A flat nonconducting disk of radius $R$ carries a total electric charge $Q$ distributed uniformly over its surface. The disk rotates about its central symmetry axis perpendicular to its plane at a constant angular frequency $\omega$. Which of the following expressions represents the magnitude of the magnetic field at the center of the disk? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Identify the target quantity and strategy	The question asks for the magnetic field at the center of a rotating charged disk, which can be modeled as a continuous series of concentric thin ring current loops.
Express the charge of a differential ring element of radius $r$ and width $dr$ \[ \sigma = \dfrac{Q}{\pi R^2} \implies dq = \sigma (2\pi r \, dr) = \dfrac{2Q r \, dr}{R^2} \]	To find the effective current of a differential element, we first need the charge $dq$ contained within that element.
Relate the charge element and rotation rate to current \[ dI = \dfrac{dq}{T} = \dfrac{dq \, \omega}{2\pi} = \left(\dfrac{2Q r \, dr}{R^2}\right) \left(\dfrac{\omega}{2\pi}\right) = \dfrac{Q \omega r \, dr}{\pi R^2} \]	As the disk rotates with period $T = 2\pi / \omega$, the charge $dq$ passes a given point once per period, creating an effective ring current $dI$.
Formulate the magnetic field contribution $dB$ at the center \[ dB = \dfrac{\mu_0 dI}{2r} = \dfrac{\mu_0}{2r} \left(\dfrac{Q \omega r \, dr}{\pi R^2}\right) = \dfrac{\mu_0 Q \omega}{2\pi R^2} dr \]	A circular current loop of radius $r$ carrying current $dI$ produces a magnetic field at its center given by $dB = \dfrac{\mu_0 dI}{2r}$.
Integrate $dB$ across the radius of the disk \[ B = \int_0^R \dfrac{\mu_0 Q \omega}{2\pi R^2} dr = \dfrac{\mu_0 Q \omega}{2\pi R^2} \int_0^R dr = \dfrac{\mu_0 Q \omega}{2\pi R} \]	Summing the contributions from all concentric rings from $r = 0$ to $r = R$ yields the total magnetic field.

Why each choice is correct or incorrect:

(A) This is the correct answer.

(B) Treats the disk as a single narrow ring of radius $R$ carrying total current $I = Q\omega / 2\pi$, yielding $B = \mu_0 I / (2R) = \mu_0 Q \omega / (4\pi R)$.

(C) Uses $dB = \mu_0 dI / r$ instead of the correct loop formula $dB = \mu_0 dI / (2r)$, forgetting the factor of $1/2$.

(D) Mistakenly calculates the area of the differential ring as $\pi r dr$ instead of $2\pi r dr$, dropping a factor of 2 in $dq$.

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Step	Reasoning
Identify the target quantity and strategy	The question asks for the magnetic field at the center of a rotating charged disk, which can be modeled as a continuous series of concentric thin ring current loops.
Express the charge of a differential ring element of radius \(r\) and width \(dr\) \[ \sigma = \dfrac{Q}{\pi R^2} \implies dq = \sigma (2\pi r \, dr) = \dfrac{2Q r \, dr}{R^2} \]	To find the effective current of a differential element, we first need the charge \(dq\) contained within that element.
Relate the charge element and rotation rate to current \[ dI = \dfrac{dq}{T} = \dfrac{dq \, \omega}{2\pi} = \left(\dfrac{2Q r \, dr}{R^2}\right) \left(\dfrac{\omega}{2\pi}\right) = \dfrac{Q \omega r \, dr}{\pi R^2} \]	As the disk rotates with period \(T = 2\pi / \omega\), the charge \(dq\) passes a given point once per period, creating an effective ring current \(dI\).
Formulate the magnetic field contribution \(dB\) at the center \[ dB = \dfrac{\mu_0 dI}{2r} = \dfrac{\mu_0}{2r} \left(\dfrac{Q \omega r \, dr}{\pi R^2}\right) = \dfrac{\mu_0 Q \omega}{2\pi R^2} dr \]	A circular current loop of radius \(r\) carrying current \(dI\) produces a magnetic field at its center given by \(dB = \dfrac{\mu_0 dI}{2r}\).
Integrate \(dB\) across the radius of the disk \[ B = \int_0^R \dfrac{\mu_0 Q \omega}{2\pi R^2} dr = \dfrac{\mu_0 Q \omega}{2\pi R^2} \int_0^R dr = \dfrac{\mu_0 Q \omega}{2\pi R} \]	Summing the contributions from all concentric rings from \(r = 0\) to \(r = R\) yields the total magnetic field.

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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