AP Physics C: Electricity & Magnetism Mastery Guide
Module: Ampère's Law, Biot-Savart Integrals & Displacement Current
1. Introduction & AP Exam Weight
Magnetostatics and electrodynamics form the core of classical field theory on the AP Physics C: Electricity & Magnetism Exam. Combined, Ampère’s Law, the Biot-Savart Law, and Displacement Current account for approximately 18%–25% of the total exam weight.
mastery of these topics requires a transition from basic algebraic field evaluation to multivariable calculus-based vector field analysis. The College Board explicitly tests your ability to: 1. Discern when spatial symmetry permits the application of Ampère’s Law versus when arbitrary geometries necessitate explicit numerical or analytical integration via the Biot-Savart Law. 2. Resolve field profiles for non-uniform current densities $J(r)$. 3. Apply Maxwell’s resolution to the continuity equation paradox via Displacement Current ($\mathbf{J}_d$), unifying time-varying electric fields with magnetic induction.
At Stanford University, advanced placement in physics demands more than formula memorization; it requires complete mathematical rigor. Achieving a Score 5 on the AP Physics C: E&M exam demonstrates readiness to bypass introductory coursework and directly engage with advanced quantitative tracks in physics and engineering.
2. Deep Concept Breakdown
2.1 The Biot-Savart Law: First Principles and Direct Integration
When continuous current distributions lack cylindrical, planar, or solenoidal symmetry, Ampère’s Law cannot isolate the magnetic field $\mathbf{B}$. You must construct line integrals using the Biot-Savart Law:
$$\mathrm{d}\mathbf{B} = \frac{\mu_0 I}{4\pi} \frac{\mathrm{d}\boldsymbol{\ell} \times \hat{\mathbf{r}}}{r^2} = \frac{\mu_0 I}{4\pi} \frac{\mathrm{d}\boldsymbol{\ell} \times \mathbf{r}}{r^3}$$
Derivation 1: Finite and Infinite Straight Current-Carrying Wire
Consider a straight wire along the z-axis carrying current $I$ from $z = -z_1$ to $z = +z_2$. We evaluate the magnetic field at a field point $P$ located at a perpendicular distance $R$ along the x-axis ($P = (R, 0, 0)$).
+z
|
| dz (Source element)
|---|------------------
| /| |
| / | |
|/ | r |
| | |
+---|-------------> x | P = (R, 0, 0)
/ | R |
/ | |
+y |
|
-z
-
Differential Element Setup: $$\mathrm{d}\boldsymbol{\ell} = \mathrm{d}z\,\hat{\mathbf{k}}$$ $$\mathbf{r} = R\,\hat{\mathbf{i}} - z\,\hat{\mathbf{k}} \implies r = \sqrt{R^2 + z^2}$$ $$\hat{\mathbf{r}} = \frac{R\,\hat{\mathbf{i}} - z\,\hat{\mathbf{k}}}{\sqrt{R^2 + z^2}}$$
-
