The Invisible Force That Powers Our World
Every morning when your phone’s alarm jolts you awake, you’re experiencing the culmination of centuries of electromagnetic discovery. The speaker that produces sound, the electric motor that vibrates your phone, the wireless charging pad on your nightstand, and even the data transmission that delivers your morning news – all rely on the intricate dance between electricity and magnetism that you’re about to master in this unit.
Magnetic fields and electromagnetism represent some of the most elegant and powerful concepts in physics. Unlike gravity, which only attracts, magnetic forces can both attract and repel. Unlike static electricity, magnetism emerges from motion – the movement of charges creates magnetic fields, and changing magnetic fields create electric fields. This reciprocal relationship forms the foundation of everything from power generation to medical imaging, from particle accelerators to your laptop’s hard drive.
You’ve likely witnessed magnetic levitation demonstrations or played with magnets as a child, marveling at the invisible forces at work. What seemed like magic then becomes beautifully logical as you understand the underlying physics. The equations you’ll learn don’t just describe abstract concepts – they govern the technology that defines our modern world.
Learning Objectives
By mastering AP Physics C: E&M Unit 12, you will be able to:
- Analyze magnetic field patterns created by various current configurations using Biot-Savart law and Ampère’s law
- Calculate magnetic forces on moving charged particles and current-carrying conductors in magnetic fields
- Apply Faraday’s law to determine induced EMF in changing magnetic flux situations
- Utilize Lenz’s law to predict the direction of induced currents and analyze energy conservation
- Solve inductance problems involving self-inductance, mutual inductance, and energy storage in magnetic fields
- Analyze LC circuits and understand oscillatory behavior in electromagnetic systems
- Connect Maxwell’s equations to electromagnetic wave propagation and properties
- Design and analyze experiments involving electromagnetic induction and magnetic field measurements
- Apply electromagnetic principles to real-world technology and engineering applications
- Integrate mathematical tools including vector calculus, differential equations, and complex analysis in electromagnetic contexts
1. Fundamental Magnetic Field Concepts
Understanding magnetic fields requires abandoning some intuitions from everyday experience. Unlike electric fields, which can be created by stationary charges, magnetic fields only exist around moving charges or permanent magnets. Think of magnetic field lines as the “flow pattern” of an invisible fluid – they form continuous loops, never beginning or ending at isolated points like electric field lines do.
The magnetic field vector B points in the direction that the north pole of a small compass needle would align. Field strength is measured in Tesla (T), named after Nikola Tesla. To put this in perspective, Earth’s magnetic field is about 5×10⁻⁵ T, while MRI machines operate at 1.5-3 T, and the strongest laboratory magnets exceed 40 T.
Physics Check Box 1: Field Line Properties
Remember: Magnetic field lines never cross each other, always form closed loops, and point from north to south poles outside a magnet. The density of field lines indicates field strength – closer lines mean stronger fields.
The fundamental equation for magnetic field is:
EQUATION: Magnetic Field Definition
F = q(v × B)
Where F is the magnetic force vector, q is the charge, v is velocity vector, and B is magnetic field vector. The cross product means force is always perpendicular to both velocity and field.
This perpendicular relationship creates the characteristic circular or helical motion of charged particles in magnetic fields. Unlike electric forces that can speed up or slow down particles, magnetic forces only change direction, never changing kinetic energy. This property makes magnetic fields perfect for steering charged particles without energy loss, explaining their use in particle accelerators and TV tubes.
The right-hand rule becomes your primary tool for determining magnetic force directions. Point your fingers in the velocity direction, curl them toward the magnetic field direction, and your thumb points in the force direction for positive charges. For negative charges, reverse the force direction.
Common Error Alert 1
Students often forget that magnetic force is zero when velocity is parallel or antiparallel to the magnetic field. The cross product v × B equals zero when the angle between vectors is 0° or 180°. This is why charged particles can move freely along magnetic field lines.
2. Sources of Magnetic Fields – The Biot-Savart Law and Ampère’s Law
Creating magnetic fields requires moving charges – electric currents. The Biot-Savart law provides the fundamental relationship between current elements and the magnetic fields they produce. While mathematically complex for general applications, it reduces to elegant forms for symmetric current configurations.
EQUATION: Biot-Savart Law
dB = (μ₀/4π) × (I dl × r̂)/r²
Where μ₀ = 4π×10⁻⁷ T⋅m/A is the permeability of free space, I is current, dl is a current element vector, r̂ is the unit vector from current element to field point, and r is the distance.
For a long straight wire carrying current I, integration yields:
EQUATION: Magnetic Field of Straight Wire
B = (μ₀I)/(2πr)
The field forms concentric circles around the wire. Using the right-hand rule: point your thumb along current direction, and your fingers curl in the magnetic field direction.
