A thin, rigid square loop of wire with side length $L$ lies in the $xy$-plane and carries a steady counterclockwise current $I$. Points $P_1$, $P_2$, and $P_3$ lie along the diagonal line connecting the center of the loop to its upper-right corner, where $P_1$ is at the center, $P_2$ is halfway along the diagonal to the corner, and $P_3$ is very close to the corner. Which of the following correctly pairs the ranking of the magnetic field magnitudes $B_1$, $B_2$, and $B_3$ at these points with the magnitude of the magnetic field $B_1$ at the center of the loop? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Advanced

Step	Reasoning
Determine the magnetic field magnitude $B_1$ at the center $P_1$ of the square loop. \[ d\vec{B} = \dfrac{\mu_0 I}{4\pi} \dfrac{d\vec{\ell} \times \hat{r}}{r^2} \]	The question requires calculating the exact field magnitude at the center by summing the vector contributions from all four straight segments of length $L$.
Integrate the Biot-Savart law along one side of length $L$ located at perpendicular distance $d = L/2$ from the center. \[ B_{\text{1 side}} = \dfrac{\mu_0 I}{4\pi (L/2)} \int_{-\pi/4}^{\pi/4} \cos\theta\,d\theta = \dfrac{\mu_0 I}{2\pi L} \left[\sin\left(\dfrac{\pi}{4}\right) – \sin\left(-\dfrac{\pi}{4}\right)\right] = \dfrac{\sqrt{2}\mu_0 I}{2\pi L} \]	To find the field from one finite straight segment subtending angles from $\theta_1 = -45^\circ$ to $\theta_2 = +45^\circ$ relative to the normal from the segment to the center.
Sum the contributions from all four sides. \[ B_1 = 4 B_{\text{1 side}} = 4 \left(\dfrac{\sqrt{2}\mu_0 I}{2\pi L}\right) = \dfrac{2\sqrt{2}\mu_0 I}{\pi L} \]	By symmetry and the right-hand rule, all four sides produce magnetic field vectors pointing out of the page at the center.
Compare the perpendicular distances from points $P_1$, $P_2$, and $P_3$ to the nearest wire segments.	The field magnitude from each wire segment scales inversely with perpendicular distance ($B \propto 1/d$). As a point moves along the diagonal from the center toward a corner, its distance to the two adjacent wire segments forming that corner decreases toward zero.
Establish the magnitude ranking $B_3 > B_2 > B_1$.	At $P_1$, all sides are at distance $L/2$. At $P_2$, two sides are at distance $L/4$, increasing their field contribution. At $P_3$, the distance to the two corner wires approaches zero, causing the field magnitude to diverge to infinity. Thus $B_3 > B_2 > B_1$.

Why each choice is correct or incorrect:

(A) This is the correct answer.

(B) Incorrectly assumes the magnetic field is largest at the center where all four sides contribute symmetrically, ignoring that approaching a wire segment increases the local field magnitude as $1/d$.

(C) Correctly ranks the magnetic field magnitudes but fails to multiply by 4 when summing the contributions of all four sides of the square loop.

(D) Incorrectly assumes the field inside a closed current loop is uniform and uses the infinite wire expression ($\sin\theta = 1$) rather than integrating over finite wire segments.

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Step	Reasoning
Determine the magnetic field magnitude \(B_1\) at the center \(P_1\) of the square loop. \[ d\vec{B} = \dfrac{\mu_0 I}{4\pi} \dfrac{d\vec{\ell} \times \hat{r}}{r^2} \]	The question requires calculating the exact field magnitude at the center by summing the vector contributions from all four straight segments of length \(L\).
Integrate the Biot-Savart law along one side of length \(L\) located at perpendicular distance \(d = L/2\) from the center. \[ B_{\text{1 side}} = \dfrac{\mu_0 I}{4\pi (L/2)} \int_{-\pi/4}^{\pi/4} \cos\theta\,d\theta = \dfrac{\mu_0 I}{2\pi L} \left[\sin\left(\dfrac{\pi}{4}\right) – \sin\left(-\dfrac{\pi}{4}\right)\right] = \dfrac{\sqrt{2}\mu_0 I}{2\pi L} \]	To find the field from one finite straight segment subtending angles from \(\theta_1 = -45^\circ\) to \(\theta_2 = +45^\circ\) relative to the normal from the segment to the center.
Sum the contributions from all four sides. \[ B_1 = 4 B_{\text{1 side}} = 4 \left(\dfrac{\sqrt{2}\mu_0 I}{2\pi L}\right) = \dfrac{2\sqrt{2}\mu_0 I}{\pi L} \]	By symmetry and the right-hand rule, all four sides produce magnetic field vectors pointing out of the page at the center.
Compare the perpendicular distances from points \(P_1\), \(P_2\), and \(P_3\) to the nearest wire segments.	The field magnitude from each wire segment scales inversely with perpendicular distance (\(B \propto 1/d\)). As a point moves along the diagonal from the center toward a corner, its distance to the two adjacent wire segments forming that corner decreases toward zero.
Establish the magnitude ranking \(B_3 > B_2 > B_1\).	At \(P_1\), all sides are at distance \(L/2\). At \(P_2\), two sides are at distance \(L/4\), increasing their field contribution. At \(P_3\), the distance to the two corner wires approaches zero, causing the field magnitude to diverge to infinity. Thus \(B_3 > B_2 > B_1\).

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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