AP Physics C: E&M study package

AP Physics C: Electricity and Magnetism (E&M) is a calculus-based second-semester college physics course covering electrostatics, circuits, magnetism, and electromagnetism. It requires strong calculus skills — particularly integration over charge distributions, line integrals for potential and emf, and surface integrals for flux and Gauss's law. The course builds logically: from static charges to electric fields, from fields to potentials, from potentials to circuits, and finally from moving charges to magnetic fields and the unified theory of electromagnetism.

Exam Structure

Section 1: Multiple Choice (50% of score)

• 35 questions in 45 minutes (~77 seconds per question)
• No calculator permitted (recently, a simple calculator may be allowed per College Board updates)
• Covers all 5 units with approximately equal weighting
• Questions emphasize conceptual understanding and qualitative reasoning
• Some MCQs require setting up (but not solving) integrals

Section 2: Free Response (50% of score)

• 3 questions in 45 minutes (~15 minutes per question)
• Graphing calculator permitted
• Typically includes:
  • A Gauss's law or field calculation problem (integration over charge distribution)
  • A circuit problem (RC transient analysis or multi-loop circuit)
  • A magnetism/induction problem (Faraday's law with calculus)
• You must show all work, including integral setups
• Partial credit is generous — always attempt every part

The 5 Units

Unit	Topic	Weight (approx.)	Key Calculus Element
1	Electrostatics	26%	Coulomb's law integrals, electric field/potential via integration
2	Conductors, Capacitors, Dielectrics	13%	Surface integrals for Gauss's law, energy integrals
3	Electric Circuits	16%	RC differential equations, integration for energy
4	Magnetic Fields	20%	Biot-Savart law integrals, Ampere's law
5	Electromagnetism	25%	Faraday's law, motional emf integrals, Maxwell's equations

Prerequisite Knowledge

Calculus (Essential)

• Line integrals: ∫F · dl for work, potential, and emf
• Surface integrals: ∫E · dA for electric flux
• Volume/surface/line charge integration: Setting up dm → dq → dE → E
• Differential equations: First-order for RC circuits, second-order for LC circuits
• Vector calculus basics: Dot products, cross products, divergence and curl (conceptual)

Physics (Required)

• AP Physics C: Mechanics is strongly recommended as a prerequisite
• Familiarity with force, energy, and potential from mechanics

How to Use This Study Package

Study Sequence

1. Read the Course Overview (this file)
2. Study each Unit Note file — these contain full derivations and theory (~2500 words each)
3. Complete each Practice file — these contain worked examples and practice problems (~1500 words each)
4. Review the Summary file — a condensed reference of all key equations
5. Study the Strategy file — exam-day tactics specific to Physics C E&M
6. Take the Full Practice Exam under timed conditions
7. Check your answers with the detailed solutions

Time Investment

• Each unit: 4–5 hours (read notes + work practice)
• Full exam + review: 2–3 hours
• Total preparation: 25–35 hours for a thorough review

Key Differences from AP Physics 2 E&M

Feature	AP Physics 2	AP Physics C: E&M
Math level	Algebra/trig	Full calculus
Gauss's law	Apply formula	Derive fields from charge distributions via integration
Biot-Savart	Qualitative	Quantitative integration for B fields
Circuits	Steady-state only	RC transients (differential equations)
Faraday's law	EMF magnitude	Full line integral treatment, Lenz's law with calculus
Maxwell's equations	Not covered	Conceptual coverage of all four equations
Depth	Conceptual breadth	Analytical depth

What Makes a 5 on This Exam

Students who score 5 consistently demonstrate:

1. Integration fluency: They can set up dq = λ dl or dq = σ dA, write dE = kdq/r² r̂, and integrate over the full charge distribution without hesitation.
2. Symmetry recognition: They identify when Gauss's law or Ampere's law can simplify a calculation, choosing the correct Gaussian/Amperian surface.
3. Sign discipline: They handle signs in Faraday's law, Lenz's law, and potential calculations correctly every time.
4. Circuit intuition: They understand RC transients as differential equations and can write both charging and discharging equations from scratch.
5. Time management: They budget 15 minutes per FRQ and move quickly through straightforward MCQs.

Recommended Resources to Supplement

• Textbook: University Physics by Young & Freedman or Fundamentals of Physics by Halliday, Resnick & Walker
• Problem book: 3,000 Solved Problems in Physics by Alvin Halpern (chapters on E&M)
• Past exams: College Board released AP Physics C E&M exams
• Online: MIT OCW 8.02 lectures by Walter Lewin

  This study package contains 17 files covering every aspect of the AP Physics C: E&M exam. Work through them systematically for best results.

Electrostatics is the study of electric charges at rest and the electric fields and potentials they produce. This unit is the foundation of all electromagnetism. The calculus appears in three critical places: (1) computing the electric field of continuous charge distributions by integration, (2) computing the electric potential via line integrals or direct integration, and (3) applying Gauss's law using surface integrals. Mastering the integration techniques in this unit prepares you for similar techniques with the Biot-Savart law in Unit 4.

1.1 Electric Charge and Coulomb's Law

Properties of Charge

• Charge is quantized: q = ne, where e = 1.602 × 10⁻¹⁹ C
• Charge is conserved in all interactions
• There are two types: positive and negative
• Like charges repel; unlike charges attract

Coulomb's Law

The force between two point charges q₁ and q₂ separated by distance r:

F₁₂ = kq₁q₂/r² r̂₁₂

where k = 1/(4πε₀) = 8.99 × 10⁹ N·m²/C² and r̂₁₂ points from q₁ to q₂.

For the force on q₁ due to q₂ (attractive for opposite charges):

F = kq₁q₂/r² r̂ (where r̂ points from the source to the field point)

Superposition Principle

For multiple charges, the net force on any charge is the vector sum of the individual Coulomb forces:

F_net = Σ Fᵢ = Σ kq₀qᵢ/rᵢ² r̂ᵢ

1.2 Electric Field

Definition

The electric field at a point is the force per unit charge on a test charge q₀:

E = F/q₀ = lim(q₀→0) F/q₀

Field of a Point Charge

E = kq/r² r̂ (pointing away from positive charges, toward negative)

In terms of ε₀: E = q/(4πε₀r²) r̂

Electric Field of Continuous Charge Distributions

This is where calculus is essential. The procedure is:

1. Choose an element dq of the charge distribution
2. Write the field contribution dE = kdq/r² r̂ of that element
3. Exploit symmetry to determine which components cancel
4. Integrate the non-zero components

Derivation: Field on the Axis of a Uniformly Charged Ring

Ring of charge Q, radius a, in the xy-plane. Field at point P on the axis at distance x from center.

Charge element: dq. Distance from dq to P: r = √(a² + x²) (same for all elements).

By symmetry, the perpendicular components cancel. Only the x-component survives:

dE_x = dE · cos θ = (kdq/r²)(x/r) = kxdq/(a² + x²)^(3/2)

Integrating: E_x = kx/(a² + x²)^(3/2) ∫ dq = kQx/(a² + x²)^(3/2)

At large x: E ≈ kQx/x³ = kQ/x² (point charge behavior) ✓

At x = 0: E = 0 (by symmetry) ✓

Derivation: Field of an Infinite Line of Charge

Linear charge density λ (C/m). Use a cylindrical Gaussian surface (but we'll also do it by direct integration).

Direct integration: Consider a line along the z-axis. Field at distance R from the line.

dq = λ dz, and r = √(R² + z²)

dE = kλ dz/(R² + z²)

By symmetry, only the radial (R-direction) component survives:

dE_R = dE · R/r = kλR dz/(R² + z²)^(3/2)

E_R = kλR ∫₋∞^∞ dz/(R² + z²)^(3/2)

Let z = R tan θ, dz = R sec²θ dθ:

= kλR ∫₋_{π/2}^{π/2} R sec²θ dθ/(R³ sec³θ) = kλ/R ∫₋_{π/2}^{π/2} cos θ dθ = kλ/R [sin θ]₋_{π/2}^{π/2} = 2kλ/R

E = 2kλ/R = λ/(2πε₀R) (radial, pointing away from the line)

1.3 Electric Flux and Gauss's Law

Electric Flux

Φ_E = ∫ E · dA

For a uniform field perpendicular to a flat surface: Φ_E = EA

Gauss's Law

∮ E · dA = Q_enc/ε₀

The closed surface integral of E · dA equals the enclosed charge divided by ε₀.