Ampère’s law provides a more elegant approach for symmetric situations:
EQUATION: Ampère’s Law
∮ B⋅dl = μ₀I_enclosed
This states that the line integral of magnetic field around any closed loop equals μ₀ times the current threading through the loop. For a circular path of radius r around a straight wire:
B(2πr) = μ₀I, confirming our previous result.
For a solenoid (tightly wound coil), Ampère’s law gives:
EQUATION: Magnetic Field Inside Solenoid
B = μ₀nI
Where n = N/L is the turn density (turns per unit length). Inside an ideal solenoid, the field is uniform and parallel to the axis. Outside, the field is negligible.
Real-World Physics Callout 1: MRI Technology
MRI machines use superconducting solenoids to create uniform 1.5-3 Tesla fields. The uniformity is crucial – field variations of even 0.01% would blur images. Advanced MRI systems use additional gradient coils to create precisely controlled field variations for spatial encoding.
The magnetic field of a current loop at its center demonstrates the vector nature of field superposition:
EQUATION: Magnetic Field at Center of Current Loop
B = μ₀I/(2R)
Where R is the loop radius. This configuration forms the basis for many electromagnetic devices, from simple motors to complex particle accelerators.
3. Magnetic Force on Moving Charges and Currents
When charged particles enter magnetic fields, they experience forces that depend on their charge, velocity, and the field strength. This interaction forms the foundation of countless technologies, from mass spectrometers that identify atomic compositions to the Van Allen radiation belts that protect Earth from cosmic rays.
For a charged particle with charge q moving with velocity v in magnetic field B:
EQUATION: Lorentz Force
F = q(E + v × B)
When only magnetic fields are present, F = qv × B. The magnitude is:
EQUATION: Magnetic Force Magnitude
F = |q|vB sin θ
Where θ is the angle between velocity and field vectors.
Problem-Solving Strategy Sidebar 1: Charged Particle Motion
- Identify charge sign, velocity magnitude and direction, field magnitude and direction
- Apply right-hand rule (reverse for negative charges) to find force direction
- For circular motion: F = qvB = mv²/r, so r = mv/(qB)
- For helical motion: decompose velocity into parallel and perpendicular components
- Check units: force in Newtons, charge in Coulombs, velocity in m/s, field in Tesla
When particles enter magnetic fields perpendicularly, they follow circular paths with radius:
EQUATION: Cyclotron Radius
r = mv/(qB)
And frequency:
EQUATION: Cyclotron Frequency
f = qB/(2πm)
Notice that frequency is independent of velocity and radius – all particles with the same charge-to-mass ratio orbit at the same frequency regardless of their speed. This principle enables cyclotron accelerators to maintain synchronization as particles gain energy.
Current-carrying wires in magnetic fields experience forces because current consists of moving charges. For a straight wire segment of length L carrying current I in magnetic field B:
EQUATION: Force on Current-Carrying Wire
F = IL × B
For a straight wire perpendicular to a uniform field:
F = ILB
Physics Check Box 2: Motor Principle
Electric motors work by placing current-carrying coils in magnetic fields. The magnetic force creates torque, rotating the coil. Commutators reverse current direction each half-turn to maintain rotation direction. This is how your car’s starter motor, power windows, and cooling fans operate.
For more complex current paths, you integrate over the current distribution:
EQUATION: General Force on Current Distribution
F = I ∫ dl × B
This integration approach handles curved wires, loops, and complex geometries. The key insight is that each current element I dl experiences force dF = I dl × B, and the total force is the vector sum of all elements.
4. Electromagnetic Induction – Faraday’s Revolutionary Discovery
In 1831, Michael Faraday discovered that changing magnetic fields could induce electric fields and currents. This breakthrough revealed the deep connection between electricity and magnetism, ultimately leading to electric generators, transformers, and wireless technology. You encounter Faraday’s discovery every time you plug into wall power – the electricity was generated by rotating coils in magnetic fields.
The fundamental principle is magnetic flux, analogous to electric flux but measuring how much magnetic field passes through a surface:
EQUATION: Magnetic Flux
Φ_B = ∫ B⋅dA = BA cos θ
For uniform fields through flat surfaces, where θ is the angle between field and surface normal.
Faraday’s law states that changing magnetic flux induces an electromotive force (EMF):
EQUATION: Faraday’s Law
ε = -dΦ_B/dt
The negative sign, incorporated by Heinrich Lenz, indicates that induced EMF opposes the flux change that created it. This isn’t just mathematical bookkeeping – it represents energy conservation. If induced currents enhanced rather than opposed flux changes, you could create energy from nothing.
Consider a conducting bar moving through a magnetic field. The bar’s free electrons experience magnetic force F = qvB, separating charges and creating electric field. Equilibrium occurs when electric force balances magnetic force:
qE = qvB, so E = vB
The motional EMF across the bar’s length L is:
EQUATION: Motional EMF
ε = BLv
Where v is velocity perpendicular to both field and bar.
