AP Physics C: Electricity & Magnetism Master Class
Module: Ampère's Law, Biot-Savart Integrals & Displacement Current
1. Introduction & AP Exam Weight
The mastery of magnetic fields generated by steady and time-varying currents represents one of the most mathematically demanding segments of the AP Physics C: Electricity & Magnetism curriculum. Content spanning the Biot-Savart Law, Ampère's Law, and the Maxwell-Ampère Law (Displacement Current) directly accounts for 15% to 25% of the total AP Exam weight.
Top-tier performance requires moving beyond plug-and-chug algebraic formulas. The College Board explicitly tests your ability to set up vector line and surface integrals, determine dynamic flux bounds, and apply differential current formulations under non-uniform conditions.
[ Microscopic Currents ] [ Symmetric Current Distribution ]
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Biot-Savart Law Ampère's Law
(Vector Cross-Product Integral) (Closed Path Integral)
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└───────────────────────┬──────────────────────┘
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Time-Varying E-Fields
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Dynamic Maxwell-Ampère Law
(Displacement Current Term)
UC Berkeley Academic Context
Earning a score of 5 on the AP Physics C: E&M exam fulfills the lower-division requirement for Physics 7B (4 units) in the UC Berkeley College of Engineering and College of Letters & Science. Exempting Physics 7B allows immediate advancement into EECS 16B (Designing Information Devices and Systems II), skipping introductory physics sequences and giving you a direct pathway into advanced hardware, circuit physics, and signal processing coursework.
2. Deep Concept Breakdown
A. The Biot-Savart Law: Differential Vector Integration
When spatial symmetry is insufficient to apply Ampère’s Law, the magnetic field $\mathbf{B}$ at a position vector $\mathbf{r}$ relative to a source element $d\boldsymbol{\ell}$ carrying current $I$ must be calculated directly via the Biot-Savart Law:
$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0 I}{4\pi} \int \frac{d\boldsymbol{\ell} \times \hat{\mathbf{r}}}{r^2} = \frac{\mu_0 I}{4\pi} \int \frac{d\boldsymbol{\ell} \times \mathbf{r}}{|\mathbf{r}|^3}$$
Formal Derivation: On-Axis Magnetic Field of a Circular Current Loop
Consider a circular wire loop of radius $R$ lying in the $xy$-plane centered at the origin, carrying a steady current $I$. We evaluate $\mathbf{B}$ at an on-axis point $P = (0, 0, z)$.
z-axis
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│ • P (0, 0, z)
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│ / |
r / | z
/ |
/ |
/ θ |
─────────────────┼─────┴───────────── y-axis
/ R
/
x-axis
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Differential Element Setup: An infinitesimal differential vector along the loop in cylindrical coordinates is: $$d\boldsymbol{\ell} = R \, d\phi \, \hat{\boldsymbol{\phi}}$$
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Position Vector and Distance: The vector pointing from the element $d\boldsymbol{\ell}$ on the loop to point $P$ is: $$\mathbf{r} = -R \, \hat{\mathbf{s}} + z \, \hat{\mathbf{k}}$$ where $\hat{\mathbf{s}}$ is the radial unit vector in the $xy$-plane. The distance magnitude is: $$r = \sqrt{R^2 + z^2}$$
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Evaluating the Vector Cross Product: $$d\boldsymbol{\ell} \times \mathbf{r} = (R \, d\phi \, \hat{\boldsymbol{\phi}}) \times (-R \, \hat{\mathbf{s}} + z \, \hat{\mathbf{k}}) = R^2 \, d\phi \, \hat{\mathbf{k}} + R z \, d\phi \, \hat{\mathbf{s}}$$
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Symmetry Analysis: By rotational symmetry around the $z$-axis, radial components integrate to zero: $\int \hat{\mathbf{s}} \, d\phi = \mathbf{0}$. Only the axial component $B_z$ survives: $$dB_z = \frac{\mu_0 I}{4\pi} \frac{R^2 \, d\phi}{(R^2 + z^2)^{3/2}}$$
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Integration: $$B_z = \frac{\mu_0 I R^2}{4\pi (R^2 + z^2)^{3/2}} \int_0^{2\pi} d\phi = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}}$$
B. Ampère's Law & Non-Uniform Current Density
Ampère’s Law relates the line integral of $\mathbf{B}$ around a closed Amperian loop $C$ to the net enclosed conduction current $I_{\text{enc}}$ passing through a surface bounded by $C$:
$$\oint_C \mathbf{B} \cdot d\boldsymbol{\ell} = \mu_0 I_{\text{enc}}$$
Symmetry Requirements
Ampère's Law is always true, but useful for calculating $\mathbf{B}$ only when high degree of spatial symmetry exists: 1. Cylindrical Symmetry: Infinite straight conductor, coaxial cable. 2. Planar Symmetry: Infinite current sheet. 3. Solenoidal/Toroidal Symmetry: Ideal solenoids, toroids.