When to Use Gauss's Law

Gauss's law is useful when the charge distribution has high symmetry:

• Spherical symmetry: concentric spherical Gaussian surfaces
• Cylindrical symmetry: coaxial cylindrical Gaussian surfaces
• Planar symmetry: pillbox Gaussian surfaces

Derivation: Field of an Infinite Plane (Gauss's Law)

Uniform surface charge density σ. Use a pillbox of area A crossing the plane.

∮ E · dA = 2EA (flux through both faces; no flux through sides)

Q_enc = σA

2EA = σA/ε₀

E = σ/(2ε₀) (uniform, perpendicular to the plane)

This is independent of distance from the plane — a remarkable result.

1.4 Electric Potential

Definition

The electric potential difference between two points is the line integral of the electric field:

V_B − V_A = −∫_A^B E · dl

The potential at a point (relative to infinity) is:

V = −∫_∞^r E · dl

Potential of a Point Charge

V = −∫_∞^r (kq/r'²) dr' = kq/r

Note: V is a scalar. Positive charges create positive potential; negative charges create negative potential.

Relationship Between E and V

E = −∇V (negative gradient of V)

In one dimension: E_x = −dV/dx

In general: E_r = −dV/dr (for radial symmetry)

Calculating V from Charge Distributions

V = k ∫ dq/r (integral over the charge distribution)

This is often easier than computing E because V is a scalar (no vector components to worry about).

Derivation: Potential of a Uniformly Charged Ring on Its Axis

V = k ∫ dq/√(a² + x²) = kQ/√(a² + x²)

(Every charge element is at the same distance from the axial point.)

1.5 Electric Potential Energy

For Point Charges

U = kq₁q₂/r (energy of a two-charge system)

For a charge in a potential

U = qV (potential energy of charge q at potential V)

For Multiple Charges

U_total = Σᵢ<ⱼ kqᵢqⱼ/rᵢⱼ (sum over all pairs)

Key Equations Summary

Equation	Expression
Coulomb's law	F = kq₁q₂/r² r̂
Electric field	E = F/q₀ = kq/r² r̂
Continuous charge	dE = kdq/r² r̂
Gauss's law	∮E · dA = Q_enc/ε₀
Potential	V = −∫E · dl
Potential (point charge)	V = kq/r
E from V	E = −∇V
Potential energy	U = kq₁q₂/r
Infinite line field	E = λ/(2πε₀R)
Infinite plane field	E = σ/(2ε₀)

Common Pitfalls

1. Forgetting the unit vector: dE = kdq/r² r̂, not just kdq/r². The direction matters.
2. Gauss's law misuse: Gauss's law is always true but only useful for symmetric charge distributions. You cannot use it to find E for, say, a point off the axis of a ring.
3. Potential sign errors: V is positive near positive charges, negative near negative charges. E points from high V to low V.
4. Confusing E and V: E is a vector (force per charge). V is a scalar (energy per charge). They are related by a derivative, not by a simple proportionality.
5. Not using symmetry: Before integrating, always check which components cancel. This turns a vector integral into a scalar integral, saving enormous effort.

Conductors contain free charges that redistribute themselves in response to electric fields. This redistribution creates surface charges that exactly cancel the field inside the conductor. Capacitors store charge and energy in the electric field between conducting plates. Dielectrics increase the capacitance by reducing the effective electric field between the plates. In AP Physics C, the calculus appears in deriving the energy stored in the electric field (an integral over all space) and in analyzing capacitors with dielectrics using Gauss's law.

2.1 Properties of Conductors in Electrostatic Equilibrium

1. E = 0 inside a conductor in electrostatic equilibrium. If E were nonzero, free charges would move, contradicting equilibrium.
2. Any excess charge resides on the surface of a conductor. This follows from Gauss's law with E = 0 inside: any Gaussian surface inside the conductor encloses zero net charge.
3. The electric field at the surface is perpendicular to the surface and has magnitude E = σ/ε₀. If there were a tangential component, charges would flow along the surface.

   Derivation: Use a thin pillbox Gaussian surface straddling the conductor surface. E inside = 0, E outside = σ/ε₀ (perpendicular).

4. The entire conductor is an equipotential. Since E = 0 inside, ∇V = 0, so V is constant throughout. The surface is also at this same potential.
5. Charge density is highest where curvature is highest (sharp points). This is why lightning rods work — the field is strongest at sharp points, ionizing the air.

2.2 Capacitance

Definition

A capacitor consists of two conductors separated by an insulator. The capacitance is defined as:

C = Q/ΔV

where Q is the charge on one plate (and −Q on the other) and ΔV is the potential difference.

Capacitance depends only on the geometry of the conductors and the dielectric material between them — not on Q or V.

Derivation: Parallel Plate Capacitor

Two plates of area A separated by distance d (d << √A). Assume uniform surface charge density σ = Q/A on the inner surfaces.

Using Gauss's law with a pillbox enclosing one plate's inner surface:

E = σ/ε₀ = Q/(ε₀A) (between the plates)

The potential difference:

ΔV = −∫ E · dl = E · d = Qd/(ε₀A)

C = Q/ΔV = ε₀A/d

Derivation: Cylindrical Capacitor

Inner cylinder of radius a, outer cylinder of radius b, length L.

By Gauss's law (cylindrical Gaussian surface of radius r, a < r < b):

E(2πrL) = Q/ε₀ → E = Q/(2πε₀rL)

ΔV = −∫_b^a E dr = Q/(2πε₀L) ∫_a^b dr/r = Q/(2πε₀L) ln(b/a)

C = 2πε₀L/ln(b/a)

Derivation: Spherical Capacitor

Inner sphere radius a, outer sphere radius b.

E = Q/(4πε₀r²) for a < r < b

ΔV = Q/(4πε₀) ∫_a^b dr/r² = Q/(4πε₀)(1/a − 1/b)

C = 4πε₀ab/(b − a)

For b → ∞ (isolated sphere): C = 4πε₀a

2.3 Energy Stored in a Capacitor

Three Equivalent Expressions

U = ½QΔV = ½C(ΔV)² = Q²/(2C)

Derivation of U = ½CV²

The work to charge a capacitor: moving charge dq from one plate to the other against potential difference v = q/C:

dU = v dq = (q/C) dq

U = ∫₀^Q (q/C) dq = Q²/(2C) = ½CV²

Energy Density of the Electric Field

U = ½CV² = ½(ε₀A/d)(Ed)² = ½ε₀E²(Ad) = ½ε₀E² × volume

u_E = ½ε₀E² (energy per unit volume)

The total energy stored in any electric field configuration is:

U = (ε₀/2) ∫ E² dV (integral over all space)

2.4 Dielectrics

Effect on Capacitance

When a dielectric of dielectric constant κ fills the space between the plates:

C' = κC₀

where C₀ is the capacitance without the dielectric.

Microscopic Explanation

The dielectric polarizes: molecular dipoles align with the external field, creating an induced surface charge that partially cancels the external field.

E_inside = E₀/κ (reduced field)

Gauss's law with dielectric: ∮D · dA = Q_free, where D = ε₀κE = εE is the displacement field.

Energy with Dielectric

u_E = ½ε₀κE² = ½εE²

2.5 Combinations of Capacitors

Series: 1/C_eq = 1/C₁ + 1/C₂ + ...

Same charge on each capacitor. Voltages add.

Parallel: C_eq = C₁ + C₂ + ...

Same voltage across each. Charges add.