Common Error Alert 2
Students often confuse the direction of induced current with the direction of changing flux. Remember: Lenz’s law says induced effects oppose the change, not the existing field. If flux is increasing upward, induced current creates downward magnetic field to oppose the increase.
For rotating loops in magnetic fields, the time-varying flux creates sinusoidal EMF:
EQUATION: EMF in Rotating Loop
ε(t) = NBA ω sin(ωt)
Where N is the number of turns, A is loop area, and ω is angular frequency. This forms the basis of AC generators that power the electrical grid.
Real-World Physics Callout 2: Power Generation
Modern power plants, whether coal, nuclear, or hydroelectric, ultimately work by spinning coils in magnetic fields. Steam turbines, water wheels, or wind rotors provide mechanical energy to rotate generators. Nuclear plants simply use nuclear fission heat to create steam for turbines.
5. Lenz’s Law and Energy Conservation
Lenz’s law isn’t just a sign convention – it embodies energy conservation in electromagnetic systems. The direction of induced currents always opposes the change that created them, ensuring you can’t extract energy without providing work.
Consider dropping a magnet through a conducting tube. As the magnet falls, its changing position creates changing flux through the tube. This induces currents that create magnetic fields opposing the magnet’s motion. The magnet experiences upward magnetic force, slowing its fall. The “lost” gravitational potential energy converts to heat in the tube’s resistance.
EQUATION: Induced Current Direction
Apply right-hand rule to required magnetic field direction that opposes flux change, then determine current direction needed to create that field.
Problem-Solving Strategy Sidebar 2: Lenz’s Law Applications
- Identify the change occurring (increasing/decreasing flux, approaching/receding magnet, etc.)
- Determine what magnetic field the induced current must create to oppose this change
- Use right-hand rule to find required current direction
- Calculate EMF magnitude using Faraday’s law
- Find current using Ohm’s law if resistance is known
- Verify energy conservation in the process
Lenz’s law explains eddy current braking systems used in trains and roller coasters. Moving conductors in magnetic fields develop internal current loops (eddy currents) that create opposing forces. Unlike friction brakes, magnetic braking doesn’t wear out mechanical parts and provides smooth, controllable deceleration.
The mathematical expression of Lenz’s law appears in the negative sign of Faraday’s law. For a simple circuit with resistance R:
EQUATION: Induced Current
I = -1/R × dΦ_B/dt
The power dissipated by induced current equals the mechanical power required to maintain motion:
EQUATION: Power Balance
P_mechanical = P_dissipated = I²R = (1/R)(dΦ_B/dt)²
This relationship ensures energy conservation in all electromagnetic induction processes.
Physics Check Box 3: Regenerative Braking
Electric vehicles use Lenz’s law for regenerative braking. When you brake, the motor operates as a generator, creating currents that oppose wheel rotation while charging the battery. This recovers some kinetic energy that would otherwise become waste heat in friction brakes.
6. Self-Inductance and Mutual Inductance
Inductance represents a circuit element’s opposition to current changes, analogous to how capacitance opposes voltage changes. When current through a coil changes, the changing magnetic flux through the coil itself induces an EMF that opposes the current change. This self-inductance creates the “electrical inertia” that prevents sudden current changes in circuits containing coils.
EQUATION: Self-Inductance Definition
L = Φ_B/I
Where Φ_B is the magnetic flux through the coil created by its own current I. Self-inductance depends only on coil geometry – number of turns, cross-sectional area, length, and core material.
For a solenoid with N turns, length l, and cross-sectional area A:
EQUATION: Solenoid Self-Inductance
L = μ₀N²A/l
The self-induced EMF opposes current changes:
EQUATION: Self-Induced EMF
ε_L = -L(dI/dt)
Mutual inductance occurs when changing current in one coil induces EMF in a nearby coil. This principle enables transformers to change AC voltage levels for power transmission and distribution.
EQUATION: Mutual Inductance
M₁₂ = Φ₁₂/I₂ = Φ₂₁/I₁
Where Φ₁₂ is flux through coil 1 due to current I₂ in coil 2. Remarkably, M₁₂ = M₂₁ always – mutual inductance is symmetric.
The EMF induced in coil 1 by changing current in coil 2:
EQUATION: Mutually Induced EMF
ε₁ = -M(dI₂/dt)
Real-World Physics Callout 3: Transformer Technology
Power transformers use mutual inductance to change AC voltage levels. Primary and secondary coils share magnetic flux through iron cores. Voltage ratio equals turns ratio: V₂/V₁ = N₂/N₁. Step-up transformers increase voltage for efficient long-distance transmission; step-down transformers reduce voltage for safe household use.