Non-Uniform Current Density Derivation
Consider a long, solid cylindrical conductor of radius $R$ carrying a non-uniform current density parallel to its axis given by:
$$J(r) = J_0 \left( \frac{r}{R} \right)^2$$
To find $B(r)$ inside the conductor ($r \le R$):
Cross-Section of Cylindrical Conductor
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│ • • • • • │
│ • Amperian │
│ / Loop •│
│ │ r│ │
│ • \─┘ •│
│ • • • • • │
└───────R───────┘
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Calculate Enclosed Current ($I_{\text{enc}}$): $$I_{\text{enc}} = \int_S \mathbf{J} \cdot d\mathbf{A} = \int_0^r J(r') \cdot (2\pi r' \, dr') = 2\pi J_0 \int_0^r \frac{r'^3}{R^2} \, dr'$$ $$I_{\text{enc}} = 2\pi J_0 \left[ \frac{r'^4}{4 R^2} \right]_0^r = \frac{\pi J_0 r^4}{2 R^2}$$
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Apply Ampère's Line Integral: By cylindrical symmetry, $\mathbf{B}$ is tangential and constant in magnitude along an Amperian loop of radius $r$: $$\oint_C \mathbf{B} \cdot d\boldsymbol{\ell} = B (2\pi r)$$
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Solve for $B(r)$: $$B (2\pi r) = \mu_0 \left( \frac{\pi J_0 r^4}{2 R^2} \right) \implies B(r) = \frac{\mu_0 J_0 r^3}{4 R^2} \quad \text{for } r \le R$$
C. Maxwell-Ampère Law & Displacement Current
Classical Ampère’s Law breaks down in dynamic circuits, such as a charging capacitor, where electric fields change over time without immediate charge movement between plates. James Clerk Maxwell solved this by introducing the Displacement Current ($I_d$):
$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$$
yielding the generalized Maxwell-Ampère Law:
$$\oint_C \mathbf{B} \cdot d\boldsymbol{\ell} = \mu_0 \left( I_{\text{enc}} + I_d \right) = \mu_0 \left( I_{\text{enc}} + \varepsilon_0 \frac{d\Phi_E}{dt} \right)$$
where $\Phi_E = \int \mathbf{E} \cdot d\mathbf{A}$ is the electric flux through a surface bounded by the contour $C$.
Charging Parallel-Plate Capacitor
┌───┐ ┌───┐
I(t) │ │ │ │ I(t)
───────────┼───┤ E(t) ├───┼───────────
│ + │ ───────► • P (r < R) │ - │
│ + │ ───────► │ - │
│ + │ ───────► Amperian Loop │ - │
└───┘ ══════════► (r) └───┘
Area A = πR²
Magnetic Field Inside a Charging Parallel-Plate Capacitor
Consider two circular parallel plates of radius $R$ carrying charge $Q(t)$, with a time-varying current $I(t) = \frac{dQ}{dt}$. Assume uniform electric field $E(t) = \frac{Q(t)}{\varepsilon_0 \pi R^2}$ between the plates.
For a circular Amperian loop of radius $r < R$ drawn parallel to and between the plates:
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Conduction Current Enclosed: $$I_{\text{enc}} = 0 \quad \text{(in vacuum/air gap)}$$
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Electric Flux Enclosed: $$\Phi_E(r) = E(t) \cdot (\pi r^2) = \frac{Q(t)}{\varepsilon_0 \pi R^2} (\pi r^2) = \frac{Q(t) r^2}{\varepsilon_0 R^2}$$
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Displacement Current Enclosed: $$I_d(r) = \varepsilon_0 \frac{d\Phi_E(r)}{dt} = \varepsilon_0 \frac{d}{dt} \left( \frac{Q(t) r^2}{\varepsilon_0 R^2} \right) = \frac{dQ}{dt} \frac{r^2}{R^2} = I(t) \frac{r^2}{R^2}$$
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Calculate Induced $B(r)$ Field: $$\oint \mathbf{B} \cdot d\boldsymbol{\ell} = \mu_0 I_d(r) \implies B(r) \cdot (2\pi r) = \mu_0 I(t) \frac{r^2}{R^2}$$ $$B(r) = \frac{\mu_0 I(t) r}{2\pi R^2} \quad \text{for } r \le R$$
3. Common AP Exam Pitfalls & Score 5 Scoring Rubric Nuances
Score 4 vs. Score 5 Performance Profile
| Concept / Task | Score 4 Response Characteristics | Score 5 Response Characteristics |
|---|---|---|
| Biot-Savart Setup | Substitutes continuous components as discrete values; drops vector directions early in derivation. | Parametrizes $d\boldsymbol{\ell}$ and $\mathbf{r}$ explicitly in vector notation; processes cross products before integration. |
| Ampère's Law Loops | States $B(2\pi r) = \mu_0 I$ without validating rotational or translational symmetry. | Formally states constant magnitude and parallel orientation of $\mathbf{B}$ relative to $d\boldsymbol{\ell}$ before evaluating the loop integral. |
| Non-Uniform $J(r)$ | Integrates $J(r)$ across 1D paths ($\int J \, dr$) instead of 2D area elements ($\int J(r) 2\pi r \, dr$). | Sets up proper differential area $dA = 2\pi r \, dr$ and integrates $J(r)$ over bounded surface regions correctly. |
| Displacement Current | Assumes total electric flux $\Phi_E$ uses the full plate radius $R$ when solving for $r < R$. | Restricts $\Phi_E(r)$ calculation strictly to the area enclosed by the specific Amperian loop radius $r$. |
Critical AP Scoring Rubric Nuances
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The Integral Sign Requirement: When AP FRQs ask you to "derive an expression," simply writing down a final standard formula without showing the initial integral step (e.g., omitting $\int \mathbf{J} \cdot d\mathbf{A}$) results in an immediate deduction of the "First Principles" point.
