A long, solid cylindrical conductor of radius $R$ carries a uniform current density $J$ parallel to its central axis. A cylindrical cavity of radius $a$ is drilled parallel to the conductor's axis, with its center displaced by a distance $d$ from the central axis of the conductor, where $d + a < R$. Which of the following expressions represents the magnitude of the magnetic field at any point inside the cavity? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Identify the strategy to determine the net magnetic field inside the cavity.	A conductor with a cavity can be modeled as a solid conductor of uniform current density $\vec{J}$ superimposed with a smaller cylinder of negative current density $-\vec{J}$ occupying the volume of the cavity.
Apply Ampère’s law to find the magnetic field vector $\vec{B}_1$ inside a solid cylinder carrying current density $\vec{J} = J\hat{k}$ at position $\vec{r}$ relative to its center. \[ \vec{B}_1(\vec{r}) = \frac{\mu_0}{2} \vec{J} \times \vec{r} \]	Ampère’s law gives $B_1(2\pi r) = \mu_0 (J \pi r^2)$, which yields $B_1 = \dfrac{1}{2}\mu_0 J r$. In vector notation, the field is perpendicular to $\vec{r}$.
Determine the magnetic field vector $\vec{B}_2$ produced by the negative current density $-\vec{J}$ filling the cavity, measured relative to the cavity center at position $\vec{r}’ = \vec{r} – \vec{d}$. \[ \vec{B}_2(\vec{r}) = -\frac{\mu_0}{2} \vec{J} \times (\vec{r} – \vec{d}) \]	The vector location of a point inside the cavity relative to the cavity axis is $\vec{r} – \vec{d}$, where $\vec{d}$ is the displacement vector from the main cylinder axis to the cavity axis.
Sum the two field contributions using the principle of superposition to find the net magnetic field inside the cavity. \[ \vec{B}_{net} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0}{2} \vec{J} \times \vec{r} – \frac{\mu_0}{2} \vec{J} \times (\vec{r} – \vec{d}) = \frac{\mu_0}{2} \vec{J} \times \vec{d} \]	The total field is the linear sum of the field of the solid cylinder and the field of the negative current in the cavity region.
Compute the magnitude of the net magnetic field. \[ B = \|\vec{B}_{net}\| = \frac{1}{2} \mu_0 J d \]	Since $\vec{J}$ is perpendicular to $\vec{d}$, the magnitude of the cross product simplifies directly.

Why each choice is correct or incorrect:

(A) Incorrectly assumes the magnetic field inside the cavity is identical to that inside a solid conductor, ignoring the cavity’s effect.

(B) This is the correct answer.

(C) Mistakenly treats the subtracted current of the cavity as a line filament located at the cavity axis rather than applying superposition of uniform volume current distributions.

(D) Incorrectly constructs a scalar difference between the outer radius R and the displacement d rather than performing vector superposition.

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Step	Reasoning
Identify the strategy to determine the net magnetic field inside the cavity.	A conductor with a cavity can be modeled as a solid conductor of uniform current density \(\vec{J}\) superimposed with a smaller cylinder of negative current density \(-\vec{J}\) occupying the volume of the cavity.
Apply Ampère’s law to find the magnetic field vector \(\vec{B}_1\) inside a solid cylinder carrying current density \(\vec{J} = J\hat{k}\) at position \(\vec{r}\) relative to its center. \[ \vec{B}_1(\vec{r}) = \frac{\mu_0}{2} \vec{J} \times \vec{r} \]	Ampère’s law gives \(B_1(2\pi r) = \mu_0 (J \pi r^2)\), which yields \(B_1 = \dfrac{1}{2}\mu_0 J r\). In vector notation, the field is perpendicular to \(\vec{r}\).
Determine the magnetic field vector \(\vec{B}_2\) produced by the negative current density \(-\vec{J}\) filling the cavity, measured relative to the cavity center at position \(\vec{r}’ = \vec{r} – \vec{d}\). \[ \vec{B}_2(\vec{r}) = -\frac{\mu_0}{2} \vec{J} \times (\vec{r} – \vec{d}) \]	The vector location of a point inside the cavity relative to the cavity axis is \(\vec{r} – \vec{d}\), where \(\vec{d}\) is the displacement vector from the main cylinder axis to the cavity axis.
Sum the two field contributions using the principle of superposition to find the net magnetic field inside the cavity. \[ \vec{B}_{net} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0}{2} \vec{J} \times \vec{r} – \frac{\mu_0}{2} \vec{J} \times (\vec{r} – \vec{d}) = \frac{\mu_0}{2} \vec{J} \times \vec{d} \]	The total field is the linear sum of the field of the solid cylinder and the field of the negative current in the cavity region.
Compute the magnitude of the net magnetic field. \[ B = \|\vec{B}_{net}\| = \frac{1}{2} \mu_0 J d \]	Since \(\vec{J}\) is perpendicular to \(\vec{d}\), the magnitude of the cross product simplifies directly.

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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