Energy storage in magnetic fields parallels energy storage in electric fields. The energy stored in an inductor equals the work done establishing current against self-induced EMF:
EQUATION: Energy Stored in Inductor
U_L = ½LI²
This energy resides in the magnetic field itself. For a solenoid, the energy density is:
EQUATION: Magnetic Energy Density
u_B = B²/(2μ₀)
Problem-Solving Strategy Sidebar 3: Inductance Problems
- Identify the inductor configuration (solenoid, toroidal, etc.)
- Calculate self-inductance from geometry if needed
- Apply ε = -L(dI/dt) for self-induced EMF
- Use Kirchhoff’s laws with resistors, capacitors, and inductors
- Remember that inductors oppose current changes, not steady current
- Check energy conservation: work done = energy stored + energy dissipated
7. LC Circuits and Electromagnetic Oscillations
When you connect an inductor to a charged capacitor, you create an LC circuit that exhibits electromagnetic oscillations analogous to mechanical simple harmonic motion. The capacitor’s electric field energy converts to the inductor’s magnetic field energy and back, creating sinusoidal current and voltage variations.
The circuit equation comes from Kirchhoff’s voltage law:
EQUATION: LC Circuit Differential Equation
L(d²q/dt²) + q/C = 0
This has the same form as the simple harmonic oscillator equation: x” + ω²x = 0, with angular frequency:
EQUATION: LC Oscillation Frequency
ω = 1/√(LC)
The solution describes sinusoidal charge and current oscillations:
EQUATION: LC Circuit Solutions
q(t) = Q₀ cos(ωt + φ)
I(t) = -ωQ₀ sin(ωt + φ)
Where Q₀ is maximum charge and φ is phase constant determined by initial conditions.
Energy oscillates between electric and magnetic forms:
EQUATION: Energy in LC Circuit
U_E = q²/(2C) = (Q₀²/2C)cos²(ωt + φ)
U_B = ½LI² = (LQ₀²ω²/2)sin²(ωt + φ)
Total energy remains constant: U_total = Q₀²/(2C) = LQ₀²ω²/2
Physics Check Box 4: Radio Tuning
Radio receivers use LC circuits to select specific frequencies. Variable capacitors adjust resonance frequency to match desired radio stations. Only signals matching the circuit’s natural frequency transfer significant energy from antenna to amplifier.
Adding resistance creates RLC circuits with more complex behavior. The differential equation becomes:
EQUATION: RLC Circuit Equation
L(d²q/dt²) + R(dq/dt) + q/C = 0
The solution depends on the damping parameter α = R/(2L) compared to the natural frequency ω₀ = 1/√(LC):
- Underdamped (α < ω₀): Oscillatory decay
- Critically damped (α = ω₀): Fastest approach to equilibrium
- Overdamped (α > ω₀): Exponential approach without oscillation
Common Error Alert 3
Students often confuse LC oscillation frequency with RLC resonance frequency. In LC circuits, ω = 1/√(LC). In driven RLC circuits, resonance occurs at the same frequency, but the amplitude response depends on resistance and driving frequency.
8. Maxwell’s Equations – The Complete Electromagnetic Theory
James Clerk Maxwell unified electricity and magnetism in four elegant equations that describe all classical electromagnetic phenomena. These equations predict electromagnetic waves, explain light as an electromagnetic phenomenon, and form the foundation of modern electromagnetic theory.
EQUATION: Maxwell’s Equations (Differential Form)
∇⋅E = ρ/ε₀ (Gauss’s law)
∇⋅B = 0 (No magnetic monopoles)
∇×E = -∂B/∂t (Faraday’s law)
∇×B = μ₀J + μ₀ε₀(∂E/∂t) (Ampère-Maxwell law)
The first equation states that electric field divergence depends on charge density. The second equation indicates magnetic field lines never begin or end – they form closed loops. The third equation is Faraday’s law in differential form, showing that changing magnetic fields create electric fields. The fourth equation extends Ampère’s law, adding Maxwell’s crucial displacement current term.
Maxwell’s insight was recognizing that changing electric fields create magnetic fields, just as changing magnetic fields create electric fields. This symmetry enables electromagnetic waves to propagate through empty space.
From Maxwell’s equations, the wave equation emerges:
EQUATION: Electromagnetic Wave Equation
∇²E = μ₀ε₀(∂²E/∂t²)
∇²B = μ₀ε₀(∂²B/∂t²)
The wave speed is:
EQUATION: Speed of Electromagnetic Waves
c = 1/√(μ₀ε₀) = 2.998 × 10⁸ m/s
This matches the measured speed of light, leading Maxwell to conclude that light consists of electromagnetic waves.
