnitude of force does the charge experience?

  1. 0.125 N

  2. 0.5 N

  3. 1 N

  4. 2 N

  5. None of the above

Original recorded key: C

Original question 4

A charged particle of 5 C and a mass of 2 kg moves with velocity (3 m/s, 0, 4 m/s) in a magnetic field (0, 2 T, 0). What will be the radius of the particles motion?

  1. 1 m

  2. 2 m

  3. 3 m

  4. 4 m

  5. None of the above

Reviewed answer and reasoning

A

The perpendicular speed is 5 m/s, so R=2·5/(5·2)=1 m.

Reviewed correction. Archived source choices and key remain readable.

Archived original question and recorded key · contains the source error

A charged particle of 5 C and a mass of 2 kg moves with velocity (3 m/s, 0, 4 m/s) in a magnetic field (0, 2 T, 0). What will be the radius of the particles motion?

  1. 1 m

  2. 2 m

  3. 3 m

  4. 4 m

  5. None of the above

Original recorded key: E

Original question 5

A charge is shot away from you on a horizontal table and moved left. If the charge was positive what direction will the magnetic field be pointing?

  1. Up

  2. Down

  3. Left

  4. Right

  5. None of the above

Reviewed answer and reasoning

B

Using right=+x, away=+y, up=+z: for positive q, +y×(−z)=−x. The field is down.

Reviewed correction. Archived source choices and key remain readable.

Archived original question and recorded key · contains the source error

A charge is shot away from you on a horizontal table and moved left. If the charge was positive what direction will the magnetic field be pointing?

  1. Up

  2. Down

  3. Left

  4. Right

  5. None of the above

Original recorded key: A

Reviewed worked examples

These authored derivations are paired with the original example identities. Added assumptions are stated.

Optional arbitrary-angle particle-force extension. For positive v and b, the stated velocity makes a 45° angle with the field. Prerequisites: vector components and a cross product; AP quantitative particle-force tasks are limited to 0°, 90° and 180°. This preserved source example makes no AP coverage claim.

Example 5.2.1 · reviewed solution

A particle of charge Q has v=(0,v,v) in B=(0,b,0). Find instantaneous force.

Authored clarification: No electric force is specified.

  1. Use magnetic force law.

    F⃗=Qv⃗×B⃗\vec F=Q\vec v\times\vec B

  2. Compute the only nonzero cross-product component.

    (v⃗×B⃗)x=0−vb=−vb(\vec v\times\vec B)_x=0-vb=-vb

  3. Multiply by signed charge.

    F⃗=(−Qvb,0,0)\vec F=(-Qvb,0,0)

F⃗=−Qvbx^\vec F=-Qvb\hat x

Check

F·v=0, so magnetic force does no work.

Example 5.2.2 · reviewed solution

A proton and electron have equal speeds perpendicular to the same uniform B. Compare their circular radii.

Authored clarification: Nonrelativistic speeds and no other forces.

  1. Equate force magnitude to centripetal requirement.

    ∣q∣vB=mv2/R|q|vB=mv^2/R

  2. Solve for radius.

    R=mv/(∣q∣B)R=mv/(|q|B)

  3. Use equal speed and equal charge magnitudes.

    Rp/Re=mp/me≈1836R_p/R_e=m_p/m_e\approx1836

The proton radius is about 1836 times the electron radius; their turn directions are opposite.

Check

Radius must be positive for both signs of charge.

Glossary and deeper reading · 4 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Magnetic field

    A magnetic field assigns a vector direction and strength to each location and predicts interactions with moving charges, currents, and magnetic materials. Its strength is measured in teslas, and the force on a moving charge need not point along the field. For example, a compass aligns with the local field while a moving charge can be deflected perpendicular to it.

    Read more: Magnetic field in detail →

  • Force

    A force describes an interaction that can change an object’s momentum. It is a vector measured in newtons, so both size and direction matter when forces are added. For example, a rope can pull a cart while the floor exerts friction in another direction.

    Read more: Force in detail →

  • Electric charge

    Electric charge is a conserved property that determines electric interactions. It can be positive or negative and is measured in coulombs; ordinary net charges arise from an imbalance of charged particles. For example, transferring electrons between initially neutral objects leaves one negatively charged and the other positively charged.

    Read more: Electric charge in detail →

  • Velocity

    Velocity describes how position changes with time and includes direction. Instantaneous velocity describes one moment, while average velocity is displacement divided by elapsed time. For example, a car turning a corner can change velocity while maintaining the same speed.

    Read more: Velocity in detail →

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