Cross Product Calculation: $$\mathrm{d}\boldsymbol{\ell} \times \mathbf{r} = (\mathrm{d}z\,\hat{\mathbf{k}}) \times (R\,\hat{\mathbf{i}} - z\,\hat{\mathbf{k}}) = R\,\mathrm{d}z\,\hat{\mathbf{j}}$$
-
Integral Formulation: $$\mathbf{B} = \frac{\mu_0 I}{4\pi} \int_{-z_1}^{z_2} \frac{R\,\mathrm{d}z}{(R^2 + z^2)^{3/2}} \hat{\mathbf{j}}$$
-
Trigonometric Substitution: Let $z = R \tan\theta \implies \mathrm{d}z = R \sec^2\theta\,\mathrm{d}\theta$, and $R^2 + z^2 = R^2\sec^2\theta$. $$\mathbf{B} = \frac{\mu_0 I}{4\pi} \int_{\theta_1}^{\theta_2} \frac{R (R \sec^2\theta\,\mathrm{d}\theta)}{R^3 \sec^3\theta} \hat{\mathbf{j}} = \frac{\mu_0 I}{4\pi R} \int_{\theta_1}^{\theta_2} \cos\theta\,\mathrm{d}\theta\,\hat{\mathbf{j}}$$ $$\mathbf{B} = \frac{\mu_0 I}{4\pi R} (\sin\theta_2 - \sin\theta_1) \hat{\mathbf{j}}$$
For an infinitely long wire ($z_1 \to -\infty, z_2 \to +\infty \implies \theta_1 = -\pi/2, \theta_2 = \pi/2$): $$\mathbf{B} = \frac{\mu_0 I}{2\pi R} \hat{\boldsymbol{\phi}}$$
Derivation 2: On-Axis Field of a Circular Loop
For a circular current loop of radius $R$ in the xy-plane centered at the origin, evaluate $\mathbf{B}(z)$ along the z-axis at $(0, 0, z)$:
$$\mathrm{d}\boldsymbol{\ell} = R\,\mathrm{d}\phi\,\hat{\boldsymbol{\phi}}$$ $$\mathbf{r} = -R\,\hat{\mathbf{r}}' + z\,\hat{\mathbf{k}} \implies r = \sqrt{R^2 + z^2}$$ $$\mathrm{d}\boldsymbol{\ell} \times \mathbf{r} = (R\,\mathrm{d}\phi\,\hat{\boldsymbol{\phi}}) \times (-R\,\hat{\mathbf{r}}' + z\,\hat{\mathbf{k}}) = R^2\mathrm{d}\phi\,\hat{\mathbf{k}} + R z\,\mathrm{d}\phi\,\hat{\mathbf{r}}'$$
By azimuthal symmetry, the radial components $\hat{\mathbf{r}}'$ integrate to zero ($\int_0^{2\pi} \hat{\mathbf{r}}'\,\mathrm{d}\phi = \mathbf{0}$). Isolating the z-component:
$$B_z = \frac{\mu_0 I}{4\pi} \int_0^{2\pi} \frac{R^2\,\mathrm{d}\phi}{(R^2 + z^2)^{3/2}} = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}}$$
$$\mathbf{B}(z) = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}} \hat{\mathbf{k}}$$
2.2 Ampère’s Law and Non-Uniform Current Densities
Ampère's Law states that for any closed Amperian loop $C$:
$$\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 I_{\text{enc}} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{A}$$
Cylindrical Conductor with Non-Uniform Current Density
Consider a long wire of radius $a$ carrying a non-uniform current density parallel to its axis:
$$\mathbf{J}(r) = J_0 \left( \frac{r}{a} \right)^n \hat{\mathbf{k}} \quad (r \le a)$$
Cross-Section of Conductive Cylinder
. - - - - - - .
. ' | ' .
' | r '
' | '
' (======*======) ' <- Amperian Loop (radius r)
' | '
' | '
' | a '
. ' | ' .
. - - - - - - .