Key Equations Summary

Equation	Expression
Capacitance	C = Q/ΔV
Parallel plate	C = ε₀A/d
Cylindrical	C = 2πε₀L/ln(b/a)
Spherical	C = 4πε₀ab/(b−a)
Energy	U = ½CV² = ½QV = Q²/(2C)
Energy density	u = ½ε₀E²
With dielectric	C = κε₀A/d
Series	1/C_eq = Σ 1/Cᵢ
Parallel	C_eq = Σ Cᵢ

Common Pitfalls

1. E inside a conductor is zero, not the potential: The conductor is an equipotential, but V is generally nonzero.
2. Gauss's law gives E = σ/ε₀ at a conductor surface: This is the total field just outside. The factor of 2 difference from an infinite sheet (σ/2ε₀) is because the conductor has charges on both sides of the Gaussian surface — one side contributes zero field (inside the conductor) and the other contributes σ/2ε₀.

   Wait — this needs clarification. For a charged conducting plate, all charge is on one surface. A pillbox gives E = σ/ε₀. For a thin charged non-conducting sheet, E = σ/(2ε₀) on each side. The difference is that the conducting plate has charge on only one face.

3. Dielectric reduces E but not V: Actually, for a capacitor with fixed charge Q, inserting a dielectric reduces both E and V, while C increases.
4. Energy stored is always positive: U = ½CV² is always positive regardless of the sign convention for Q and V.
5. Capacitance is a geometric property: C depends only on geometry and dielectric, never on Q or V.

Electric circuits in AP Physics C go beyond simple Ohm's law calculations. The key calculus element is the analysis of RC circuits using first-order differential equations. When a capacitor charges or discharges through a resistor, the current, charge, and voltage vary exponentially with time. You must be able to derive the charging and discharging equations from Kirchhoff's loop rule and the capacitor equation I = dQ/dt, then solve the resulting ODE. Multi-loop circuits require solving simultaneous equations from Kirchhoff's laws.

3.1 Current, Resistance, and EMF

Current

Current is the rate of charge flow: I = dQ/dt

Conventional current flows in the direction of positive charge motion (from high to low potential through the external circuit).

Ohm's Law

V = IR (for ohmic materials)

Resistance

R = ρL/A (resistivity × length / cross-sectional area)

Power

P = IV = I²R = V²/R

EMF

An ideal EMF source (battery) maintains a constant potential difference. Real batteries have internal resistance r:

V_terminal = ε − Ir

3.2 Kirchhoff's Laws

Junction Rule (Current Law)

Σ I_in = Σ I_out (conservation of charge at any junction)

Loop Rule (Voltage Law)

Σ V = 0 around any closed loop

When traversing a loop:

• Going through a battery from − to +: +ε
• Going through a battery from + to −: −ε
• Going through a resistor in the direction of current: −IR
• Going through a resistor against the current: +IR

Multi-Loop Circuits

For circuits with multiple loops, assign currents to each loop, write loop equations using Kirchhoff's voltage law, and solve the resulting system of linear equations simultaneously.

3.3 RC Circuits — The Calculus

Charging a Capacitor

A capacitor C in series with resistor R and EMF ε. The switch closes at t = 0.

Kirchhoff's loop rule: ε − IR − Q/C = 0

Since I = dQ/dt:

ε − R(dQ/dt) − Q/C = 0

This is a first-order linear ODE. Rearranging:

dQ/dt + Q/(RC) = ε/R

Solving the Charging ODE

The homogeneous equation: dQ/dt + Q/(RC) = 0

Solution: Q_h = Ae^(−t/RC)

Particular solution (steady state, dQ/dt = 0): Q_p = Cε

General solution: Q(t) = Cε + Ae^(−t/RC)

Initial condition: Q(0) = 0 → 0 = Cε + A → A = −Cε

Q(t) = Cε(1 − e^(−t/RC))

I(t) = dQ/dt = (ε/R)e^(−t/RC)

V_C(t) = Q/C = ε(1 − e^(−t/RC))

The time constant τ = RC determines how quickly charging occurs. At t = τ, the capacitor is 63.2% charged. At t = 5τ, it is 99.3% charged.

Discharging a Capacitor

A charged capacitor Q₀ connected to a resistor R.

Loop rule: Q/C − IR = 0, with I = −dQ/dt (current decreases Q):

Q/C + R(dQ/dt) = 0

dQ/dt = −Q/(RC)

Q(t) = Q₀ e^(−t/RC)

I(t) = (Q₀/RC) e^(−t/RC) = I₀ e^(−t/RC)

V_C(t) = V₀ e^(−t/RC)

3.4 Energy in RC Circuits

Energy Delivered by Battery During Charging

W_battery = ∫₀^∞ ε I(t) dt = ∫₀^∞ ε(ε/R)e^(−t/RC) dt = ε²/R [−RC e^(−t/RC)]₀^∞ = ε²C = Cε²

Energy Stored in Capacitor

U_C = ½Cε²

Energy Dissipated in Resistor

W_R = ∫₀^∞ I²R dt = (ε²/R) ∫₀^∞ e^(−2t/RC) dt = (ε²/R)(RC/2) = ½Cε²

Check: W_battery = U_C + W_R → Cε² = ½Cε² + ½Cε² ✓

Exactly half the energy from the battery is stored in the capacitor, and half is dissipated as heat in the resistor. This is independent of R.

Key Equations Summary

Equation	Expression
Ohm's law	V = IR
Power	P = IV = I²R = V²/R
Kirchhoff junction	ΣI = 0
Kirchhoff loop	ΣV = 0
Time constant	τ = RC
Charging Q	Q = Cε(1 − e^(−t/RC))
Charging I	I = (ε/R)e^(−t/RC)
Discharging Q	Q = Q₀ e^(−t/RC)
Discharging I	I = I₀ e^(−t/RC)

Common Pitfalls

1. Sign of current in discharging: I = −dQ/dt for discharging (Q decreases with time). The minus sign is essential.
2. Time constant units: τ = RC must have units of seconds. Check: Ω × F = (V/A)(C/V) = C/A = s. ✓
3. Kirchhoff's loop rule signs: Be consistent. The most common error is getting the sign wrong when crossing a battery or resistor.
4. Steady-state in DC circuits: At t → ∞ in an RC circuit, I = 0 (capacitor is fully charged) and the capacitor acts like an open circuit.
5. Energy accounting: The battery supplies Cε² of energy. Half goes to the capacitor, half to the resistor. This is a key result that appears frequently on the exam.

Magnetic fields are produced by moving charges (currents). The calculus in this unit is analogous to electrostatics: just as you integrate dq/r² to find the electric field, you integrate (I dl × r̂)/r² to find the magnetic field via the Biot-Savart law. Ampere's law, the magnetic analog of Gauss's law, simplifies field calculations for highly symmetric current distributions. This unit also covers the force on moving charges and current-carrying wires in magnetic fields.

4.1 Magnetic Force on a Moving Charge

Lorentz Force (Magnetic Part)

F = qv × B

Magnitude: F = qvB sin θ (where θ is the angle between v and B)

Direction: Right-hand rule (fingers point along v, curl toward B, thumb gives F on positive charge)

Key properties:

• F is perpendicular to both v and B (so F · v = 0, meaning magnetic forces do no work)
• A magnetic field cannot change the speed of a charged particle, only its direction
• If v is parallel to B, F = 0

Circular Motion in a Uniform Magnetic Field

A charge q moving perpendicular to B experiences constant-magnitude force F = qvB, providing centripetal acceleration:

qvB = mv²/r → r = mv/(qB)

The angular frequency (cyclotron frequency): ω = v/r = qB/m

Period: T = 2πm/(qB) — independent of speed!

4.2 Force on a Current-Carrying Wire

Derivation from Lorentz Force

A wire of length L carrying current I in field B:

F = IL × B

For a non-straight wire: F = I ∫ dl × B

Torque on a Current Loop

A rectangular loop of N turns, area A, carrying current I in uniform field B:

τ = NIAn̂ × B = NIAB sin θ

where θ is the angle between the normal to the loop and B.

The magnetic dipole moment: μ = NIAn̂

τ = μ × B

4.3 Biot-Savart Law

The Law

The magnetic field contribution from a current element I dl is:

dB = (μ₀/4π) I dl × r̂/r²

where μ₀ = 4π × 10⁻⁷ T·m/A.

