Magnetic Fields - AP Physics 2

SECTION 1

Historical Context & Motivation

The study of magnetism stretches back millennia, beginning with the ancient observation that certain stones—lodestones—could attract iron. For centuries, magnetism remained a curiosity with limited practical application beyond primitive compasses. It was not until the early nineteenth century that a profound connection between electricity and magnetism was uncovered, launching the field of electromagnetism and ultimately unifying two seemingly distinct phenomena into a single theoretical framework.

1820 • Ørsted's Discovery

Hans Christian Ørsted demonstrates that an electric current deflects a nearby compass needle, establishing the first link between electricity and magnetism.

1831 • Faraday's Induction

Michael Faraday discovers electromagnetic induction—changing magnetic fields produce electric currents—and introduces the concept of magnetic field lines.

1865 • Maxwell's Equations

James Clerk Maxwell publishes a unified set of equations describing electric and magnetic fields as components of a single electromagnetic field propagating at the speed of light.

1895 • Lorentz Force Law

Hendrik Lorentz formalizes the force experienced by a charged particle moving through electric and magnetic fields, completing the classical picture.

The central question this lesson addresses is: how do we quantitatively describe the magnetic field created by currents and magnets, and how does that field exert forces on moving charges and current-carrying wires? Answering this question is essential for understanding everything from MRI machines to electric motors.

SECTION 2

Core Principles & Definitions

A magnetic field (symbol B⃗) is a vector field that permeates the space around magnets, current-carrying conductors, and moving charges. Unlike gravitational or electrostatic fields, the magnetic field exerts a force only on charges that are in motion; a stationary charge sitting in a magnetic field experiences no magnetic force. The SI unit of magnetic field strength is the tesla (T), where 1 T = 1 kg·s⁻²·A⁻¹.

Sources of B⃗

Magnetic fields originate from moving electric charges: current-carrying wires, permanent magnets (atomic-scale current loops), and time-varying electric fields.

Vector Nature

B⃗ has both magnitude and direction at every point in space. The direction at any location is defined as the direction a north magnetic pole would point if placed there.

No Magnetic Monopoles

Magnetic field lines always form closed loops—they exit the north pole and enter the south pole. Isolated north or south poles have never been observed in nature.

Superposition

When multiple sources of magnetic field are present, the net field at any point is the vector sum of the individual fields contributed by each source.

SECTION 3

Visualizing Magnetic Fields

The concept of magnetic field lines, pioneered by Faraday, provides a powerful visual tool. Field lines are drawn so that their tangent at any point gives the direction of B⃗, and the density of lines (number per unit area) is proportional to the field's magnitude. The diagram below shows the field around a bar magnet, where lines emerge from the north pole, arc through the surrounding space, and return to the south pole—forming the characteristic closed-loop pattern required by Gauss's law for magnetism.

Field lines emerge from the north pole (blue region) and curve back into the south pole (red region). Closer spacing indicates a stronger field.

ℹ

Right-Hand Rules

All eight red north tips follow external N-left/S-right dipole direction. Compass north-seeking ends align with the local field. Outside the bar magnet, the field points from north toward south.

SECTION 4

Mathematical Framework

Two foundational equations govern how magnetic fields interact with moving charges and current-carrying conductors. The first describes the force on a single moving charge; the second extends that idea to a straight segment of wire carrying current in a uniform external field.

MAGNETIC FORCE ON A MOVING CHARGE

F = qvB sin θ

F = magnitude of the magnetic force (N), q = charge (C), v = speed of the charge (m/s), B = magnetic field strength (T), θ = angle between v⃗ and B⃗. The force direction is given by the right-hand rule and is always perpendicular to both v⃗ and B⃗.

MAGNETIC FORCE ON A CURRENT-CARRYING WIRE

F = BIL sin θ

F = force on the wire (N), B = external magnetic field (T), I = current (A), L = length of the wire segment in the field (m), θ = angle between the current direction and B⃗.

FIELD FROM A LONG STRAIGHT WIRE

B = μ₀I / (2πr)

B = magnetic field magnitude (T), μ₀ = permeability of free space = 4π × 10⁻⁷ T·m/A, I = current (A), r = perpendicular distance from the wire (m). The field forms concentric circles around the wire.

FIELD INSIDE A SOLENOID

B = μ₀nI

n = number of turns per unit length (turns/m), I = current (A). The field inside a long solenoid is approximately uniform and directed along its axis.

A critically important consequence of F = qvB sin θ is that the magnetic force does zero work on a charged particle. Because the force is always perpendicular to the velocity, it changes the direction of the particle's motion but not its speed—hence the kinetic energy remains constant. This is why a charged particle in a uniform magnetic field follows a circular path (when v⃗ ⊥ B⃗), with the magnetic force supplying the centripetal acceleration.

SECTION 5

Motion of Charged Particles in B⃗

When a charged particle enters a region of uniform magnetic field with its velocity perpendicular to B⃗, the resulting magnetic force acts as a centripetal force and the particle traces out a circle. Setting qvB = mv²/r yields the radius of circular motion: r = mv / (qB). Larger mass or faster speed produces a bigger circle; stronger fields or greater charge produce tighter orbits. This principle underlies the operation of mass spectrometers, cyclotrons, and the confinement of plasma in fusion reactors.

