A long wire carrying a steady current $I$ lies in a plane and is bent into two straight sections connected by a semi-circular arc of radius $R$. The straight sections lie along a horizontal line passing through point $P$, which is the center of curvature of the semi-circular arc. Which of the following gives the magnitude of the net magnetic field at point $P$ and provides the correct physical justification for the contribution from the straight sections?(a) $B = \dfrac{\mu_0 I}{2R} + \dfrac{\mu_0 I}{\pi R}$, because both the arc and the two straight sections contribute to the net magnetic field at point $P$. , (b) $B = \dfrac{\mu_0 I}{2R}$, because the straight sections create equal and opposite magnetic fields that cancel each other at point $P$. , (c) $B = \dfrac{\mu_0 I}{4R} + \dfrac{\mu_0 I}{2\pi R}$, because each straight section acts as a semi-infinite wire contributing to the field at point $P$. , (d) $B = \dfrac{\mu_0 I}{4R}$, because current elements in the straight sections are collinear with point $P$, yielding $d\vec{\ell} \times \hat{r} = 0$. :

A long wire carrying a steady current $I$ lies in a plane and is bent into two straight sections connected by a semi-circular arc of radius $R$. The straight sections lie along a horizontal line passing through point $P$, which is the center of curvature of the semi-circular arc. Which of the following gives the magnitude of the net magnetic field at point $P$ and provides the correct physical justification for the contribution from the straight sections? • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

MCQ

Mathematical

Beginner

Step	Reasoning
Identify the target quantity and apply the Biot-Savart law to determine the contribution from the straight wire sections. \[d\vec{B} = \dfrac{\mu_0 I}{4\pi} \dfrac{d\vec{\ell} \times \hat{r}}{r^2}\]	The question asks for the net magnetic field at point $P$ and the reasoning for the straight wire contribution. The Biot-Savart law relates current elements $d\vec{\ell}$ to the magnetic field vector $d\vec{B}$.
Evaluate the cross product $d\vec{\ell} \times \hat{r}$ for the straight sections. \[d\vec{\ell} \times \hat{r} = 0 \implies B_{\text{straight}} = 0\]	Point $P$ lies directly on the axis of both straight sections, meaning the length element $d\vec{\ell}$ is parallel or antiparallel to the unit vector $\hat{r}$ pointing toward point $P$.
Calculate the magnetic field contribution from the semi-circular arc. \[B_{\text{arc}} = \dfrac{\mu_0 I}{4\pi R^2} \int_0^{\pi R} d\ell = \dfrac{\mu_0 I}{4\pi R^2} (\pi R) = \dfrac{\mu_0 I}{4R}\]	Along the semi-circular arc, $d\vec{\ell}$ is perpendicular to $\hat{r}$ at all points, so $\|d\vec{\ell} \times \hat{r}\| = d\ell$. The distance to point $P$ is constant at $r = R$.
Sum the contributions to find the total magnetic field magnitude at point $P$. \[B_{\text{net}} = B_{\text{straight}} + B_{\text{arc}} = 0 + \dfrac{\mu_0 I}{4R} = \dfrac{\mu_0 I}{4R}\]	By the principle of superposition, the net field is the vector sum of the contributions from the straight sections and the arc.

Why each choice is correct or incorrect:

(A) Incorrectly treats the wire as a full circle and an infinite straight wire, failing to apply $d\vec{\ell} \times \hat{r} = 0$ for straight segments collinear with the observation point.

(B) Uses the field expression for a full circular loop rather than a semi-circle, and incorrectly claims the zero net contribution of straight segments is due to vector cancellation.

(C) Calculates the correct arc field but incorrectly adds the field of semi-infinite wires, which would only apply if point $P$ were at distance $R$ perpendicular to the straight segments.

(D) This is the correct answer.

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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

Step	Reasoning
Identify the target quantity and apply the Biot-Savart law to determine the contribution from the straight wire sections. \[d\vec{B} = \dfrac{\mu_0 I}{4\pi} \dfrac{d\vec{\ell} \times \hat{r}}{r^2}\]	The question asks for the net magnetic field at point \(P\) and the reasoning for the straight wire contribution. The Biot-Savart law relates current elements \(d\vec{\ell}\) to the magnetic field vector \(d\vec{B}\).
Evaluate the cross product \(d\vec{\ell} \times \hat{r}\) for the straight sections. \[d\vec{\ell} \times \hat{r} = 0 \implies B_{\text{straight}} = 0\]	Point \(P\) lies directly on the axis of both straight sections, meaning the length element \(d\vec{\ell}\) is parallel or antiparallel to the unit vector \(\hat{r}\) pointing toward point \(P\).
Calculate the magnetic field contribution from the semi-circular arc. \[B_{\text{arc}} = \dfrac{\mu_0 I}{4\pi R^2} \int_0^{\pi R} d\ell = \dfrac{\mu_0 I}{4\pi R^2} (\pi R) = \dfrac{\mu_0 I}{4R}\]	Along the semi-circular arc, \(d\vec{\ell}\) is perpendicular to \(\hat{r}\) at all points, so \(\|d\vec{\ell} \times \hat{r}\| = d\ell\). The distance to point \(P\) is constant at \(r = R\).
Sum the contributions to find the total magnetic field magnitude at point \(P\). \[B_{\text{net}} = B_{\text{straight}} + B_{\text{arc}} = 0 + \dfrac{\mu_0 I}{4R} = \dfrac{\mu_0 I}{4R}\]	By the principle of superposition, the net field is the vector sum of the contributions from the straight sections and the arc.

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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Unit 1 - Vectors and Kinematics

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