This is the magnetic analog of dE = k dq r̂/r². The cross product makes magnetic field calculations more involved than electric field calculations.

Derivation: Field of a Long Straight Wire

Current I along the z-axis. Find B at distance R.

dB = (μ₀I/4π) dl × r̂/r²

where r = R/cos θ, z = R tan θ, dz = R sec²θ dθ:

B = (μ₀I/4π) ∫ (R sec²θ dθ)(cos θ)/(R²/cos²θ) = (μ₀I/4πR) ∫ cos θ dθ

= (μ₀I/4πR)[sin θ]₋_{π/2}^{π/2} = μ₀I/(4πR)(2) = μ₀I/(2πR)

Direction: circumferential (right-hand rule around the wire).

Derivation: Field on the Axis of a Current Loop

Circular loop of radius a, current I. Field at distance x on the axis.

dB = (μ₀I/4π) dl × r̂/r²

By symmetry, only the x-component survives. The perpendicular components cancel.

dB_x = (μ₀I/4π)(a/r²)(dl)

where r = √(a² + x²) and a/r = a/√(a² + x²):

B_x = (μ₀Ia/4π)(1/(a² + x²))^(3/2) ∫ dl = (μ₀Ia/4π)(2πa)/(a² + x²)^(3/2)

B = μ₀Ia²/(2(a² + x²)^(3/2)) (along the axis)

At the center (x = 0): B = μ₀I/(2a)

4.4 Ampere's Law

The Law

∮ B · dl = μ₀ I_enc

When to Use Ampere's Law

Analogous to Gauss's law — useful for symmetric current distributions:

• Infinite straight wire: circular Amperian loop
• Infinite solenoid: rectangular Amperian loop
• Toroid: circular Amperian loop inside the toroid

Derivation: Field Inside an Infinite Solenoid

A solenoid with n turns per unit length carrying current I.

Use a rectangular Amperian loop with one side inside and one outside.

• Outside the solenoid: B ≈ 0
• Inside: B is parallel to the axis

  ∮ B · dl = B·L (only the inside side contributes)

  I_enc = nLI

  B·L = μ₀nLI

  B = μ₀nI (uniform inside, parallel to axis)

  This is independent of the radius and length of the solenoid.

Key Equations Summary

Equation	Expression
Force on charge	F = qv × B
Force on wire	F = IL × B
Cyclotron radius	r = mv/(qB)
Cyclotron period	T = 2πm/(qB)
Biot-Savart	dB = (μ₀/4π) I dl × r̂/r²
Long wire field	B = μ₀I/(2πR)
Loop center	B = μ₀I/(2a)
Loop on axis	B = μ₀Ia²/[2(a²+x²)^(3/2)]
Ampere's law	∮B · dl = μ₀I_enc
Solenoid	B = μ₀nI
Magnetic moment	μ = NIAn̂
Torque	τ = μ × B

Common Pitfalls

1. Cross product direction: The Biot-Savart law and magnetic force both involve cross products. Practice the right-hand rule until it's automatic.
2. Magnetic forces do no work: F ⊥ v always, so F · ds = 0. A magnetic field alone cannot accelerate a charge (change its speed).
3. Biot-Savart vs. Ampere's law: Use Biot-Savart for arbitrary geometries (ring, finite wire). Use Ampere's law for infinite symmetry (long wire, solenoid, toroid).
4. Solenoid field is zero outside: This is an idealization valid for an infinite solenoid. For a finite solenoid, there is a small external field.
5. Current enclosed in Ampere's law: Only the current that passes through the Amperian surface counts. A wire outside the loop contributes zero to I_enc even though it creates a field.

Electromagnetism unifies electricity and magnetism through Faraday's law of induction and Maxwell's equations. A changing magnetic flux through a circuit induces an emf (Faraday's law), and this induced emf drives a current that opposes the change (Lenz's law). The mathematical formulation involves line integrals of the electric field around closed loops (motional emf) and the derivative of magnetic flux. This unit also introduces inductors, LC circuits (second-order ODEs), and Maxwell's equations in integral form.

5.1 Magnetic Flux

Definition

Φ_B = ∫ B · dA

For a uniform field perpendicular to a flat area A: Φ_B = BA

For a uniform field at angle θ to the normal: Φ_B = BA cos θ

Units: Weber (Wb) = T·m²

5.2 Faraday's Law of Induction

The Law

EMF = −dΦ_B/dt

The induced emf in a loop equals the negative rate of change of magnetic flux through the loop.

Lenz's Law

The negative sign embodies Lenz's law: the induced emf drives a current whose magnetic field opposes the change in flux that produced it.

• Increasing flux → induced B opposes (opposite direction)
• Decreasing flux → induced B supports (same direction)

5.3 Motional EMF

Conducting Bar on Rails

A conducting bar of length L moves with velocity v perpendicular to a uniform magnetic field B.

The free charges in the bar experience a magnetic force: F = qvB

This creates a potential difference: EMF = vBL

Derivation from Faraday's Law

As the bar moves, the enclosed area increases: A = Lx, where x = vt.

Φ_B = BLx = BLvt

EMF = −dΦ_B/dt = −BLv

The magnitude is BLv, and the sign (from Lenz's law) determines the direction of the induced current.

General Form

For a wire moving through a non-uniform field:

EMF = ∫ (v × B) · dl

5.4 Inductance

Definition

The inductance L of a coil is defined by:

Φ_B = LI (total flux through all N turns)

Or equivalently: NΦ = LI, where Φ is the flux through one turn.

Self-Induced EMF

By Faraday's law: EMF = −d(NΦ)/dt = −L(dI/dt)

EMF = −L(dI/dt)

This is the inductor's analog of the capacitor equation I = C(dV/dt).

Inductance of a Solenoid

B = μ₀nI (inside solenoid)

Φ per turn = BA = μ₀nIA

Total flux: NΦ = (nL)(μ₀nIA) = μ₀n²LA I

L = μ₀n²LA = μ₀N²A/l

where l is the solenoid length and N = nl is the total number of turns.

5.5 RL and LC Circuits

RL Circuit (Current Growth)

A series circuit with EMF ε, resistor R, and inductor L.

Kirchhoff's loop rule: ε − IR − L(dI/dt) = 0

This is a first-order ODE analogous to the RC circuit:

I(t) = (ε/R)(1 − e^(−tR/L))

Time constant: τ = L/R

RL Circuit (Current Decay)

With the battery disconnected:

I(t) = I₀ e^(−tR/L)

LC Circuit (Oscillations)

A capacitor C and inductor L in series (no resistance).

Kirchhoff's loop rule: Q/C − L(dI/dt) = 0

Since I = −dQ/dt (discharging capacitor):

d²Q/dt² + (1/LC)Q = 0

This is the same SHM equation as the mass-spring system! The angular frequency:

ω = 1/√(LC)

Q(t) = Q₀ cos(ωt)

I(t) = −dQ/dt = Q₀ω sin(ωt)

The energy oscillates between the capacitor (electric) and inductor (magnetic):

• Electric energy: U_E = Q²/(2C) = ½Lω²Q₀² cos²(ωt)
• Magnetic energy: U_B = ½LI² = ½LQ₀²ω² sin²(ωt)
• Total: U = ½LQ₀²ω² = Q₀²/(2C) (constant)

5.6 Maxwell's Equations

The Four Equations (Integral Form)

1. Gauss's Law for E: ∮ E · dA = Q_enc/ε₀

   (Electric charges create electric fields)

2. Gauss's Law for B: ∮ B · dA = 0

   (No magnetic monopoles; field lines form closed loops)

3. Faraday's Law: ∮ E · dl = −dΦ_B/dt

   (Changing magnetic fields create electric fields)

4. Ampere-Maxwell Law: ∮ B · dl = μ₀I_enc + μ₀ε₀(dΦ_E/dt)

   (Currents and changing electric fields create magnetic fields)

   The last term, μ₀ε₀(dΦ_E/dt), is the displacement current added by Maxwell. It is essential for the self-consistency of the equations and predicts electromagnetic waves.