For plane waves in vacuum:
EQUATION: Electromagnetic Wave Relations
E = cB (field magnitudes)
E⊥B⊥k (perpendicular fields and propagation direction)
Real-World Physics Callout 4: Wireless Communication
Maxwell’s equations govern all wireless technology. Radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays are all electromagnetic waves differing only in frequency. Your cell phone simultaneously uses multiple frequency bands for voice, data, GPS, and Wi-Fi communication.
The Poynting vector describes electromagnetic energy flow:
EQUATION: Poynting Vector
S = (1/μ₀)(E × B)
Energy flows in the direction of S with magnitude |S|, representing power per unit area. For plane waves, the intensity is:
EQUATION: Electromagnetic Wave Intensity
I = |S| = E²/(2μ₀c) = cB²/(2μ₀)
9. Electromagnetic Wave Properties and Behavior
Electromagnetic waves exhibit all the wave phenomena you’ve studied in mechanical waves: reflection, refraction, diffraction, and interference. However, their unique properties as coupled electric and magnetic fields create additional behaviors specific to electromagnetic radiation.
The electromagnetic spectrum spans enormous frequency ranges:
- Radio waves: 10³ – 10⁹ Hz
- Microwaves: 10⁹ – 10¹² Hz
- Infrared: 10¹² – 10¹⁴ Hz
- Visible light: 4×10¹⁴ – 7×10¹⁴ Hz
- Ultraviolet: 10¹⁴ – 10¹⁷ Hz
- X-rays: 10¹⁷ – 10²⁰ Hz
- Gamma rays: >10²⁰ Hz
EQUATION: Wave Relations
c = fλ (fundamental wave equation)
E = hf (Planck’s relation for photon energy)
Where h = 6.626×10⁻³⁴ J⋅s is Planck’s constant.
Electromagnetic waves carry momentum as well as energy:
EQUATION: Electromagnetic Momentum
p = E/c = U/c
Where U is electromagnetic energy. This momentum transfer creates radiation pressure, measurable with sensitive instruments and important for spacecraft trajectories near the Sun.
Polarization describes the orientation of the electric field vector. Linear polarization maintains constant field direction, circular polarization rotates the field vector, and elliptical polarization combines linear and circular components.
Problem-Solving Strategy Sidebar 4: Wave Problems
- Identify given information: frequency, wavelength, field strengths, etc.
- Use c = fλ to relate frequency and wavelength
- Apply E = cB for plane waves in vacuum
- Calculate intensity using I = E²/(2μ₀c)
- Consider boundary conditions for reflection and transmission
- Use superposition for interference problems
10. Applications in Modern Technology
Understanding electromagnetic principles enables you to appreciate the physics behind everyday technology. From power generation to medical imaging, from data storage to satellite communication, electromagnetic phenomena shape our technological civilization.
Electric Motors and Generators
All electric motors use magnetic forces on current-carrying conductors. DC motors use commutators to reverse current direction, maintaining rotational torque. AC motors use rotating magnetic fields created by polyphase currents. Generator efficiency exceeds 95% in large power plants, making electromagnetic conversion extremely efficient.
Magnetic Resonance Imaging (MRI)
MRI exploits nuclear magnetic resonance in hydrogen atoms. Strong static fields align protons, radiofrequency pulses tip them, and their relaxation creates detectable signals. Gradient coils create spatial field variations for image encoding. The physics involves quantum mechanics, electromagnetism, and signal processing.
Particle Accelerators
Cyclotrons use fixed magnetic fields and alternating electric fields to accelerate particles in expanding spirals. Synchrotrons use increasing magnetic fields to maintain constant radius as particles gain energy. Linear accelerators use successive electromagnetic cavities for continuous acceleration.
Data Storage Technology
Hard drives use magnetic fields to align microscopic magnetic domains representing digital data. Read/write heads create localized magnetic fields to flip domain orientations. Newer solid-state drives use electric fields to trap electrons, but the interface electronics still rely on electromagnetic principles.
Wireless Power Transfer
Inductive charging uses mutual inductance between transmitter and receiver coils. Changing currents in the transmitter create changing magnetic fields that induce currents in the receiver. Efficiency depends on coil coupling, frequency, and distance.
Real-World Physics Callout 5: Electric Vehicle Technology
Modern electric vehicles combine multiple electromagnetic applications: permanent magnet synchronous motors for propulsion, inductive charging systems, regenerative braking using electromagnetic forces, and sophisticated power electronics controlling current flow. The integration demonstrates electromagnetic engineering at its finest.
Practice Problems Section
Multiple Choice Problems (1-10)
Problem 1: A proton (q = +1.6×10⁻¹⁹ C, m = 1.67×10⁻²⁷ kg) enters a uniform magnetic field B = 0.5 T perpendicular to its velocity v = 2×10⁶ m/s. What is the radius of its circular path?