-
Calculate Total Enclosed Current $I_{\text{enc}}(r)$ for $r \le a$: $$I_{\text{enc}}(r) = \int_0^r \mathbf{J}(r') \cdot \mathrm{d}\mathbf{A}' = \int_0^r J_0 \left( \frac{r'}{a} \right)^n (2\pi r'\,\mathrm{d}r')$$ $$I_{\text{enc}}(r) = \frac{2\pi J_0}{a^n} \int_0^r (r')^{n+1} \mathrm{d}r' = \frac{2\pi J_0}{a^n (n+2)} r^{n+2}$$
-
Apply Ampère’s Law ($r \le a$): By cylindrical symmetry, $\mathbf{B} = B_\phi(r) \hat{\boldsymbol{\phi}}$, and $\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = B_\phi(r) (2\pi r)$. $$B_\phi(r) (2\pi r) = \mu_0 \left( \frac{2\pi J_0 r^{n+2}}{a^n (n+2)} \right)$$ $$B_\phi(r) = \frac{\mu_0 J_0}{a^n (n+2)} r^{n+1}$$
-
External Field ($r > a$): Total enclosed current $I_{\text{total}} = I_{\text{enc}}(a) = \frac{2\pi J_0 a^2}{n+2}$. $$B_\phi(r) = \frac{\mu_0 I_{\text{total}}}{2\pi r} = \frac{\mu_0 J_0 a^2}{(n+2)r}$$
2.3 Displacement Current & The Generalized Maxwell-Ampère Law
In static conditions, $\nabla \cdot \mathbf{J} = 0$. However, considering the charge continuity equation:
$$\nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t} = 0$$
Taking the divergence of the static Ampère’s Law ($\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$) yields $\nabla \cdot (\nabla \times \mathbf{B}) = \mu_0 (\nabla \cdot \mathbf{J})$. Since the divergence of any curl is identically zero ($\nabla \cdot (\nabla \times \mathbf{A}) \equiv 0$), static Ampère's Law implies $\nabla \cdot \mathbf{J} = 0$, violating conservation of charge when $\frac{\partial \rho}{\partial t} \neq 0$.
Maxwell resolved this by adding the displacement current density $\mathbf{J}_d$:
$$\mathbf{J}_d = \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$
Maxwell-Ampère Law (Integral Form):
$$\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \left( I_{\text{enc}} + I_d \right) = \mu_0 I_{\text{enc}} + \mu_0 \epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t}$$
where $\Phi_E = \iint_S \mathbf{E} \cdot \mathrm{d}\mathbf{A}$ is the electric flux through a surface bounded by contour $C$.
Parallel-Plate Capacitor Charging
Surface S1 (Flat disk)
| |
I ----> |=========| ----> I
| | | | |
| |E| | |
| V V V |
|=========|
| |
\_______/
Surface S2 (Bulging surface)
Bounded by SAME Amperian Loop C!
Field Inside a Charging Circular Parallel-Plate Capacitor
Consider circular plates of radius $R$, driven by a time-varying current $I(t)$, producing a uniform electric field $E(t) = \frac{Q(t)}{\epsilon_0 \pi R^2}$ between the plates.
-
Electric Flux through a concentric circle of radius $r < R$: $$\Phi_E(r) = E(t) \cdot (\pi r^2) = \frac{Q(t)}{\epsilon_0 \pi R^2} (\pi r^2) = \frac{Q(t) r^2}{\epsilon_0 R^2}$$
-
Displacement Current Enclosed ($I_d$): $$I_d(r) = \epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t} = \epsilon_0 \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{Q(t) r^2}{\epsilon_0 R^2} \right) = \frac{\mathrm{d}Q}{\mathrm{d}t} \frac{r^2}{R^2} = I(t) \frac{r^2}{R^2}$$
-
Induced Magnetic Field ($r < R$): $$\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 I_d(r) \implies B(r) (2\pi r) = \mu_0 I(t) \frac{r^2}{R^2}$$ $$B(r) = \frac{\mu_0 I(t) r}{2\pi R^2} \quad (r < R)$$
2.4 Computational Vector Modeling: Biot-Savart Numerical Engine
In modern electrodynamics, complex geometries require numerical integration. The following production-grade Python script computes the magnetic field vector $\mathbf{B}$ along the axis of an arbitrary 3D current loop via Biot-Savart discretization.
import numpy as np
import matplotlib.pyplot as plt
def compute_biot_savart_loop(I: float, radius: float, z_eval: np.ndarray, num_segments: int = 1000) -> np.ndarray:
"""
Computes the magnetic field B_z along the z-axis of a circular loop using Biot-Savart discretization.