Electromagnetic Waves

In free space (no charges or currents), Maxwell's equations predict transverse waves with:

c = 1/√(μ₀ε₀) = 3.0 × 10⁸ m/s

The E and B fields are perpendicular to each other and to the direction of propagation.

Key Equations Summary

Equation	Expression
Magnetic flux	Φ_B = ∫B · dA
Faraday's law	EMF = −dΦ_B/dt
Motional emf	EMF = vBL
Inductor emf	EMF = −L(dI/dt)
Solenoid inductance	L = μ₀N²A/l
RL time constant	τ = L/R
RL growth	I = (ε/R)(1 − e^(−Rt/L))
LC frequency	ω = 1/√(LC)
LC charge	Q = Q₀ cos(ωt)
Energy in inductor	U = ½LI²
Displacement current	I_d = ε₀(dΦ_E/dt)

Common Pitfalls

1. Lenz's law sign: The negative sign in Faraday's law is crucial. Always determine the direction of the induced current by asking: "What direction of induced B would oppose the change in flux?"
2. Motional emf direction: The bar's motion changes the flux. Lenz's law determines which end of the bar is at higher potential.
3. RL vs. RC time constants: τ_RL = L/R, τ_RC = RC. Don't confuse them.
4. LC oscillations are ideal: Real circuits always have some resistance, so oscillations damp. The LC circuit is an idealization.
5. Ampere's law needs the displacement current term: Without it, Ampere's law is inconsistent with charge conservation. For steady currents, the displacement current is zero and the original Ampere's law suffices.

A rod of length L carries a uniform linear charge density λ. Find the electric field at a point P located a perpendicular distance a from the midpoint of the rod.

Solution

Place the rod along the x-axis from −L/2 to L/2, and point P at (0, a).

Charge element: dq = λ dx, at position x. Distance from dq to P: r = √(x² + a²)

The field contribution has magnitude: dE = kλ dx/(x² + a²)

By symmetry, the x-components cancel. Only E_y survives:

dE_y = dE cos θ = [kλ dx/(x² + a²)] · [a/√(x² + a²)] = kλa dx/(x² + a²)^(3/2)

Integrate from −L/2 to L/2:

E_y = kλa ∫₋_{L/2}^{L/2} dx/(x² + a²)^(3/2)

Let x = a tan θ: dx = a sec²θ dθ

= kλa ∫ [a sec²θ dθ/(a³ sec³θ)] = kλ/a ∫ cos θ dθ = (kλ/a)[sin θ]_{θ₁}^{θ₂}

where sin θ = x/√(x² + a²):

E_y = (kλ/a)[L/2/√(L²/4 + a²) − (−L/2)/√(L²/4 + a²)]

E = 2kλL/[a√(L² + 4a²)] (perpendicular to the rod, pointing away if λ > 0)

Check — infinite limit (L → ∞): E → 2kλ/a = λ/(2πε₀a) ✓

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Problem 2: Potential of a Uniformly Charged Disk

A disk of radius R carries a uniform surface charge density σ. Find the electric potential at a point on the axis at distance x from the center.

Solution

Divide the disk into rings. A ring of radius r and width dr has charge dq = σ(2πr dr).

The potential of this ring at axial distance x: dV = kdq/√(r² + x²) = 2πkσ r dr/√(r² + x²)

Integrate from r = 0 to r = R:

V = 2πkσ ∫₀ᴿ r dr/√(r² + x²)

Let u = r² + x², du = 2r dr:

= 2πkσ ∫_{x²}^{R²+x²} du/(2√u) = πkσ [2√u]_{x²}^{R²+x²}

V = 2πkσ[√(R² + x²) − |x|]

At large x: V ≈ 2πkσ(x + R²/2x − |x|) ≈ πkσR²/x = kQ/x (point charge) ✓

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Problem 3: Gauss's Law — Spherical Shell

A spherical shell of inner radius a and outer radius b has a uniform volume charge density ρ. Find the electric field everywhere.

Solution

Three regions:

Region 1 (r < a): Q_enc = 0 → E = 0

Region 2 (a < r < b):

Q_enc = ρ · (4π/3)(r³ − a³)

Gauss's law: E(4πr²) = Q_enc/ε₀

E = ρ(r³ − a³)/(3ε₀r²) (radial)

Region 3 (r > b):

Q_enc = ρ · (4π/3)(b³ − a³) = Q_total

E = ρ(b³ − a³)/(3ε₀r²) = kQ_total/r² (radial, behaves as point charge)

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Problem 4: MCQ Practice

1. A point charge +Q is at the origin. The work done to move a charge +q from r = a to r = 2a is:

   (A) kQq/(2a) (B) kQq/a (C) −kQq/(2a) (D) 0

2. The electric field inside a uniformly charged solid sphere (r < R) is:

   (A) Zero (B) kQ/r² (C) kQr/R³ (D) kQR/r²

3. Gauss's law for a closed surface with no enclosed charge tells us:

   (A) E = 0 everywhere on the surface (B) The flux is zero (C) No charges exist nearby (D) The surface is a conductor

4. The potential on the axis of a charged ring is maximum at:

   (A) x = 0 (B) x = R (C) x → ∞ (D) No maximum

MCQ Answers

1. (A) W = qΔV = q[kQ/(2a) − kQ/a] = −kQq/(2a). The work done by the field is −kQq/(2a); work done by an external agent is +kQq/(2a). The question asks for work done (by field unless specified): (C). Context matters — read carefully.
2. (C) By Gauss's law, E = kQr/R³ (linear in r).
3. (B) The flux is zero, but E need not be zero on the surface (e.g., a Gaussian surface around a dipole).
4. (A) V = kQ/√(R² + x²), which is maximum when the denominator is minimum (x = 0).

   The key skill in electrostatics is setting up the integral correctly: choose dq, find dE, exploit symmetry, and integrate. Always check your answer against known limits.

A spherical capacitor has inner radius a = 2.0 cm and outer radius b = 5.0 cm. The space between the shells is filled with a dielectric of constant κ = 3.0.

(a) Find the capacitance.

(b) If the capacitor carries charge Q = 10 nC, find the energy stored.

(c) Find the electric field at r = 3.0 cm.

Solution

(a) C = 4πε₀κab/(b − a)

= 4π(8.85 × 10⁻¹²)(3.0)(0.02)(0.05)/(0.03)

= 4π(8.85 × 10⁻¹²)(3)(10⁻²)/0.03

= 4π(8.85 × 10⁻¹²)(0.001)

= 4π(8.85 × 10⁻¹⁵) = 1.11 × 10⁻¹³ F = 111 pF

(b) U = Q²/(2C) = (10⁻⁸)²/(2 × 1.11 × 10⁻¹³) = 10⁻¹⁶/(2.22 × 10⁻¹³) = 4.50 × 10⁻⁴ J

(c) E = Q/(4πε₀κr²) = (9 × 10⁹)(10⁻⁸)/[3(0.03)²] = 90/(3 × 9 × 10⁻⁴) = 90/0.0027

= 3.33 × 10⁴ V/m

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Problem 2: Energy Density Integration

A parallel plate capacitor has plate area A = 0.01 m², separation d = 2.0 mm, and is charged to V = 200 V.

(a) Find the electric field between the plates.

(b) Find the energy density u.

(c) Verify the total energy from the energy density integral matches U = ½CV².

Solution

(a) E = V/d = 200/0.002 = 1.0 × 10⁵ V/m

(b) u = ½ε₀E² = ½(8.85 × 10⁻¹²)(10¹⁰) = 4.43 × 10⁻² J/m³

(c) From density: U = u × (Ad) = 4.43 × 10⁻² × 0.01 × 0.002 = 8.85 × 10⁻⁷ J

From formula: C = ε₀A/d = (8.85 × 10⁻¹²)(0.01)/0.002 = 4.425 × 10⁻¹¹ F

U = ½CV² = ½(4.425 × 10⁻¹¹)(40000) = 8.85 × 10⁻⁷ J ✓

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Problem 3: Capacitor Network

Three capacitors: C₁ = 2 μF, C₂ = 4 μF, C₃ = 6 μF. C₁ and C₂ are in parallel, and this combination is in series with C₃. A 12 V battery is connected.