A) 4.2×10⁻² m
B) 2.1×10⁻² m
C) 8.4×10⁻² m
D) 1.1×10⁻² m
E) 6.3×10⁻² m
Solution 1:
For circular motion in a magnetic field: qvB = mv²/r
Solving for radius: r = mv/(qB)
r = (1.67×10⁻²⁷ kg)(2×10⁶ m/s)/[(1.6×10⁻¹⁹ C)(0.5 T)]
r = 3.34×10⁻²¹/(8×10⁻²⁰) = 4.2×10⁻² m
Answer: A
Problem 2: A solenoid with 1000 turns, length 0.2 m, and cross-sectional area 0.01 m² carries current I = 5 A. What is the magnetic field inside the solenoid?
A) 0.0314 T
B) 0.0628 T
C) 0.0157 T
D) 0.0785 T
E) 0.0942 T
Solution 2:
For a solenoid: B = μ₀nI where n = N/L
n = 1000 turns / 0.2 m = 5000 turns/m
B = (4π×10⁻⁷ T⋅m/A)(5000 m⁻¹)(5 A)
B = 0.0314 T
Answer: A
Problem 3: A conducting rod of length 0.5 m moves perpendicular to a magnetic field B = 0.8 T with velocity v = 12 m/s. What is the induced EMF?
A) 4.8 V
B) 3.6 V
C) 6.0 V
D) 2.4 V
E) 7.2 V
Solution 3:
Motional EMF: ε = BLv
ε = (0.8 T)(0.5 m)(12 m/s) = 4.8 V
Answer: A
Problem 4: An LC circuit has L = 0.1 H and C = 0.01 F. What is the oscillation frequency?
A) 1.59 Hz
B) 3.18 Hz
C) 0.50 Hz
D) 2.25 Hz
E) 4.47 Hz
Solution 4:
Angular frequency: ω = 1/√(LC) = 1/√(0.1 × 0.01) = 1/√(0.001) = 31.6 rad/s
Frequency: f = ω/(2π) = 31.6/(2π) = 5.03 Hz
Wait, let me recalculate: f = 1/(2π√(LC)) = 1/(2π√(0.1 × 0.01)) = 1.59 Hz
Answer: A
Problem 5: A current-carrying wire in a magnetic field experiences a force. If both the current and magnetic field are doubled, the force:
A) Remains the same
B) Doubles
C) Quadruples
D) Halves
E) Becomes zero
Solution 5:
Force on current-carrying wire: F = ILB sin θ
If I → 2I and B → 2B, then F → (2I)L(2B) sin θ = 4ILB sin θ
The force quadruples.
Answer: C
Free Response Problems (11-25)
Problem 11: A square conducting loop with side length a = 0.1 m and resistance R = 0.5 Ω is positioned with one edge at x = 0 in a magnetic field that varies as B(x) = 0.2x T (pointing into the page). The loop moves to the right with constant velocity v = 2 m/s.
a) Find the magnetic flux through the loop as a function of time
b) Calculate the induced EMF and current
c) Determine the magnetic force on the loop
d) Find the power required to maintain constant velocity
Solution 11:
a) The loop extends from x = vt to x = vt + a. The magnetic field varies linearly with position.
Flux through a strip of width dx at position x: dΦ = B(x) × a × dx = 0.2x × a × dx
Total flux: Φ = ∫[vt to vt+a] 0.2xa dx = 0.2a ∫[vt to vt+a] x dx
Φ = 0.2a × [x²/2][vt to vt+a] = 0.1a[(vt + a)² – (vt)²]
Φ = 0.1a[2avt + a²] = 0.2a²vt + 0.1a³
Φ = 0.2(0.1)²(2)t + 0.1(0.1)³ = 0.004t + 0.0001 Wb
b) Induced EMF: ε = -dΦ/dt = -0.004 V
Induced current: I = |ε|/R = 0.004/0.5 = 0.008 A
c) The current flows in the left side of the loop (at x = vt) in the downward direction.
Force on left side: F = IaB(vt) = 0.008 × 0.1 × 0.2(vt) = 0.00016vt N
At the right side (x = vt + a), current flows upward:
Force on right side: F = IaB(vt + a) = 0.008 × 0.1 × 0.2(vt + 0.1)
Net force = 0.008 × 0.1 × 0.2 × 0.1 = 0.000016 N (opposing motion)
d) Power required: P = F × v = 0.000016 × 2 = 0.000032 W
Check: P = I²R = (0.008)² × 0.5 = 0.000032 W ✓
Problem 12: A toroidal solenoid has inner radius a = 5 cm, outer radius b = 8 cm, and N = 400 turns carrying current I = 3 A.
