AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the target principle for induced current in terms of magnetic flux rate of change and motional EMF.
\[ I(t) = \dfrac{\mathcal{E}_{\text{net}}}{R} = \dfrac{1}{R} \left| \dfrac{d\Phi_B}{dt} \right| \]
The question asks for the physical origin of the constant current plateau \(+I_0\) during the interval \(T < t < 2T\).
Calculate the magnetic flux \(\Phi_B(t)\) through the loop when it is entirely inside the field region \(0 \le x \le 2L\).
\[ \Phi_B(t) = \int_{v_0 t – L}^{v_0 t} \left( B_0 \dfrac{x}{L} \right) w \, dx = \dfrac{B_0 w}{2L} \left[ (v_0 t)^2 – (v_0 t – L)^2 \right] = B_0 w v_0 t – \dfrac{1}{2} B_0 w L \]
During \(T \le t \le 2T\), the front edge is at position \(x_f = v_0 t\) and the rear edge is at \(x_b = v_0 t – L\). Integrating \(B(x) = B_0 \dfrac{x}{L}\) over the loop length yields the enclosed flux.
Differentiate the flux with respect to time to find the net induced EMF and current during this interval.
\[ \mathcal{E} = \dfrac{d\Phi_B}{dt} = B_0 w v_0 \implies I = \dfrac{B_0 w v_0}{R} = +I_0 \]
Faraday’s law connects the time derivative of magnetic flux to the induced EMF.
Relate the constant net EMF to the motional EMF along the leading and trailing edges of the loop.
\[ \mathcal{E}_{\text{net}} = \mathcal{E}_f – \mathcal{E}_b = [B(x_f) – B(x_b)] w v_0 = \left[ B_0 \dfrac{v_0 t}{L} – B_0 \dfrac{v_0 t – L}{L} \right] w v_0 = B_0 w v_0 \]
Motional EMF provides an equivalent physical framework: the leading edge experiences an upward EMF \(\mathcal{E}_f = B(x_f) w v_0\) and the trailing edge experiences an upward EMF \(\mathcal{E}_b = B(x_b) w v_0\).
Conclude why the difference in magnetic field strength produces the constant plateau. Because the spatial field gradient \(\dfrac{dB}{dx} = \dfrac{B_0}{L}\) is constant, the difference in magnetic field strength between the front and rear edges separated by distance \(L\) is strictly constant: \(\Delta B = B_0\). This produces a constant net motional EMF \(\mathcal{E}_{\text{net}} = B_0 w v_0\) and a constant current \(+I_0\).

Why each choice is correct or incorrect:

(A) Incorrect. Magnetic flux increases continuously at a rate of \(\dfrac{d\Phi_B}{dt} = B_0 w v_0\) because the loop moves into regions of higher field strength.

(B) Incorrect. A constant rate of area increase in a uniform field yields a constant current; here, current increases linearly because \(B(x)\) varies linearly with position, making \(\Phi_B \propto t^2\) and \(I \propto t\).

(C) Incorrect. The magnetic field vector points into the page everywhere in the region; the current reverses direction because flux changes from increasing to decreasing as the loop exits.

(D) This is the correct answer. The constant field difference \(\Delta B = B_0\) between the leading and trailing edges produces a constant net motional EMF \(\mathcal{E}_{\text{net}} = B_0 w v_0\).

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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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