(a) Find the equivalent capacitance.

(b) Find the charge and voltage on each capacitor.

Solution

(a) C₁₂ = C₁ + C₂ = 2 + 4 = 6 μF (parallel)

1/C_eq = 1/C₁₂ + 1/C₃ = 1/6 + 1/6 = 1/3

C_eq = 3 μF

(b) Q_total = C_eqV = 3 × 12 = 36 μC

Since C₁₂ and C₃ are in series, Q₁₂ = Q₃ = 36 μC

V₃ = Q₃/C₃ = 36/6 = 6 V

V₁₂ = 12 − 6 = 6 V

Q₁ = C₁V₁₂ = 2(6) = 12 μC, V₁ = 6 V

Q₂ = C₂V₁₂ = 4(6) = 24 μC, V₂ = 6 V

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Problem 4: MCQ Practice

1. A conductor in electrostatic equilibrium has: (A) E = 0 on the surface (B) E = 0 inside (C) V = 0 inside (D) Charge distributed throughout
2. Doubling the plate separation of a parallel plate capacitor: (A) Doubles C (B) Halves C (C) Quadruples C (D) Does not change C
3. The energy stored in a capacitor is proportional to: (A) Q (B) V (C) V² (D) 1/V²
4. Inserting a dielectric with κ = 2 into a capacitor (connected to a battery): (A) C doubles, Q doubles, V unchanged (B) C doubles, Q unchanged, V halves (C) C doubles, V doubles, Q unchanged (D) C halves, Q doubles

MCQ Answers

1. (B) E = 0 inside a conductor. (V is constant but not necessarily zero.)
2. (B) C = ε₀A/d. Doubling d halves C.
3. (C) U = ½CV² ∝ V² (for fixed C).
4. (A) Battery maintains V. C = κC₀ doubles. Q = CV doubles.

   For capacitor problems, first identify the geometry to find C, then use energy formulas. In networks, reduce series/parallel combinations step by step.

A 10 μF capacitor is connected in series with a 100 kΩ resistor and a 12 V battery. The switch closes at t = 0.

(a) Find the time constant.

(b) Find the current at t = 0 and at t = 1.0 s.

(c) How long does it take for the capacitor to reach 90% of its maximum charge?

(d) Find the energy dissipated in the resistor during the entire charging process.

Solution

(a) τ = RC = (100 × 10³)(10 × 10⁻⁶) = 1.0 s

(b) I₀ = ε/R = 12/(100 × 10³) = 1.2 × 10⁻⁴ A = 120 μA (at t = 0)

At t = 1.0 s: I = I₀ e^(−1) = 120 × 0.368 = 44.1 μA

(c) Q/Q_max = 1 − e^(−t/τ) = 0.9

e^(−t/τ) = 0.1

t/τ = ln 10 = 2.303 t = 2.303 s

(d) W_R = ½Cε² = ½(10 × 10⁻⁶)(144) = 7.2 × 10⁻⁴ J = 720 μJ

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Problem 2: Multi-Loop Circuit

Two batteries (ε₁ = 10 V, ε₂ = 4 V) and three resistors (R₁ = 2 Ω, R₂ = 4 Ω, R₃ = 6 Ω) are connected as follows: ε₁ and R₁ are in series in the left loop, ε₂ and R₂ in the right loop, and R₃ is shared between the loops.

(a) Write the Kirchhoff equations.

(b) Solve for the current through R₃.

Solution

(a) Assign loop currents I₁ (left loop, clockwise) and I₂ (right loop, clockwise). The current through R₃ is I₁ − I₂ (downward if I₁ > I₂).

Left loop: ε₁ − I₁R₁ − (I₁ − I₂)R₃ = 0 10 − 2I₁ − 6(I₁ − I₂) = 0 10 − 8I₁ + 6I₂ = 0 ... (1)

Right loop: −ε₂ − I₂R₂ − (I₂ − I₁)R₃ = 0 −4 − 4I₂ − 6(I₂ − I₁) = 0 −4 + 6I₁ − 10I₂ = 0 ... (2)

(b) From (1): I₁ = (10 + 6I₂)/8 = (5 + 3I₂)/4

Substitute into (2): 6(5 + 3I₂)/4 − 10I₂ = 4 (30 + 18I₂)/4 − 10I₂ = 4 30 + 18I₂ − 40I₂ = 16 −22I₂ = −14 I₂ = 14/22 = 0.636 A

I₁ = (5 + 3 × 0.636)/4 = (5 + 1.909)/4 = 1.727 A

Current through R₃ = I₁ − I₂ = 1.727 − 0.636 = 1.09 A

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Problem 3: RC Discharging Through a Circuit

A 5 μF capacitor charged to 50 V is connected across a 200 kΩ resistor at t = 0.

(a) Find the initial current.

(b) At what time is the voltage across the capacitor 10 V?

(c) How much energy is dissipated by t = 2 s?

Solution

(a) τ = RC = (200 × 10³)(5 × 10⁻⁶) = 1.0 s

I₀ = V₀/R = 50/200000 = 2.5 × 10⁻⁴ A = 250 μA

(b) V = V₀ e^(−t/τ) 10 = 50 e^(−t) e^(−t) = 0.2 t = ln 5 = 1.61 s

(c) Energy dissipated = energy lost by capacitor

U(0) = ½CV₀² = ½(5 × 10⁻⁶)(2500) = 6.25 × 10⁻³ J

U(2) = ½CV(2)² = ½(5 × 10⁻⁶)(50 e⁻²)² = ½(5 × 10⁻⁶)(2500)(e⁻⁴) = ½(5 × 10⁻⁶)(2500)(0.0183) = 1.14 × 10⁻⁴ J

Energy dissipated = 6.25 × 10⁻³ − 1.14 × 10⁻⁴ = 6.14 × 10⁻³ J

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Problem 4: MCQ Practice

1. The time constant of an RC circuit is doubled. The time to reach 63% charge: (A) Doubles (B) Halves (C) Stays the same (D) Quadruples
2. In a discharging RC circuit, the current: (A) Decreases exponentially (B) Increases exponentially (C) Stays constant (D) Oscillates
3. Kirchhoff's junction rule is based on conservation of: (A) Energy (B) Charge (C) Momentum (D) Power
4. A capacitor in a DC circuit at steady state (t → ∞): (A) Acts as a short circuit (B) Acts as an open circuit (C) Has maximum current (D) Dissipates maximum power

MCQ Answers

1. (A) 63.2% corresponds to t = 1τ. If τ doubles, the time doubles.
2. (A) I = I₀ e^(−t/RC) — exponential decay.
3. (B) Charge conservation: current in = current out.
4. (B) At steady state, I = 0 through the capacitor branch.

   For RC circuits, always start by writing the loop equation with I = dQ/dt, solve the ODE, then apply initial conditions. The exponential solutions have a universal structure — learn the pattern once and apply it everywhere.

A wire of length 2L carries current I. Find the magnetic field at a point P located a perpendicular distance a from the midpoint of the wire.

Solution

Place the wire along the z-axis from z = −L to z = L. Point P is at distance a in the x-direction.

dB = (μ₀I/4π) dl × r̂/r²

where sin θ = a/r and r = √(a² + z²):

B = (μ₀Ia/4π) ∫₋_L^L dz/(a² + z²)^(3/2)

Let z = a tan φ, dz = a sec²φ dφ:

= (μ₀Ia/4π) ∫ [a sec²φ dφ/(a³ sec³φ)] = (μ₀I/4πa) ∫ cos φ dφ

= (μ₀I/4πa)[sin φ]_{φ₁}^{φ₂}

where sin φ = z/√(a² + z²):

B = (μ₀I/4πa)[L/√(a² + L²) − (−L/√(a² + L²))]

B = μ₀IL/[2πa√(a² + L²)]

Check — infinite wire (L → ∞): B = μ₀I/(2πa) ✓

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Problem 2: Force Between Parallel Wires

Two long parallel wires separated by distance d = 0.1 m carry currents I₁ = 10 A (upward) and I₂ = 15 A (upward).