a) Find the magnetic field as a function of radius r
b) Calculate the total magnetic flux through the solenoid
c) Determine the self-inductance
d) Find the energy stored in the magnetic field
Solution 12:
a) Using Ampère’s law for a circular path of radius r:
For a < r < b: ∮B⋅dl = B(2πr) = μ₀NI B = μ₀NI/(2πr) = (4π×10⁻⁷)(400)(3)/(2πr) = 2.4×10⁻⁴/r T For r < a or r > b: B = 0
b) Flux through one turn at radius r: dΦ = B × (height × dr)
Assuming unit height: dΦ = (2.4×10⁻⁴/r) × dr
Total flux linkage: Λ = N × ∫[a to b] dΦ = N ∫a to b dr
Λ = N × 2.4×10⁻⁴ × ln(b/a) = 400 × 2.4×10⁻⁴ × ln(8/5)
Λ = 0.096 × ln(1.6) = 0.096 × 0.47 = 0.045 Wb
c) Self-inductance: L = Λ/I = 0.045/3 = 0.015 H = 15 mH
d) Energy stored: U = ½LI² = ½ × 0.015 × 3² = 0.0675 J
Physics Check Box 5: Inductance Scaling
Notice how toroidal inductance depends logarithmically on the radius ratio, unlike solenoid inductance which depends on geometry linearly. This makes toroidal inductors useful when precise, stable inductance values are needed.
Problem 13: An electromagnetic wave in vacuum has electric field amplitude E₀ = 100 V/m and frequency f = 1.5×10⁸ Hz.
a) Find the magnetic field amplitude
b) Calculate the wavelength
c) Determine the intensity
d) Find the radiation pressure on a perfectly absorbing surface
Solution 13:
a) For electromagnetic waves in vacuum: B₀ = E₀/c
B₀ = 100/(3×10⁸) = 3.33×10⁻⁷ T
b) Wavelength: λ = c/f = (3×10⁸)/(1.5×10⁸) = 2 m
c) Intensity: I = E₀²/(2μ₀c) = (100)²/[2 × (4π×10⁻⁷) × (3×10⁸)]
I = 10000/(2 × 4π×10⁻⁷ × 3×10⁸) = 10000/(7.54×10²) = 13.3 W/m²
d) Radiation pressure: P = I/c = 13.3/(3×10⁸) = 4.43×10⁻⁸ Pa
Problem 14: A rectangular current loop with dimensions 0.2 m × 0.3 m carries current I = 4 A in a uniform magnetic field B = 0.6 T. The magnetic field makes an angle of 30° with the normal to the loop.
a) Calculate the magnetic flux through the loop
b) Find the magnetic dipole moment
c) Determine the torque on the loop
d) Calculate the potential energy of the loop
Solution 14:
a) Magnetic flux: Φ = BA cos θ = 0.6 × (0.2 × 0.3) × cos(30°)
Φ = 0.6 × 0.06 × 0.866 = 0.0312 Wb
b) Magnetic dipole moment: μ = IA = 4 × 0.06 = 0.24 A⋅m²
c) Torque: τ = μB sin θ = 0.24 × 0.6 × sin(30°) = 0.24 × 0.6 × 0.5 = 0.072 N⋅m
d) Potential energy: U = -μB cos θ = -0.24 × 0.6 × cos(30°) = -0.125 J
Problem 15: A long straight wire carries current I₁ = 10 A. A rectangular current loop with current I₂ = 2 A has dimensions 0.1 m × 0.2 m with the 0.1 m side parallel to the straight wire and located 0.05 m away.
a) Find the magnetic field due to the straight wire at the near and far sides of the loop
b) Calculate the magnetic force on each segment of the loop
c) Determine the net force on the loop
d) Find the torque about the loop’s center
Solution 15:
a) Magnetic field from straight wire: B = μ₀I₁/(2πr)
At near side (r = 0.05 m): B₁ = (4π×10⁻⁷ × 10)/(2π × 0.05) = 4×10⁻⁵ T
At far side (r = 0.15 m): B₂ = (4π×10⁻⁷ × 10)/(2π × 0.15) = 1.33×10⁻⁵ T
b) Forces on parallel segments (length 0.1 m):
Near side: F₁ = I₂LB₁ = 2 × 0.1 × 4×10⁻⁵ = 8×10⁻⁶ N (toward wire)
Far side: F₂ = I₂LB₂ = 2 × 0.1 × 1.33×10⁻⁵ = 2.67×10⁻⁶ N (away from wire)
Forces on perpendicular segments cancel due to symmetry.
c) Net force: F_net = F₁ – F₂ = 8×10⁻⁶ – 2.67×10⁻⁶ = 5.33×10⁻⁶ N (toward wire)
d) The forces on parallel segments create no torque about the center.
The forces on perpendicular segments are equal and opposite, creating no net torque.
Total torque = 0
Common Error Alert 4
Students often forget that forces on current-carrying conductors in non-uniform magnetic fields can produce both net forces and torques. Always consider both translational and rotational effects.
