A long, solid cylindrical conductor of radius $R$ carries a uniform current $I$ distributed evenly over its cross section. A particle of mass $m$ and positive charge $q$ moves parallel to the central axis of the cylinder with constant speed $v_0$. The particle is evaluated at two locations: point 1 at a radial distance $r_1 = \dfrac{R}{2}$ inside the conductor, and point 2 at a radial distance $r_2 = 2R$ outside the conductor. What is the ratio of the magnitude of the spatial rate of change of the particle's acceleration with respect to radial position at point 1 to that at point 2, \(\dfrac{|da/dr|_{r_1}}{|da/dr|_{r

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Relate the spatial rate of change of acceleration $\left\|\dfrac{da}{dr}\right\|$ to the spatial rate of change of the magnetic field $\left\|\dfrac{dB}{dr}\right\|$. \[ \left\| \dfrac{da}{dr} \right\| = \dfrac{q v_0}{m} \left\| \dfrac{dB}{dr} \right\| \]	The question asks for the ratio of $\left\|\dfrac{da}{dr}\right\|$ at two points. Because velocity is perpendicular to the radial magnetic field and speed $v_0$ is constant, the acceleration magnitude is $a(r) = \dfrac{q v_0 B(r)}{m}$.
Use Ampère’s law to find $B(r)$ inside the conductor ($r < R$) and evaluate its derivative at $r_1 = \dfrac{R}{2}$. \[ B_{\text{in}}(2\pi r) = \mu_0 I \left(\dfrac{\pi r^2}{\pi R^2}\right) \implies B_{\text{in}}(r) = \dfrac{\mu_0 I r}{2\pi R^2} \]	Point 1 is inside the conductor, so the enclosed current depends on the ratio of cross-sectional areas.
Differentiate $B_{\text{in}}(r)$ with respect to $r$ to find $\left\|\dfrac{da}{dr}\right\|_{r_1}$. \[ \left\| \dfrac{dB_{\text{in}}}{dr} \right\| = \dfrac{\mu_0 I}{2\pi R^2} \implies \left\| \dfrac{da}{dr} \right\|_{r_1} = \dfrac{\mu_0 q v_0 I}{2\pi m R^2} \]	Taking the derivative yields a constant spatial gradient inside the conductor.
Use Ampère’s law to find $B(r)$ outside the conductor ($r > R$) and evaluate its derivative at $r_2 = 2R$. \[ B_{\text{out}}(r) = \dfrac{\mu_0 I}{2\pi r} \implies \left\| \dfrac{dB_{\text{out}}}{dr} \right\| = \dfrac{\mu_0 I}{2\pi r^2} \]	Point 2 is outside the conductor, where the entire current $I$ is enclosed.
Evaluate $\left\|\dfrac{da}{dr}\right\|_{r_2}$ at $r_2 = 2R$. \[ \left\| \dfrac{da}{dr} \right\|_{r_2} = \dfrac{\mu_0 q v_0 I}{2\pi m (2R)^2} = \dfrac{\mu_0 q v_0 I}{8\pi m R^2} \]	Substituting $r_2 = 2R$ gives the derivative magnitude outside.
Compute the ratio of the two spatial derivatives. \[ \dfrac{\left\| \dfrac{da}{dr} \right\|_{r_1}}{\left\| \dfrac{da}{dr} \right\|_{r_2}} = \dfrac{\dfrac{\mu_0 q v_0 I}{2\pi m R^2}}{\dfrac{\mu_0 q v_0 I}{8\pi m R^2}} = \dfrac{1/2}{1/8} = 4 \]	Dividing the result at point 1 by the result at point 2 yields the requested ratio.

Why each choice is correct or incorrect:

(A) Incorrect. Evaluates the ratio of the accelerations $a(r_1)/a(r_2) = 1$ instead of the ratio of the derivatives of acceleration with respect to position.

(B) Incorrect. Treats the derivative of $1/r$ as proportional to $1/r$ rather than $1/r^2$, leading to a factor of 2 instead of 4.

(C) Correct. Correctly applies Ampère’s law in both regions, computes the spatial derivatives of the magnetic field, and takes their ratio.