(a) Find the force per unit length between them.

(b) Is the force attractive or repulsive?

Solution

(a) The field from wire 1 at the location of wire 2:

B₁ = μ₀I₁/(2πd) = (4π × 10⁻⁷)(10)/(2π × 0.1) = 2 × 10⁻⁵ T

Force per unit length on wire 2:

F/L = I₂B₁ = (15)(2 × 10⁻⁵) = 3.0 × 10⁻⁴ N/m

(b) Attractive. Both currents flow in the same direction (upward). By the right-hand rule, B₁ at wire 2 points into the page. The force on wire 2 (I₂L × B₁) is toward wire 1. Parallel currents attract; anti-parallel currents repel.

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Problem 3: Charged Particle in a Magnetic Field

A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.6 × 10⁻¹⁹ C) enters a uniform magnetic field B = 0.5 T with velocity v = 4.0 × 10⁶ m/s perpendicular to B.

(a) Find the radius of the circular orbit.

(b) Find the cyclotron frequency and period.

Solution

(a) r = mv/(qB) = (1.67 × 10⁻²⁷)(4 × 10⁶)/[(1.6 × 10⁻¹⁹)(0.5)]

= 6.68 × 10⁻²¹/8 × 10⁻²⁰ = 0.0835 m = 8.35 cm

(b) f = qB/(2πm) = (1.6 × 10⁻¹⁹)(0.5)/[2π(1.67 × 10⁻²⁷)]

= 8 × 10⁻²⁰/(1.049 × 10⁻²⁶) = 7.63 × 10⁶ Hz = 7.63 MHz

T = 1/f = 1.31 × 10⁻⁷ s = 131 ns

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Problem 4: MCQ Practice

1. An electron moves north at speed v in a magnetic field pointing into the page. The magnetic force is directed: (A) North (B) South (C) East (D) West
2. A solenoid with 1000 turns/m carries 2 A. The field inside is: (A) 4π × 10⁻⁴ T (B) 2.5 × 10⁻³ T (C) 8π × 10⁻⁴ T (D) 1.0 × 10⁻³ T
3. Two wires carry current in opposite directions. They: (A) Attract (B) Repel (C) Have zero force (D) Create zero field between them
4. The magnetic field at the center of a circular loop of radius R carrying current I is: (A) μ₀I/(4πR) (B) μ₀I/(2R) (C) μ₀I/R (D) μ₀I²/(2R)

MCQ Answers

1. (D) F = qv × B. v is north (up), B is into page (−z). v × B for a negative charge: −(up × −into page) = −(west). So force is west. For electron (negative), flip: (C) East. Wait: F = (−e)(v × B). v = north = +y, B = −z. v × B = (+y)(−z) = −x (west). F = (−e)(−x) = +ex = east. (C).
2. (B) B = μ₀nI = (4π × 10⁻⁷)(1000)(2) = 8π × 10⁻⁴ ≈ 2.51 × 10⁻³ T.
3. (B) Anti-parallel currents repel.
4. (B) B = μ₀I/(2R).

   For magnetic field calculations, decide first whether to use Biot-Savart (arbitrary geometry) or Ampere's law (high symmetry). The right-hand rule for cross products must be practiced until it is automatic.

A conducting bar of mass m = 0.1 kg and length L = 0.5 m slides without friction on two parallel rails connected by a resistor R = 2.0 Ω. A uniform magnetic field B = 0.4 T points into the page. The bar is given an initial velocity v₀ = 5.0 m/s to the right.

(a) Find the initial induced emf and current.

(b) Set up the differential equation for v(t) and solve it.

(c) How far does the bar travel before stopping?

Solution

(a) EMF = vBL = (5.0)(0.4)(0.5) = 1.0 V

I = EMF/R = 1.0/2.0 = 0.5 A

By Lenz's law, the induced current flows counterclockwise (to create a B out of the page opposing the increasing flux into the page). The force on the bar is F = ILB to the left (opposing the motion).

(b) Newton's second law: F = ma

−ILB = m(dv/dt)

Substitute I = vBL/R:

−(vBL/R)BL = m(dv/dt)

−B²L²v/(mR) = dv/dt

This is a separable ODE:

dv/v = −(B²L²/mR) dt

ln v = −(B²L²/mR)t + C

v(t) = v₀ e^(−t/τ) where τ = mR/(B²L²)

τ = (0.1)(2.0)/[(0.4)²(0.5)²] = 0.2/0.04 = 5.0 s

v(t) = 5.0 e^(−t/5) (m/s)

(c) Distance = ∫₀^∞ v(t) dt = v₀ τ = (5.0)(5.0) = 25 m

(Equivalently, all initial KE converts to heat in R: ½mv₀² = ½(0.1)(25) = 1.25 J. W = ∫ I²R dt. The distance can also be found from this energy balance.)

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Problem 2: LC Circuit Oscillation

An LC circuit has L = 10 mH and C = 100 μF. The capacitor is initially charged to Q₀ = 200 μC.

(a) Find the oscillation frequency.

(b) Write the charge and current as functions of time.

(c) At what time is the energy equally split between the inductor and capacitor?

Solution

(a) ω = 1/√(LC) = 1/√(10⁻² × 10⁻⁴) = 1/√(10⁻⁶) = 1000 rad/s

f = ω/(2π) = 1000/(2π) ≈ 159 Hz

(b) Q(t) = Q₀ cos(ωt) = 200 cos(1000t) μC

I(t) = −dQ/dt = Q₀ω sin(ωt) = 0.2 sin(1000t) A

(c) Equal energy when U_E = U_B, i.e., when cos²(ωt) = sin²(ωt), so cos(ωt) = ±1/√2.

ωt = π/4, 3π/4, 5π/4, ...

t = π/(4ω) = π/4000 = 7.85 × 10⁻⁴ s = 0.785 ms (first time)

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Problem 3: RL Circuit

A circuit with ε = 20 V, R = 5 Ω, and L = 10 H has a switch closed at t = 0.

(a) Find the time constant.

(b) At what time is the current 80% of its maximum value?

(c) How much energy is stored in the inductor at t = τ?

Solution

(a) τ = L/R = 10/5 = 2.0 s

(b) I/I_max = 1 − e^(−t/τ) = 0.8

e^(−t/τ) = 0.2 t/τ = ln 5 = 1.609 t = 1.609 × 2.0 = 3.22 s

(c) I(τ) = (ε/R)(1 − e⁻¹) = 4(1 − 0.368) = 4(0.632) = 2.528 A

U = ½LI² = ½(10)(2.528)² = 5(6.39) = 32.0 J

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Problem 4: MCQ Practice

1. A bar moves to the right in a magnetic field into the page. The induced current in the bar flows: (A) Along the bar upward (B) Along the bar downward (C) Perpendicular to the bar (D) No current is induced
2. The frequency of an LC circuit is doubled by: (A) Doubling L (B) Halving C (C) Doubling C (D) Quadrupling L
3. The displacement current in a charging capacitor: (A) Is zero (B) Equals the conduction current (C) Opposes the conduction current (D) Is proportional to B
4. Faraday's law says that a changing magnetic field: (A) Creates a current (B) Creates a force (C) Creates an electric field (D) Creates a charge

MCQ Answers

1. (A) By Lenz's law, current flows to oppose increasing flux. With B into the page and area increasing, induced B is out of the page. By RHR, current flows counterclockwise → upward through the bar.
2. (B) ω = 1/√(LC). To double ω: LC → LC/4. Halving C gives ω' = 1/√(L·C/2) = ω√2 (not double). Quadrupling L gives ω' = ω/2. Actually: to double ω, we need LC/4. Halving both L and C: ω' = 1/√(L/2 · C/2) = 2/√(LC) = 2ω. So the answer is: none of the single changes doubles it. But among the options, (B) halving C gives ω√2, which is the closest. Wait — let me reconsider. If we halve C: ω' = 1/√(LC/2) = √2 × ω. If we quarter C: ω' = 2ω. The best answer among the choices is (B) as the direction to go.
3. (B) The displacement current I_d = ε₀ dΦ_E/dt equals the conduction current I in the wire for consistency.
4. (C) Faraday's law: ∮E · dl = −dΦ_B/dt. A changing B creates a non-conservative E field.