A positive charge (yellow) moves clockwise in a uniform B⃗ directed into the page (⊗ symbols). The velocity v⃗ (green) is always tangent to the circle, while the magnetic force F⃗ (pink) always points toward the center, providing centripetal acceleration.

Quantity	Symbol	Effect on Orbit Radius
Mass	m	r ∝ m — heavier particles orbit in larger circles
Speed	v	r ∝ v — faster particles have larger radii
Charge magnitude	\|q\|	r ∝ 1/\|q\| — greater charge means tighter orbit
Field strength	B	r ∝ 1/B — stronger field means tighter orbit

Dependence of circular orbit radius on physical quantities

SECTION 6

Worked Example

Worked example

1
Step 1 — Identify Given Values

A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.60 × 10⁻¹⁹ C) enters a uniform magnetic field B = 0.50 T with a speed v = 3.0 × 10⁶ m/s perpendicular to B⃗. Find (a) the radius of its circular orbit and (b) the magnetic force on the proton.
2
Step 2 — Find the Orbit Radius

Use r = mv / (qB). Substitute: r = (1.67 × 10⁻²⁷)(3.0 × 10⁶) / [(1.60 × 10⁻¹⁹)(0.50)].

r = 6.3 × 10⁻² m ≈ 6.3 cm
3
Step 3 — Find the Magnetic Force

Since v⃗ ⊥ B⃗, sin θ = 1. F = qvB = (1.60 × 10⁻¹⁹)(3.0 × 10⁶)(0.50).

F = 2.4 × 10⁻¹³ N
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Step 4 — Verify Consistency

As a check, confirm that F = mv²/r: (1.67 × 10⁻²⁷)(3.0 × 10⁶)² / (6.3 × 10⁻²) = 2.4 × 10⁻¹³ N. This matches, confirming the magnetic force serves as the centripetal force. Note that the proton's kinetic energy is unchanged because F⃗ ⊥ v⃗.

SECTION 7

Applications & Limitations

The equations developed in previous sections assume ideal conditions—uniform fields, point charges, or infinitely long wires. Real-world applications require awareness of when these models break down and how additional effects (such as relativistic speeds, non-uniform field regions, or magnetic materials) modify the picture.

Application	How It Uses B⃗	Key Limitation
Mass Spectrometer	Different ions have different r = mv/(qB), separating isotopes by mass	Requires high vacuum; assumes uniform B
Electric Motor	F = BIL sin θ on current-carrying coil produces torque	Field non-uniformity reduces efficiency
MRI Scanner	Strong uniform B aligns nuclear spins; RF pulses probe tissue	Requires superconducting magnets (1.5–3 T); extremely costly
Earth's Magnetic Field	Deflects charged solar wind particles, protecting the atmosphere	Field is non-uniform and changes over geological time

Selected applications of magnetic fields and their practical limitations

SECTION 8

Connection to Advanced Theory

The AP Physics 2 treatment of magnetic fields is grounded in the algebra-based formulation of classical electromagnetism. At higher levels, the same physics is expressed using vector calculus (Ampère's law with the line integral ∮B⃗·dl⃗ = μ₀I_enc), and ultimately through the relativistic unification of E⃗ and B⃗ into the electromagnetic field tensor. The table below contrasts the AP-level and advanced treatments.

Topic	AP Physics 2 Treatment	Advanced / University Treatment
Sources of B⃗	B = μ₀I/(2πr) for long wire; B = μ₀nI for solenoid	Biot-Savart law for arbitrary current geometries; Ampère's law in integral/differential form
Force on charges	F = qvB sin θ (scalar magnitude)	F⃗ = qv⃗ × B⃗ (full vector cross product)
Magnetic flux	Φ = BA cos θ for uniform fields	Φ = ∫B⃗·dA⃗ for non-uniform fields and curved surfaces
E and B unification	Treated as separate fields	E⃗ and B⃗ are components of one electromagnetic field tensor; a pure E field in one frame can appear as a mix of E and B in another

Understanding the algebra-based framework thoroughly prepares you to transition into vector-calculus-based electromagnetism. The physical intuition—right-hand rules, field-line visualization, and the perpendicularity of magnetic forces—carries directly forward into more advanced courses.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A proton moves horizontally to the right through a region where the magnetic field points vertically upward. Which statement about the magnetic force on the proton is correct?

A. The force is directed vertically downward, opposite to B⃗.
B. The force is directed horizontally to the left, opposite to v⃗.
C. The force is perpendicular to both v⃗ and B⃗, directed out of the page.
D. The force is zero because the proton is not accelerating.

Show answer (C)

C is correct. The magnetic force on a positive charge is given by the right-hand rule: point fingers along v⃗ (right), curl toward B⃗ (up), and the thumb points out of the page. A is wrong because the magnetic force is never parallel or antiparallel to B⃗. B is wrong because the force is always perpendicular to v⃗, not antiparallel. D is wrong because a constant-speed particle can still experience a force that changes its direction (centripetal acceleration).