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Vector Cross-Product Sign Conventions: In Biot-Savart derivations, points are explicitly assigned for correctly identifying component directions via right-hand rules or explicit vector operations (e.g., $d\boldsymbol{\ell} \times \hat{\mathbf{r}}$). Dropping negative signs without geometrical justification loses accuracy points.
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Bound Distinction in Flux Integrals: Do not confuse the physical boundary of a conductor/plate ($R$) with the variable radius of an Amperian contour ($r$).
- Inside ($r < R$): Upper integral limit is $r$.
- Outside ($r > R$): Upper integral limit is $R$ because current density drops to zero beyond $R$.
4. UC Berkeley Placement Pathway
Course Exemption: Physics 7B
Achieving a score of 5 on the AP Physics C: E&M exam awards 4.0 semester units and grants a complete course waiver for Physics 7B (Physics for Scientists and Engineers: Heat, Electricity, and Magnetism) at UC Berkeley.
[ AP Physics C: E&M (Score: 5) ] ──► Exemption: [ UC Berkeley Physics 7B (4 Units) ]
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Prerequisites Met / Accelerated
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[ EECS 16B Enrollment ]
Direct Acceleration into EECS 16B
- EECS 16B Focus: Designing Information Devices and Systems II heavily incorporates continuous electromagnetic dynamic modeling, vector field theory, time-varying inductors, high-speed capacitance modeling, and dynamic wave transmission lines.
- Academic Advantage: Waiving Physics 7B allows engineering students (EECS, Mechanical Engineering, BioE) to enroll in EECS 16B during their first year. This unlocks advanced upper-division sequences early, including EECS 105 (Microelectronic Devices and Circuits) and EECS 117 (Electromagnetics for Engineering).
5. High-Yield Practice Problem
Problem Statement
A coaxial cable consists of a solid inner conductor of radius $a$ and a thin, outer cylindrical shell of radius $c$. The space between $a$ and $c$ is filled with vacuum.
Coaxial Cable Cross-Section
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/ c \
/ ┌─────┐ \
│ / a \ │
│ │ • J(r) │ │
│ \ / │
\ └─────┘ /
\ /
└───────────┘
The solid inner conductor carries a total forward current $I_0$ distributed non-uniformly according to the current density:
$$J(r) = C r \quad \text{for } 0 \le r \le a$$
where $C$ is a positive constant and $r$ is the radial distance from the central axis. The thin outer shell carries a return current $I_0$ uniformly distributed across its surface at $r = c$ in the opposite direction.
Simultaneously, a circular parallel-plate capacitor with radius $a$ situated in an adjacent circuit module is being charged by a current $I(t) = I_0 e^{-t/\tau}$.
Questions
- (a) Determine the constant $C$ in terms of $I_0$ and $a$.
- (b) Using Ampère's Law, derive an expression for the magnitude of the magnetic field $B(r)$ in the region $0 \le r \le a$.
- (a) Derive an expression for $B(r)$ in the region $a < r < c$.
- (d) Inside the external parallel-plate capacitor (radius $a$), derive the magnitude of the induced magnetic field $B_{\text{cap}}(r, t)$ at a distance $r < a$ from its central axis as a function of time $t$.