(D) Incorrect. Erroneously multiplies the derivative inside by $r_1$, failing to recognize that the derivative of a linear function is constant.

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Step	Reasoning
Relate the spatial rate of change of acceleration \(\left\|\dfrac{da}{dr}\right\|\) to the spatial rate of change of the magnetic field \(\left\|\dfrac{dB}{dr}\right\|\). \[ \left\| \dfrac{da}{dr} \right\| = \dfrac{q v_0}{m} \left\| \dfrac{dB}{dr} \right\| \]	The question asks for the ratio of \(\left\|\dfrac{da}{dr}\right\|\) at two points. Because velocity is perpendicular to the radial magnetic field and speed \(v_0\) is constant, the acceleration magnitude is \(a(r) = \dfrac{q v_0 B(r)}{m}\).
Use Ampère’s law to find \(B(r)\) inside the conductor (\(r < R\)) and evaluate its derivative at \(r_1 = \dfrac{R}{2}\). \[ B_{\text{in}}(2\pi r) = \mu_0 I \left(\dfrac{\pi r^2}{\pi R^2}\right) \implies B_{\text{in}}(r) = \dfrac{\mu_0 I r}{2\pi R^2} \]	Point 1 is inside the conductor, so the enclosed current depends on the ratio of cross-sectional areas.
Differentiate \(B_{\text{in}}(r)\) with respect to \(r\) to find \(\left\|\dfrac{da}{dr}\right\|_{r_1}\). \[ \left\| \dfrac{dB_{\text{in}}}{dr} \right\| = \dfrac{\mu_0 I}{2\pi R^2} \implies \left\| \dfrac{da}{dr} \right\|_{r_1} = \dfrac{\mu_0 q v_0 I}{2\pi m R^2} \]	Taking the derivative yields a constant spatial gradient inside the conductor.
Use Ampère’s law to find \(B(r)\) outside the conductor (\(r > R\)) and evaluate its derivative at \(r_2 = 2R\). \[ B_{\text{out}}(r) = \dfrac{\mu_0 I}{2\pi r} \implies \left\| \dfrac{dB_{\text{out}}}{dr} \right\| = \dfrac{\mu_0 I}{2\pi r^2} \]	Point 2 is outside the conductor, where the entire current \(I\) is enclosed.
Evaluate \(\left\|\dfrac{da}{dr}\right\|_{r_2}\) at \(r_2 = 2R\). \[ \left\| \dfrac{da}{dr} \right\|_{r_2} = \dfrac{\mu_0 q v_0 I}{2\pi m (2R)^2} = \dfrac{\mu_0 q v_0 I}{8\pi m R^2} \]	Substituting \(r_2 = 2R\) gives the derivative magnitude outside.
Compute the ratio of the two spatial derivatives. \[ \dfrac{\left\| \dfrac{da}{dr} \right\|_{r_1}}{\left\| \dfrac{da}{dr} \right\|_{r_2}} = \dfrac{\dfrac{\mu_0 q v_0 I}{2\pi m R^2}}{\dfrac{\mu_0 q v_0 I}{8\pi m R^2}} = \dfrac{1/2}{1/8} = 4 \]	Dividing the result at point 1 by the result at point 2 yields the requested ratio.

Kinematics	Forces
\(\Delta x = v_i t + \frac{1}{2} at^2\)	\(F = ma\)
\(v = v_i + at\)	\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)	\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)	\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\)

Circular Motion	Energy
\(F_c = \frac{mv^2}{r}\)	\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)	\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)	\(KE_i + PE_i = KE_f + PE_f\)
	\(W = Fd \cos\theta\)

Momentum	Torque 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 Motion	Fluids
\(F = -kx\)	\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)	\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)	\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)	\(F_b = \rho V g\)
\(a = -\omega^2 x\)	\(A_1v_1 = A_2v_2\)

Constant	Description
[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

Variable	SI 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]

Variable	Derived 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]

Prefix	Symbol	Power of Ten
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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AP Physics

Unit 1 - Vectors and Kinematics

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Advanced

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