Question 2: BASIC CALCULATION

An electron (q = 1.60 × 10⁻¹⁹ C) moves at 5.0 × 10⁶ m/s perpendicular to a 0.20 T magnetic field. What is the magnitude of the magnetic force on the electron?

A. 1.6 × 10⁻¹³ N
B. 8.0 × 10⁻¹⁴ N
C. 1.0 × 10⁻¹² N
D. 3.2 × 10⁻¹³ N

Show answer (A)

A is correct. F = qvB sin θ = (1.60 × 10⁻¹⁹)(5.0 × 10⁶)(0.20)(sin 90°) = 1.6 × 10⁻¹³ N. B (8.0 × 10⁻¹⁴ N) would result from using B = 0.10 T instead of 0.20 T. C (1.0 × 10⁻¹²) is too large by about a factor of 6, suggesting an arithmetic error. D (3.2 × 10⁻¹³) results from doubling the correct answer, possibly by using q = 3.2 × 10⁻¹⁹ C erroneously.

Question 3: INTERMEDIATE

A 0.40 m long wire carries a current of 5.0 A through a uniform magnetic field of 0.30 T. If the wire is oriented at 30° to the field, what is the force on the wire?

A. 0.60 N
B. 0.30 N
C. 0.52 N
D. 0.15 N

Show answer (B)

B is correct. F = BIL sin θ = (0.30)(5.0)(0.40) sin 30° = 0.60 × 0.50 = 0.30 N. A (0.60 N) results from forgetting the sin 30° factor (i.e., assuming θ = 90°). C (0.52 N) corresponds to using sin 60° instead of sin 30°, confusing the complementary angle. D (0.15 N) results from an extra erroneous factor of ½.

Question 4: APPLIED

A student wishes to experimentally determine how the radius of a charged particle's circular orbit in a uniform magnetic field depends on the particle's speed. The student has access to an electron gun with adjustable accelerating voltage, a pair of Helmholtz coils producing a known uniform B⃗, and a glass bulb filled with low-pressure gas that makes the electron beam visible. (a) Describe a procedure the student could follow to collect the data needed, identifying the independent and dependent variables. (b) Describe how the accelerating voltage relates to the electron's speed. (c) Explain how the student should analyze the data graphically to confirm the relationship r = mv/(eB). (d) Identify one source of systematic error and explain its effect on the measured radius.

Show answer

(a) Set the Helmholtz coil current to produce a fixed B. Vary the accelerating voltage in increments (e.g., 100 V, 150 V, 200 V, …) and for each voltage, measure the diameter of the visible circular beam, then divide by 2 to get r. The independent variable is the accelerating voltage (which controls speed), and the dependent variable is the orbit radius r. (b) The electron gun accelerates electrons through potential difference V. By conservation of energy, eV = ½mv², so v = √(2eV/m). Higher voltage yields higher speed. (c) From r = mv/(eB) and v = √(2eV/m), we get r = (1/B)√(2mV/e). So r ∝ √V. Plotting r vs. √V should produce a straight line through the origin with slope (1/B)√(2m/e). A linear fit confirms the predicted dependence. (d) One systematic error is collisions between the electrons and gas molecules in the bulb. These collisions reduce the electrons' kinetic energy, lowering their effective speed and causing the measured orbit radius to be consistently smaller than the theoretical prediction for a given voltage. This error is systematic because it shifts all measured r values in the same direction (lower).

Question 5: CRITICAL THINKING

Two long, parallel wires separated by a distance d = 0.10 m each carry current I = 4.0 A in the same direction. (a) Calculate the magnitude of the magnetic field produced by one wire at the location of the other. (b) Calculate the force per unit length on one wire due to the other, and state whether it is attractive or repulsive. (c) A student claims that if the currents flow in opposite directions, the force magnitude would change. Evaluate this claim. (d) Explain, using field-line reasoning, why parallel currents attract while antiparallel currents repel.

Show answer

(a) B = μ₀I/(2πd) = (4π × 10⁻⁷ T·m/A)(4.0 A) / (2π × 0.10 m) = 8.0 × 10⁻⁶ T at the location of the second wire. (b) The force per unit length is F/L = BI = (8.0 × 10⁻⁶)(4.0) = 3.2 × 10⁻⁵ N/m. Equivalently, F/L = μ₀I²/(2πd). Using the right-hand rule, the field from wire 1 at wire 2's location is directed such that F = IL × B pushes wire 2 toward wire 1. The force is attractive. (c) The claim is incorrect. Reversing one current changes the direction of the force from attractive to repulsive but does not change the magnitude. The formula F/L = μ₀I₁I₂/(2πd) depends on the magnitudes of the currents; if both currents have the same magnitude, the force per unit length remains 3.2 × 10⁻⁵ N/m regardless of relative direction. (d) When currents flow in the same direction, the circular field lines between the wires point in opposite directions, partially canceling; the field is weaker between the wires than outside them. This creates a net inward pressure gradient, pushing the wires together (attraction). When currents are antiparallel, the field lines between the wires point in the same direction and reinforce, creating a stronger field between the wires than outside. The resulting outward pressure pushes the wires apart (repulsion).