A long, solid cylindrical conductor of radius $R$ carries a non-uniform current directed along its central axis. The current density inside the conductor varies with radial distance $r$ from the axis according to $J(r) = J_0 \left(1 - \dfrac{r}{R}\right)$ for $r \le R$, where $J_0$ is a positive constant. Which of the following expressions gives the magnitude of the magnetic field $B(r)$ inside the conductor at a distance \(r

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Identify Ampère’s law as the tool connecting the magnetic field along a closed circular path to the enclosed current. \[ \oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}} \]	The question asks for the magnetic field magnitude $B(r)$ inside a cylindrically symmetric current distribution.
Evaluate the line integral around a concentric circular path of radius $r < R$. \[ \oint \vec{B} \cdot d\vec{\ell} = B(r) \cdot 2\pi r \]	Cylindrical symmetry implies that the magnetic field is tangential and uniform in magnitude along the path of radius $r$.
Set up and integrate the enclosed current $I_{\text{enc}}(r)$ using thin cylindrical shells of area $dA = 2\pi r’ \, dr’$. \[ \begin{align} I_{\text{enc}}(r) &= \int_0^r J(r’) \, dA = \int_0^r J_0 \left(1 – \dfrac{r’}{R}\right) (2\pi r’) \, dr’ \\ &= 2\pi J_0 \int_0^r \left(r’ – \dfrac{r’^2}{R}\right) dr’ \\ &= 2\pi J_0 \left[ \dfrac{r^2}{2} – \dfrac{r^3}{3R} \right] \end{align} \]	Because current density depends on radial position $r’$, total enclosed current requires integrating $J(r’) \, dA$ from $r’ = 0$ to $r’ = r$.
Substitute $I_{\text{enc}}(r)$ back into Ampère’s law and solve for $B(r)$. \[ \begin{align} B(r) \cdot 2\pi r &= \mu_0 \cdot 2\pi J_0 \left( \dfrac{r^2}{2} – \dfrac{r^3}{3R} \right) \\ B(r) &= \mu_0 J_0 r \left( \dfrac{1}{2} – \dfrac{r}{3R} \right) = \dfrac{\mu_0 J_0 r}{2}\left(1 – \dfrac{2r}{3R}\right) \end{align} \]	Equating the line integral from Step 2 to $\mu_0 I_{\text{enc}}(r)$ allows solving directly for the field magnitude.

Why each choice is correct or incorrect:

(A) Incorrectly assumes that the enclosed current is given by multiplying the local current density at radius $r$ by the total cross-sectional area, $I_{\text{enc}} = J(r) \pi r^2$, instead of integrating $J(r’) \, dA$.

(B) Omits the factor of $r’$ in the differential area element by incorrectly using $dA = 2\pi \, dr’$ rather than $dA = 2\pi r’ \, dr’$.

(C) Evaluates $\int_0^r r’ \, dr’$ incorrectly as $r^2$ rather than $\dfrac{r^2}{2}$, dropping the factor of $\dfrac{1}{2}$ during integration.

(D) This is the correct answer.

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Step	Reasoning
Identify Ampère’s law as the tool connecting the magnetic field along a closed circular path to the enclosed current. \[ \oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}} \]	The question asks for the magnetic field magnitude \(B(r)\) inside a cylindrically symmetric current distribution.
Evaluate the line integral around a concentric circular path of radius \(r < R\). \[ \oint \vec{B} \cdot d\vec{\ell} = B(r) \cdot 2\pi r \]	Cylindrical symmetry implies that the magnetic field is tangential and uniform in magnitude along the path of radius \(r\).
Set up and integrate the enclosed current \(I_{\text{enc}}(r)\) using thin cylindrical shells of area \(dA = 2\pi r’ \, dr’\). \[ \begin{align} I_{\text{enc}}(r) &= \int_0^r J(r’) \, dA = \int_0^r J_0 \left(1 – \dfrac{r’}{R}\right) (2\pi r’) \, dr’ \\ &= 2\pi J_0 \int_0^r \left(r’ – \dfrac{r’^2}{R}\right) dr’ \\ &= 2\pi J_0 \left[ \dfrac{r^2}{2} – \dfrac{r^3}{3R} \right] \end{align} \]	Because current density depends on radial position \(r’\), total enclosed current requires integrating \(J(r’) \, dA\) from \(r’ = 0\) to \(r’ = r\).
Substitute \(I_{\text{enc}}(r)\) back into Ampère’s law and solve for \(B(r)\). \[ \begin{align} B(r) \cdot 2\pi r &= \mu_0 \cdot 2\pi J_0 \left( \dfrac{r^2}{2} – \dfrac{r^3}{3R} \right) \\ B(r) &= \mu_0 J_0 r \left( \dfrac{1}{2} – \dfrac{r}{3R} \right) = \dfrac{\mu_0 J_0 r}{2}\left(1 – \dfrac{2r}{3R}\right) \end{align} \]	Equating the line integral from Step 2 to \(\mu_0 I_{\text{enc}}(r)\) allows solving directly for the field magnitude.

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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