Parameters:
I (float): Current in Amperes
radius (float): Radius of the loop in meters
z_eval (np.ndarray): Array of z-coordinates along the axis (meters)
num_segments (int): Number of discrete current elements d_l
Returns:
np.ndarray: B_z magnetic field strength in Tesla
"""
mu_0 = 4 * np.pi * 1e-7
d_phi = 2 * np.pi / num_segments
phi = np.linspace(0, 2 * np.pi - d_phi, num_segments)
# Source elements dl on the loop: r_source = [R cos(phi), R sin(phi), 0]
# dl = [-R sin(phi) dphi, R cos(phi) dphi, 0]
dl = np.stack([-radius * np.sin(phi) * d_phi,
radius * np.cos(phi) * d_phi,
np.zeros(num_segments)], axis=1)
r_source = np.stack([radius * np.cos(phi), radius * np.sin(phi), np.zeros(num_segments)], axis=1)
B_z = np.zeros(len(z_eval))
for idx, z in enumerate(z_eval):
r_field = np.array([0.0, 0.0, z])
r_vec = r_field - r_source # Vector from dl to field point
r_mag = np.linalg.norm(r_vec, axis=1)
# dB = (mu_0 * I / (4 * pi)) * (dl x r_vec) / |r|^3
cross_prod = np.cross(dl, r_vec)
dB = (mu_0 * I / (4 * np.pi)) * cross_prod / (r_mag[:, np.newaxis]**3)
B_z[idx] = np.sum(dB[:, 2]) # Sum z-components
return B_z
if __name__ == "__main__":
z_arr = np.linspace(-0.2, 0.2, 200)
I_test = 5.0 # 5 Amperes
R_test = 0.05 # 5 centimeters
# Numerical computation
Bz_num = compute_biot_savart_loop(I_test, R_test, z_arr)
# Analytical computation for validation
Bz_analytic = (4 * np.pi * 1e-7 * I_test * R_test**2) / (2 * (R_test**2 + z_arr**2)**(1.5))
# Verify relative error is below threshold
max_err = np.max(np.abs(Bz_num - Bz_analytic) / Bz_analytic)
print(f"Calculation Complete. Maximum relative error vs Analytical solution: {max_err:.2e}")
3. Common AP Exam Pitfalls & Score 5 Scoring Rubric Nuances
Critical Pitfalls & Concept Traps
- Misapplication of Ampère’s Law to Non-Symmetric Systems:
- Pitfall: Attempting to use $\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 I_{\text{enc}} \implies B(2\pi r) = \mu_0 I$ for a finite wire segment or a square loop.
-
Score 5 Correction: Ampère’s Law is always true, but it is only useful for isolating $B$ when spatial symmetry guarantees $|\mathbf{B}|$ is constant along the chosen integration contour. If symmetry fails, you must integrate using the Biot-Savart Law.
-
Ignoring Enclosed Surface Boundary Dynamics for Displacement Current:
- Pitfall: Failing to recognize that $I_{\text{enc}}$ and $I_d$ are defined for a surface $S$ bounded by the contour $C$.
-
Score 5 Correction: For a flat surface intersecting a capacitor plate, $I_{\text{enc}} = I$ and $I_d = 0$. For a curved "ballooning" surface extending between the capacitor plates bounded by the same contour, $I_{\text{enc}} = 0$ and $I_d = I$. Both choices yield identical values for $\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell}$.
-
Neglecting Differential Area Elements in Integrals:
- Pitfall: Substituting $I_{\text{enc}} = J \cdot A = J \pi r^2$ when $J(r)$ is a function of radius.
- Score 5 Correction: Write $I_{\text{enc}} = \int \mathbf{J} \cdot \mathrm{d}\mathbf{A} = \int J(r') 2\pi r' \mathrm{d}r'$ explicitly before evaluating limits.
