AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the magnetic field produced by the long straight wire at a distance \(r\).
\[ B(r) = \dfrac{\mu_0 I_1}{2\pi r} \]
The net force on the loop depends on the magnetic field created by current \(I_1\) at the position of each segment.
Analyze the magnetic forces on the two perpendicular segments of length \(b\).
\[ F_{\text{perp}} = \int_d^{d+b} I_2 B(r) \, dr = \int_d^{d+b} I_2 \left(\dfrac{\mu_0 I_1}{2\pi r}\right) dr = \dfrac{\mu_0 I_1 I_2}{2\pi} \ln\left(\dfrac{d+b}{d}\right) \]
To find the net force on the loop, we must evaluate the vector sum of forces on all four sides. The field varies along these segments, requiring integration.
Determine the net contribution from the two perpendicular segments.
\[ \vec{F}_{\text{top}} + \vec{F}_{\text{bottom}} = \vec{0} \]
By the right-hand rule, current flows in opposite horizontal directions on the top and bottom sides, so their magnetic forces are equal in magnitude and opposite in direction.
Calculate the magnetic forces on the two parallel segments of length \(a\).
\[ F_{\text{near}} = I_2 a B(d) = \dfrac{\mu_0 I_1 I_2 a}{2\pi d} \\
F_{\text{far}} = I_2 a B(d+b) = \dfrac{\mu_0 I_1 I_2 a}{2\pi (d+b)} \]
Each parallel segment is at a uniform distance from the wire (\(r = d\) and \(r = d+b\)), so the field along each segment is constant.
Subtract the opposing parallel force magnitudes to find the net magnetic force.
\[ F_{\text{net}} = F_{\text{near}} – F_{\text{far}} = \dfrac{\mu_0 I_1 I_2 a}{2\pi d} – \dfrac{\mu_0 I_1 I_2 a}{2\pi (d+b)} = \dfrac{\mu_0 I_1 I_2 a}{2\pi} \left(\dfrac{1}{d} – \dfrac{1}{d+b}\right) = \dfrac{\mu_0 I_1 I_2 a b}{2\pi d (d+b)} \]
Because the currents in the two parallel segments flow in opposite directions, the magnetic forces on them point in opposite radial directions.

Why each choice is correct or incorrect:

(A) Calculates only the force on the nearest parallel segment \(F_{\text{near}}\), ignoring the opposing magnetic force on the far segment.

(B) Approximates \(d+b \approx d\) in the denominator or applies a derivative field gradient expression at \(r = d\) instead of evaluating the finite difference between the forces at \(r = d\) and \(r = d+b\).

(C) This is the correct answer.

(D) Evaluates the force on a single perpendicular segment via integration \(\int_d^{d+b} \frac{\mu_0 I_1 I_2}{2\pi r} dr\) and mistakes it for the net force on the loop, failing to recognize that the forces on the two perpendicular segments cancel out.

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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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