Magnetic Fields and Electromagnetism

Four topics on a force that acts only on charge in motion and never along the field. Magnetic fields and why their lines close on themselves, moving charges steered in circles and helices by a force that does no work, the fields of currents from the straight wire out to the Biot-Savart element integral, and Ampere's law, which is always true and hands you a field only when the symmetry survives.

AP exam 10-20%4 topics

Topics

12.1

Magnetic Fields

Magnetism and Moving Charges

Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law

Ampère's Law

Lesson

Diagnostic

after diagnostic

Key forms For every problem in this unit

Magnetic force on a charge

F = q v × B, magnitude |q| v B sinθ. PERPENDICULAR to both v and B, never along B

No motion, no force

v = 0 gives F = 0 however strong B is. Only the component of v perpendicular to B counts

Right-hand rule

fingers along v, curl toward B, thumb gives v × B. Flip the result ONCE for a negative charge, never twice

Zero work

(v × B)·v = 0, so B alone never changes speed or kinetic energy. It bends paths only

Circular orbit

r = m v⊥ / (|q| B). Speed is in the NUMERATOR: faster means a bigger circle

Cyclotron period

T = 2πm / (|q| B), ω = |q| B / m. Speed-independent, which is why a fixed frequency stays in step

Helix

v parallel to B is untouched, v perpendicular circles. Pitch = v∥ T, constant

Velocity selector

crossed E and B pass only v = E / B. Charge and mass both cancel; faster particles bend toward the magnetic force

Force on a current

dF = I dℓ × B, straight segment in uniform B: F = I L B sinθ. A neutral wire still feels it

Magnetic moment

μ = N I A along the loop normal, by curling the fingers along the current

Torque on a loop

τ = μ × B, magnitude N I A B sinθ. Uniform B gives ZERO net force. Maximum torque when the plane CONTAINS B

Dipole energy

U = −μ·B. Stable with μ along B, which is the plane perpendicular to B

Permeability of free space

μ₀ = 4π × 10⁻⁷ T·m/A

Biot-Savart law

dB = (μ₀ / 4π) I dℓ × r̂ / r². Each element has its OWN r and its own direction; integrate as vectors

Element that gives nothing

dℓ parallel to r̂ makes the cross product zero, so a segment aimed straight at the field point contributes 0

Long straight wire

B = μ₀ I / (2πr), on CIRCLES around the wire. Grip it with the right thumb along I. Tangential, never radial, and 1/r not 1/r²

Finite straight segment

B = μ₀ I (sinθ₂ − sinθ₁) / (4πd), angles taken to the perpendicular from the field point

Center of a loop

B = μ₀ I / (2R) along the axis. Every element contributes the SAME way: they add, they do not cancel

Arc of angle φ

B = μ₀ I φ / (4πR) at the center of curvature. Radial leads contribute nothing

On the axis of a loop

B = μ₀ I R² / [2(R² + z²)³⁄²], falling as 1/z³ far away: a magnetic dipole, not 1/r²

Two parallel wires

F / L = μ₀ I₁ I₂ / (2πd). SAME direction ATTRACTS, opposite repels: the reverse of the charge rule

Superposing wire fields

get each field's direction from its own grip rule FIRST. Same-direction currents cancel at the midpoint, opposite ones double

Ampere's law

∮ B·dℓ = μ₀ I(enclosed), around ANY closed path. B in the integrand is the TOTAL field, outside currents included

When it hands you B

only when symmetry makes B constant in magnitude and everywhere tangent to the loop: infinite wire, solenoid, toroid

Enclosed current

a SIGNED census: right-hand fingers along the travel direction, piercings along the thumb positive. N turns give N I

Inside a thick wire

uniform density: I(enclosed) = I r² / R², so B = μ₀ I r / (2πR²), rising linearly, then 1/r outside

Ideal solenoid

B = μ₀ n I inside, uniform and along the axis; about zero outside. n is turns per unit LENGTH

Toroid

B = μ₀ N I / (2πr) inside the windings, not uniform across the cross-section. Outside: I(enclosed) = 0, so B = 0

No magnetic monopoles

∮ B·dA = 0 through EVERY closed surface: field lines close on themselves and never end on a pole

Unit 12 tools

Challenge bank

1 / 60

60 open-ended problems.

Read the question, work it out, then flip the card to compare your reasoning to the worked solution. Mark each card so you can return to the ones that still bite.

Question

Tap card to reveal explanation

Worked solution

Tap card to return to question

Cumulative assessment

Test the unit.

Twenty mixed items drawn from across all 4 topics, with guaranteed misconception-code coverage. Identifies which misconceptions still bite when you cannot see which topic the question came from.

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Check what stuck.

Units 8 through 12, drawn evenly so earlier units get the same share as this one. Twenty questions or a full 42-question section, your choice. Even coverage means this is a retention check rather than a score estimate.

20 or 42questions

25topics

92codes covered

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