Score 4 vs. Score 5 Solution Nuances
To illustrate how AP Readers grade these questions, consider a standard prompt: “A long cylindrical conductor of radius $R$ carries a non-uniform current density $J(r) = C r$. Find the magnetic field $B(r)$ inside the conductor ($r < R$).”
| AP Exam Dimension | Score 4 Performance (Competent) | Score 5 Performance (Mastery) |
|---|---|---|
| Mathematical Setup | $I = J A \implies I = (C r) (\pi r^2) = C \pi r^3$ $\oint B \cdot dl = \mu_0 I$ $B (2\pi r) = \mu_0 C \pi r^3 \implies B = \frac{\mu_0 C r^2}{2}$ |
Sets up explicit differential integration: $\mathrm{d}A' = 2\pi r'\,\mathrm{d}r'$ $I_{\text{enc}}(r) = \int_0^r (C r')(2\pi r'\,\mathrm{d}r') = 2\pi C \int_0^r (r')^2 \mathrm{d}r' = \frac{2\pi C r^3}{3}$ |
| Symmetry Argument | Assumes $B$ is constant along a radial circle without justification. | Explicitly states: "By cylindrical symmetry, $\mathbf{B}$ is azimuthal and constant in magnitude along a concentric circular Amperian path of radius $r$." |
| Final Calculation & Vector Form | $B = \frac{\mu_0 C r^2}{2}$ (Incorrect factor due to missing calculus, missing vector/direction context) |
$\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = B(2\pi r) = \mu_0 \left(\frac{2\pi C r^3}{3}\right)$ $\mathbf{B}(r) = \frac{\mu_0 C r^2}{3} \hat{\boldsymbol{\phi}}$ |
4. Stanford University Placement Pathway
Exemption Criteria & Academic Progression
Achieving a Score 5 on the AP Physics C: Electricity & Magnetism exam provides direct academic credit and placement advantages at Stanford University:
[ AP Physics C: E&M Score 5 ]
|
v
[ Waive PHYSICS 43 ] ---> Waives 4 Units of Introductory Physics Requirements
|
+-----------------------------------+
| |
v v
[ TRACK 1: Physics Majors ] [ TRACK 2: Engineering / EE ]
PHYSICS 63: Advanced Mechanics EE 101A: Circuits II / Electromagnetics
and Electromagnetism EE Core Foundations
| |
v v
Quantum Mechanics & Photonics RF, Integrated Circuits & Electrodynamics
- Exempted Course: PHYSICS 43 (Electricity and Magnetism, 4 units).
- Degree Advancement: Fulfills the foundational E&M requirement for all majors within the School of Engineering (e.g., Electrical Engineering, Mechanical Engineering) and the Department of Physics.
- Accelerated Track Enrollment:
- Physics/Applied Physics Majors: Eligible to enroll directly in PHYSICS 63 (Advanced Mechanics and Electromagnetism), bypassing introductory physics sequences entirely.
- Electrical Engineering (EE) Majors: Accelerates entry into the EE Core (EE 101A, EE 102A) and advanced coursework in electrodynamics, micro-photonics, and high-frequency RF systems.
Strategic Academic Advantage
At Stanford, skipping PHYSICS 43 frees 4 units during your freshman year. This allows high-achieving students to begin undergraduate research early at facilities like the Stanford Linear Accelerator Center (SLAC) or the Ginzton Laboratory, while taking honors-level field theory courses as a first-year student.
5. High-Yield Practice Problem & Step-by-Step Solution Checklist
Problem Statement
A coaxial system consists of a long, solid inner conductor of radius $a$ and a thin, outer cylindrical shell of radius $c$ ($c > a$).
- Within the inner conductor ($r \le a$), the volume current density is non-uniform and directed along the $+z$-axis, given by: $$\mathbf{J}(r) = J_0 \left(1 - \frac{r}{a}\right) \hat{\mathbf{k}}$$
- The outer cylindrical shell carries a uniform total return current $I_{\text{return}}$ in the $-z$-direction.
- Simultaneously, the region $a < r < b$ (where $b < c$) contains a parallel-plate capacitor-like segment with a time-varying radial electric field given by $\mathbf{E}(r, t) = \left( \frac{E_0 a}{r} e^{-\alpha t} \right) \hat{\mathbf{r}}$.
