AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the target quantity and express magnetic flux in terms of the magnetic field inside a toroid.
\[ L = \dfrac{N}{I} \int_a^b B(r) h(r) \, dr \]
Self-inductance is defined as \(L = \dfrac{N \Phi_1}{I}\), where \(\Phi_1\) is the magnetic flux through a single turn.
Apply Ampère’s law to determine the radial dependence of the magnetic field inside each toroid.
\[ \oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}} \implies B(r) \cdot 2\pi r = \mu_0 N I \implies B(r) = \dfrac{\mu_0 N I}{2\pi r} \]
To compare the flux integrals, we need the functional form of \(B(r)\) as a function of distance \(r\) from the central axis.
Compare the height function \(h_1(r)\) of Toroid 1 to those of Toroids 2 and 3.
\[ h_2(r) + h_3(r) = h \left(\dfrac{b-r}{b-a}\right) + h \left(\dfrac{r-a}{b-a}\right) = h = h_1(r) \]
At every radial distance \(r \in [a, b]\), the height of Toroid 1 is \(h_1(r) = h\), while Toroid 2 has height \(h_2(r) = h \left(\dfrac{b-r}{b-a}\right)\) and Toroid 3 has height \(h_3(r) = h \left(\dfrac{r-a}{b-a}\right)\).
Determine the relationship between \(L_1\), \(L_2\), and \(L_3\).
\[ L_1 = L_2 + L_3 \implies L_1 > L_2 \text{ and } L_1 > L_3 \]
Because \(h_1(r) = h_2(r) + h_3(r)\) at every point in the domain, the flux integrals satisfy \(\Phi_1 = \Phi_2 + \Phi_3\), which implies \(L_1 = L_2 + L_3\).
Compare \(L_2\) and \(L_3\) based on the spatial distribution of cross-sectional area.
\[ B(r_{\text{near } a}) > B(r_{\text{near } b}) \implies L_2 > L_3 \]
Because \(B(r) \propto \dfrac{1}{r}\), the magnetic field is strictly stronger near \(r = a\) than near \(r = b\). Toroid 2 places more cross-sectional height near \(r = a\), whereas Toroid 3 places more cross-sectional height near \(r = b\).
Combine the partial inequalities into a complete ranking.
\[ L_1 > L_2 > L_3 \]
Combining \(L_1 = L_2 + L_3\) (which establishes \(L_1 > L_2\)) with \(L_2 > L_3\) yields the final ordering.

Why each choice is correct or incorrect:

(A) Incorrect. This reverses the physical ranking by assuming magnetic field strength increases with radius and ignoring that Toroid 1 has height equal to the sum of Toroids 2 and 3 at every radius.

(B) Incorrect. While correctly identifying that \(L_1\) is the largest because \(h_1(r) = h_2(r) + h_3(r)\), this choice incorrectly assumes that placing cross-sectional area near the outer boundary \(r = b\) yields greater flux than placing it near \(r = a\).

(C) Incorrect. This choice assumes that concentrating area in the high-field region enables Toroid 2 to exceed Toroid 1 in inductance, overlooking that Toroid 1 has equal or greater height than Toroid 2 at all radii.

(D) Correct. Toroid 1 has height \(h_1(r) = h_2(r) + h_3(r)\) at every radius, so \(L_1 = L_2 + L_3\), making \(L_1\) the largest. Furthermore, because \(B(r) \propto 1/r\) is strongest near \(r = a\), Toroid 2 (which has larger height near \(r = a\)) encloses more flux than Toroid 3 (which has larger height near \(r = b\)), giving \(L_2 > L_3\).

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

Metric Prefixes

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

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