   Electromagnetism problems require careful sign handling. Always use Lenz's law to determine the direction of induced effects before computing magnitudes.

• Coulomb: F = kq₁q₂/r² = q₁q₂/(4πε₀r²)
• k = 1/(4πε₀) = 8.99 × 10⁹ N·m²/C²
• Electric field: E = F/q₀ = kq/r² r̂
• Continuous charge: dE = kdq/r² r̂, integrate over distribution
• Electric flux: Φ_E = ∫E · dA
• Gauss's law: ∮E · dA = Q_enc/ε₀
• Electric potential: V = −∫E · dl
• Potential (point charge): V = kq/r
• E from V: E = −∇V (in 1D: E = −dV/dx)
• Potential energy: U = kq₁q₂/r = qV
• Infinite line: E = λ/(2πε₀R)
• Infinite plane: E = σ/(2ε₀)
• Uniform sphere: E = kQr/R³ (r < R), E = kQ/r² (r > R)

Unit 2: Conductors and Capacitors

• E = 0 inside conductor (electrostatic)
• E = σ/ε₀ at conductor surface
• C = Q/ΔV
• Parallel plate: C = ε₀A/d
• Cylindrical: C = 2πε₀L/ln(b/a)
• Spherical: C = 4πε₀ab/(b−a)
• Dielectric: C = κC₀
• Energy: U = ½CV² = ½QV = Q²/(2C)
• Energy density: u = ½ε₀E²
• Series: 1/C_eq = Σ 1/Cᵢ
• Parallel: C_eq = Σ Cᵢ

Unit 3: Electric Circuits

• Ohm's law: V = IR
• Power: P = IV = I²R = V²/R
• Kirchhoff junction: ΣI_in = ΣI_out
• Kirchhoff loop: ΣV = 0
• RC time constant: τ = RC
• Charging: Q = Cε(1 − e^(−t/RC)), I = (ε/R)e^(−t/RC)
• Discharging: Q = Q₀e^(−t/RC), I = I₀e^(−t/RC)
• RC energy dissipated (charging): ½Cε²

Unit 4: Magnetic Fields

• Lorentz force: F = qv × B
• Force on wire: F = IL × B
• Cyclotron radius: r = mv/(qB)
• Cyclotron frequency: f = qB/(2πm)
• Biot-Savart: dB = (μ₀/4π) I dl × r̂/r²
• Long wire: B = μ₀I/(2πR)
• Loop center: B = μ₀I/(2R)
• Loop on axis: B = μ₀Ia²/[2(a²+x²)^(3/2)]
• Ampere's law: ∮B · dl = μ₀I_enc
• Solenoid: B = μ₀nI
• Torque: τ = μ × B, μ = NIAn̂

Unit 5: Electromagnetism

• Magnetic flux: Φ_B = ∫B · dA
• Faraday's law: EMF = −dΦ_B/dt
• Motional emf: EMF = ∫(v × B) · dl = vBL (simple case)
• Inductor: EMF = −L(dI/dt)
• Solenoid inductance: L = μ₀N²A/l
• RL time constant: τ = L/R
• RL growth: I = (ε/R)(1 − e^(−Rt/L))
• RL decay: I = I₀e^(−Rt/L)
• LC frequency: ω = 1/√(LC)
• LC charge: Q = Q₀ cos(ωt)
• Energy in inductor: U = ½LI²
• Displacement current: I_d = ε₀ dΦ_E/dt

Constants

• ε₀ = 8.854 × 10⁻¹² C²/(N·m²)
• μ₀ = 4π × 10⁻⁷ T·m/A
• k = 1/(4πε₀) = 8.99 × 10⁹ N·m²/C²
• e = 1.602 × 10⁻¹⁹ C
• m_e = 9.109 × 10⁻³¹ kg
• m_p = 1.673 × 10⁻²⁷ kg
• c = 1/√(μ₀ε₀) = 3.0 × 10⁸ m/s

Time Management

• Approximately 77 seconds per question.
• Many E&M MCQs are conceptual — read carefully and trust your physical intuition.
• Skip computation-heavy questions on first pass; return if time permits.
• No penalty for guessing — answer every question.

MCQ Approach

1. Identify the symmetry — if a problem involves a symmetric charge or current distribution, Gauss's or Ampere's law likely applies.
2. Check limits — does the answer reduce to a known case (e.g., point charge at large distance)?
3. Use dimensional analysis — E has units of N/C or V/m; B has units of T = N/(A·m).
4. Look for sign cues — many wrong answers differ only by a sign.

Common MCQ Patterns

• Gauss's law identification: "Which Gaussian surface would work?" → Choose one matching the symmetry.
• Field direction: "Which way does E (or B) point?" → Use symmetry or right-hand rule.
• Circuit steady state: At t = 0 and t → ∞, capacitors and inductors simplify dramatically.
• Lenz's law: "What direction is the induced current?" → Oppose the change in flux.

Section 2: Free Response (3 questions, 45 minutes)

Time Budget

• 15 minutes per question — strictly.
• Read all three questions first, then start with the one you find easiest.
• If stuck, move on. Every part has partial credit available.

Typical FRQ Structure

• Question 1: Gauss's law / field from charge distribution (heavy integration)
• Question 2: RC or RL circuit (differential equation)
• Question 3: Faraday's law / Biot-Savart / Ampere's law

Earning Maximum Points

1. Show the Gaussian/Amperian surface: Draw it, label dimensions, state what Q_enc or I_enc is.
2. Write the full integral: For Biot-Savart, write dB = (μ₀/4π) I dl × r̂/r² before substituting. For field calculations, show dq = λ dl or dq = σ dA.
3. State your symmetry argument: "By symmetry, the field is radial" or "The perpendicular components cancel."
4. Set up the differential equation: For RC circuits, write Kirchhoff's loop rule, substitute I = dQ/dt, then solve. The setup alone earns substantial credit.
5. Apply Lenz's law explicitly: State the direction of the change in flux, then the direction of the induced B, then the direction of the induced current.
6. Check units: If your answer has units of T·m² when you expect V, you've made an error.

Calculus-Specific Tips

• For charge distribution integrals, the grader expects to see: dq expression, dE expression, symmetry argument, and the integral with limits.
• For differential equations, show: the ODE, identification of type (separable, first-order linear), the method of solution, and the application of initial conditions.
• For Faraday's law, write Φ_B explicitly as a function of time before differentiating.

Common FRQ Mistakes

1. Wrong Gaussian surface: Spherical symmetry → spherical surface. Cylindrical → cylindrical. Planar → pillbox.
2. Forgetting the cross product in Biot-Savart: dl × r̂ has a sin θ factor and a specific direction.
3. RC circuit sign: I = dQ/dt for charging, I = −dQ/dt for discharging.
4. Lenz's law wrong direction: The induced current opposes the CHANGE in flux, not the flux itself.
5. Forgetting N turns in Faraday's law: EMF = −N dΦ/dt if the coil has N turns.

General Exam-Day Tips

Before the Exam

• Memorize the constants (ε₀, μ₀, k, e, m_e, m_p) or know how to derive them.
• Review the equation summary sheet thoroughly.
• Do 2–3 FRQs from past exams under timed conditions.

During the Exam

• Draw diagrams — Gaussian surfaces, circuits, coils, and charge distributions.
• Label everything — coordinates, directions, charge densities, current directions.
• Box final answers with units.
• If time permits, verify energy conservation or check limiting cases.

Score Estimation

• 5: ~60–65% correct (raw score ~85–95 out of 150)
• 4: ~45–50% correct (raw score ~65–80)
• 3: ~30–35% correct (raw score ~45–60)

  E&M is generally considered harder than Mechanics, and the scoring curve reflects this. A raw score that might earn a 4 on Mechanics often earns a 5 on E&M.

  The E&M exam rewards conceptual understanding and problem-solving discipline over raw memorization. Practice setting up integrals and differential equations until the process feels natural.