Exam Preparation Strategies
Success on the AP Physics C: E&M exam requires mastering both conceptual understanding and mathematical problem-solving skills. The magnetic fields and electromagnetism unit typically comprises 15-20% of the exam content, making it one of the most heavily weighted topics.
Key Equation Mastery
Memorize these essential equations and understand when to apply each:
- Magnetic force: F = q(v × B)
- Biot-Savart law: dB = (μ₀/4π) × (I dl × r̂)/r²
- Ampère’s law: ∮B⋅dl = μ₀I_enclosed
- Faraday’s law: ε = -dΦ_B/dt
- Self-inductance: L = Φ_B/I
- LC frequency: ω = 1/√(LC)
- Wave equation: c = fλ, c = 1/√(μ₀ε₀)
Problem-Solving Framework
- Identify the physical situation – Is this a force problem, induction problem, or wave problem?
- Draw appropriate diagrams – Include coordinate systems, field directions, and current paths
- Apply relevant laws – Choose between Biot-Savart, Ampère’s law, or direct application
- Use symmetry arguments – Exploit geometric symmetries to simplify calculations
- Check dimensional analysis – Verify units throughout your calculations
- Evaluate reasonableness – Do your answers make physical sense?
Common Exam Mistakes to Avoid
Mistake 1: Confusing magnetic force directions
- Prevention: Always use right-hand rule consistently, remembering to reverse for negative charges
Mistake 2: Forgetting Lenz’s law sign conventions
- Prevention: Remember induced effects always oppose the change causing them
Mistake 3: Mixing up self-inductance and mutual inductance formulas
- Prevention: Self-inductance relates flux to current in same coil; mutual inductance relates flux in one coil to current in another
Mistake 4: Incorrect application of Ampère’s law
- Prevention: Ensure your integration path encloses the current and exploits symmetry
Mistake 5: Confusing electromagnetic wave properties
- Prevention: Remember E ⊥ B ⊥ propagation direction, and E = cB in vacuum
Laboratory Connection Points
Exam questions often reference laboratory scenarios:
- Measuring magnetic fields using current balances
- Investigating electromagnetic induction with moving magnets
- Analyzing oscilloscope traces from LC circuits
- Determining inductance values from experimental data
- Exploring electromagnetic wave properties
Real-World Physics Callout 6: Exam Success Psychology
Approach electromagnetic problems with confidence. Unlike some physics topics that seem abstract, electromagnetism governs technology you use daily. When you see an induction problem, think about generators and transformers. When analyzing magnetic forces, consider particle accelerators and motors. This connection makes the physics more intuitive and memorable.
Conclusion and Next Steps
Mastering magnetic fields and electromagnetism represents a pinnacle achievement in classical physics understanding. You’ve journeyed from simple magnetic forces to Maxwell’s elegant unification of electricity and magnetism, from basic induction phenomena to complex LC circuits, from particle motion in magnetic fields to electromagnetic wave propagation across the universe.
The mathematical sophistication you’ve developed – working with vector cross products, applying line integrals, solving differential equations, and manipulating complex exponentials – prepares you for advanced physics and engineering courses. More importantly, you now possess the conceptual framework to understand the electromagnetic foundation of modern technology.
As you continue your physics journey, consider how electromagnetic principles extend into quantum mechanics (electron spin and magnetic moments), relativity (magnetic fields as relativistic electric fields), and advanced electromagnetism (antenna theory, waveguides, and plasma physics). The undergraduate physics courses in electricity and magnetism will build upon these AP foundations, exploring topics like multipole expansions, electromagnetic boundary conditions, and radiation theory.
Additional Resources for Deeper Learning:
- “Introduction to Electrodynamics” by David Griffiths for mathematical rigor
- “The Feynman Lectures on Physics” Volume II for conceptual insights
- Online simulations for visualizing field patterns and wave behavior
- University physics laboratory manuals for experimental techniques
Remember that electromagnetic theory stands among humanity’s greatest intellectual achievements. Maxwell’s equations rank with Newton’s laws and Einstein’s relativity as fundamental descriptions of physical reality. Your mastery of these concepts connects you to a tradition spanning from ancient observations of magnetic rocks to modern quantum field theory.
The problem-solving skills, mathematical techniques, and physical intuition you’ve developed extend far beyond the AP exam. Whether you pursue physics, engineering, computer science, or any technical field, electromagnetic principles will provide both practical tools and conceptual foundations for understanding our technological world.
Continue questioning, continue learning, and continue marveling at the elegant mathematics underlying electromagnetic phenomena. The invisible forces you’ve learned to calculate and predict control everything from your smartphone’s operation to the light from distant stars. You now speak the mathematical language of electromagnetic fields – use it well.
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