Coaxial Cross-Section
. - - - - .
. ' Outer ' . <- Radius c (Return Current -I_tot)
' Shell '
' . - - - . '
' ' Region ' ' <- Region a < r < b (Time-varying E-field)
' ' a < r < b ' '
' ' . - - . ' '
' ' ' Inner ' ' ' <- Radius a (Non-uniform J(r))
' ' ' Wire ' ' '
' ' . - - . ' '
' ' ' '
' ' ' '
' ' - - - - ' '
' '
' . . '
. - - - - - .
Questions:
- (a) Derive an expression for the total current $I_{\text{tot}}$ passing through the inner conductor.
- (b) Derive an expression for the magnetic field magnitude $B(r)$ as a function of radial distance $r$ for $r \le a$.
- (c) Find the displacement current density $\mathbf{J}_d(r, t)$ and total displacement current $I_d(t)$ inside the region $a < r < b$ across a cylinder of length $L$.
- (d) Derive the total magnetic field magnitude $B(r, t)$ in the region $a < r < b$, taking into account both the enclosed inner conductor current and the displacement current within length $L$.
Free-Response Grading Rubric & Solution Checklist
Part (a): Total Current $I_{\text{tot}}$ in Inner Conductor [3 Points]
-
Step 1: Set up the integral with the proper area element. [1 Point] $$I_{\text{tot}} = \iint_{S} \mathbf{J} \cdot \mathrm{d}\mathbf{A} = \int_0^a J_0 \left(1 - \frac{r}{a}\right) (2\pi r\,\mathrm{d}r)$$
-
Step 2: Expand and integrate explicitly. [1 Point] $$I_{\text{tot}} = 2\pi J_0 \int_0^a \left(r - \frac{r^2}{a}\right) \mathrm{d}r = 2\pi J_0 \left[ \frac{r^2}{2} - \frac{r^3}{3a} \right]0^a$$ $$I{\text{tot}} = 2\pi J_0 \left( \frac{a^2}{2} - \frac{a^2}{3} \right) = 2\pi J_0 \left( \frac{a^2}{6} \right)$$
-
Step 3: State final value clearly. [1 Point] $$I_{\text{tot}} = \frac{\pi J_0 a^2}{3}$$
Part (b): Magnetic Field $B(r)$ for $r \le a$ [4 Points]
-
Step 1: Apply Ampère's Law with symmetry justification. [1 Point] "Due to cylindrical symmetry, the magnetic field is azimuthal ($\mathbf{B} = B_\phi \hat{\boldsymbol{\phi}}$) and constant along a circular Amperian loop of radius $r$." $$\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = B(r) (2\pi r)$$
-
Step 2: Evaluate $I_{\text{enc}}(r)$ for $r \le a$. [1 Point] $$I_{\text{enc}}(r) = 2\pi J_0 \int_0^r \left(r' - \frac{(r')^2}{a}\right) \mathrm{d}r' = 2\pi J_0 \left( \frac{r^2}{2} - \frac{r^3}{3a} \right)$$
-
Step 3: Equate and solve for $B(r)$. [1 Point] $$B(r) (2\pi r) = \mu_0 \left[ 2\pi J_0 \left( \frac{r^2}{2} - \frac{r^3}{3a} \right) \right]$$ $$B(r) = \mu_0 J_0 \left( \frac{r}{2} - \frac{r^2}{3a} \right)$$
-
Step 4: Verify boundary conditions. [1 Point] At $r = 0 \implies B(0) = 0$. At $r = a \implies B(a) = \mu_0 J_0 \left( \frac{a}{2} - \frac{a}{3} \right) = \frac{\mu_0 J_0 a}{6} = \frac{\mu_0 I_{\text{tot}}}{2\pi a}$. (Matches external wire formula).