Step-by-Step Solution & Marking Scheme
Part (a): Determine Constant $C$
To find $C$, integrate $J(r)$ over the inner conductor's cross-sectional area and set equal to $I_0$:
$$I_0 = \int_S \mathbf{J} \cdot d\mathbf{A} = \int_0^a (C r) (2\pi r \, dr) = 2\pi C \int_0^a r^2 \, dr$$ $$I_0 = 2\pi C \left[ \frac{r^3}{3} \right]_0^a = \frac{2\pi C a^3}{3}$$ $$C = \frac{3 I_0}{2\pi a^3}$$
Part (b): Derive $B(r)$ for $0 \le r \le a$
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Find $I_{\text{enc}}(r)$: $$I_{\text{enc}}(r) = \int_0^r (C r') (2\pi r' \, dr') = 2\pi C \frac{r^3}{3}$$ Substitute $C = \frac{3 I_0}{2\pi a^3}$: $$I_{\text{enc}}(r) = 2\pi \left( \frac{3 I_0}{2\pi a^3} \right) \frac{r^3}{3} = I_0 \left( \frac{r}{a} \right)^3$$
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Apply Ampère’s Law: $$\oint \mathbf{B} \cdot d\boldsymbol{\ell} = \mu_0 I_{\text{enc}}$$ $$B(r) \cdot (2\pi r) = \mu_0 I_0 \left( \frac{r}{a} \right)^3$$ $$B(r) = \frac{\mu_0 I_0 r^2}{2\pi a^3}$$
Part (c): Derive $B(r)$ for $a < r < c$
For $a < r < c$, the circular Amperian loop encloses the entire current of the inner conductor ($I_0$). The outer return current at $r = c$ is outside this loop.
$$\oint \mathbf{B} \cdot d\boldsymbol{\ell} = \mu_0 I_{\text{enc}} \implies B(r) \cdot (2\pi r) = \mu_0 I_0$$ $$B(r) = \frac{\mu_0 I_0}{2\pi r}$$
Part (d): Find Induced $B_{\text{cap}}(r, t)$ in Charging Capacitor
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Displacement Current Density Formulation: The total dynamic charging current entering the capacitor is $I(t) = I_0 e^{-t/\tau}$. Assuming uniform electric field expansion across the capacitor plates of radius $a$: $$I_{d,\text{enc}}(r, t) = I(t) \cdot \left( \frac{\pi r^2}{\pi a^2} \right) = I_0 e^{-t/\tau} \left( \frac{r}{a} \right)^2$$
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Apply Maxwell-Ampère Law: $$\oint \mathbf{B} \cdot d\boldsymbol{\ell} = \mu_0 I_{d,\text{enc}}(r, t)$$ $$B_{\text{cap}}(r, t) \cdot (2\pi r) = \mu_0 I_0 e^{-t/\tau} \frac{r^2}{a^2}$$ $$B_{\text{cap}}(r, t) = \frac{\mu_0 I_0 r}{2\pi a^2} e^{-t/\tau}$$
Official College Board Style Grading Rubric (15 Points Total)
| Part | Description | Points |
|---|---|---|
| (a) | 1 Point: Correct integral setup $\int J \, dA = \int J(r) 2\pi r \, dr$. 1 Point: Correct integration bounds ($0$ to $a$). 1 Point: Correct algebraic solution for $C$. |
3 Points |
| (b) | 1 Point: Correct expression for enclosed current $I_{\text{enc}}(r)$ in terms of $r$. 1 Point: Applying Ampère's Law left-hand side substitution $\oint \mathbf{B} \cdot d\boldsymbol{\ell} = B(2\pi r)$. 1 Point: Correct final derived expression for $B(r)$. |
3 Points |
| (c) | 1 Point: Correctly identifying that $I_{\text{enc}} = I_0$. 1 Point: Correct final derived expression for $B(r)$. |
2 Points |
| (d) | 1 Point: Stating or applying Maxwell-Ampère relation $\oint \mathbf{B} \cdot d\boldsymbol{\ell} = \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$. 1 Point: Correct calculation of flux $\Phi_E(r)$ or direct current ratio for $r < a$. 1 Point: Correct substitution of $I(t) = I_0 e^{-t/\tau}$. 1 Point: Correct explicit final formula for $B_{\text{cap}}(r, t)$. |
4 Points |
| Total | AP Physics C: E&M Free-Response Precision Standard | 12 Points |
6. AP Exam Execution Checklist
- [ ] Verify Limits: Did you integrate non-uniform density over $dA = 2\pi r \, dr$ instead of $dr$?
- [ ] Check Boundaries: Does $B(r)$ match at $r = a$ across both internal and external expressions? $$\left. B_{\text{internal}}(a) = \frac{\mu_0 I_0}{2\pi a} \quad \right| \quad B_{\text{external}}(a) = \frac{\mu_0 I_0}{2\pi a} \quad \checkmark$$
- [ ] Confirm Units: Verify that displacement current $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$ carries units of Amperes ($\text{A}$).
- [ ] State Symmetry: Explicitly state that "by rotational symmetry, $\mathbf{B}$ is constant along the circular Amperian path" to secure full points on AP FRQ rubrics.