Part (c): Displacement Current $\mathbf{J}_d(r, t)$ and $I_d(t)$ [4 Points]
-
Step 1: Apply definition of displacement current density. [1 Point] $$\mathbf{J}_d(r, t) = \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} = \epsilon_0 \frac{\partial}{\partial t} \left( \frac{E_0 a}{r} e^{-\alpha t} \hat{\mathbf{r}} \right)$$ $$\mathbf{J}_d(r, t) = -\frac{\epsilon_0 \alpha E_0 a}{r} e^{-\alpha t} \hat{\mathbf{r}}$$
-
Step 2: Identify the surface for computing total displacement current. [1 Point] For a cylindrical surface of radius $r$ and length $L$ between $a$ and $b$, the outward differential area vector is $\mathrm{d}\mathbf{A} = (2\pi r L)\hat{\mathbf{r}}$.
-
Step 3: Compute $I_d(t)$ across the cylindrical area. [2 Points] $$I_d(t) = \iint \mathbf{J}_d \cdot \mathrm{d}\mathbf{A} = \left( -\frac{\epsilon_0 \alpha E_0 a}{r} e^{-\alpha t} \right) (2\pi r L)$$ $$I_d(t) = -2\pi \epsilon_0 \alpha E_0 a L e^{-\alpha t}$$ (Note: The negative sign indicates displacement current directed radially inward).
Part (d): Generalized Field $B(r, t)$ for $a < r < b$ [4 Points]
-
Step 1: Formulate Maxwell-Ampère Law for a circular path of radius $r$. [1 Point] $$\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \left( I_{\text{conduction, enc}} + I_{d, \text{enc}} \right)$$
-
Step 2: Identify enclosed currents accurately. [1 Point]
- Conduction current enclosed = Total current of inner wire = $I_{\text{tot}} = \frac{\pi J_0 a^2}{3}$.
-
Electric flux through the flat circular disk bounded by Amperian loop of radius $r$: Since $\mathbf{E}$ is entirely radial ($\mathbf{E} \parallel$ surface of disk), $\mathbf{E} \cdot \mathrm{d}\mathbf{A}_{\text{disk}} = 0$. Therefore, zero electric flux passes through the flat disk area bounded by the loop.
-
Step 3: Resolve physical surface choice nuance. [1 Point] Because the electric field is radial, no electric flux passes through the circular surface bounded by the coaxial loop at a constant z-plane! Thus, $\Phi_E = 0$ across the flat circular enclosed surface. $$I_d = \epsilon_0 \frac{\mathrm{d}\Phi_E}{\mathrm{d}t} = 0 \quad \text{(through the flat circular disk area)}$$
-
Step 4: Calculate final magnetic field. [1 Point] $$\oint \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = B(r) (2\pi r) = \mu_0 I_{\text{tot}}$$ $$B(r) = \frac{\mu_0 I_{\text{tot}}}{2\pi r} = \frac{\mu_0 J_0 a^2}{6 r}$$
(Self-Correction/Verification Note for Score 5 Candidates: The radial displacement current derived in Part (c) flows radially between $r=a$ and $r=b$. By symmetry, radial currents create no azimuthal magnetic field $\mathbf{B}\phi$. Thus, only axial currents $I_z$ contribute to azimuthal $\mathbf{B}\phi$ via Ampère loops in the xy-plane!)
Final Review Summary Matrix
+-------------------+---------------------------------------+---------------------------------------+
| Feature | Ampère's Law | Biot-Savart Law |
+-------------------+---------------------------------------+---------------------------------------+
| Mathematical Form | \oint B · dl = μ₀ I_enc | dB = (μ₀ I / 4π) (dl × r̂) / r² |
| Applicability | Universal (Useful only with symmetry) | Universal (Works for any geometry) |
| Primary AP Targets| Coax cables, infinite wires, solenoids| Wire segments, circular loops, axes |
| Key Variable | Enclosed current density J(r) | Differential element dl × r̂ |
+-------------------+---------------------------------------+---------------------------------------+