\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[margin=1in]{geometry}
\usepackage{xcolor}
\usepackage[most]{tcolorbox}
\usepackage{amsmath,amssymb}
\usepackage{fancyhdr}
\usepackage{graphicx}
\usepackage[hidelinks]{hyperref}
\definecolor{accent}{HTML}{7D3C98}
\definecolor{softbg}{HTML}{F5EEF8}
\pagestyle{fancy}
\fancyhf{}
\lhead{\textcolor{accent}{\textbf{First Last}}}
\chead{\textbf{Physics 101 -- Homework 7}}
\rhead{\today}
\cfoot{\thepage}
\newtcolorbox{problem}[1]{
enhanced, colback=softbg, colframe=accent, boxrule=0.8pt, arc=3pt,
title={\textbf{Problem #1}}, coltitle=white, fonttitle=\bfseries
}
\newtcolorbox{solution}{
enhanced, colback=white, colframe=accent!50, boxrule=0.6pt, arc=3pt,
title={\textbf{Solution}}, coltitle=white, colbacktitle=accent!80, fonttitle=\bfseries
}
\begin{document}
\begin{center}
\begin{tcolorbox}[colback=accent,colframe=accent,arc=4pt,width=\linewidth]
\centering\color{white}\Huge\bfseries Physics 101 -- Homework 7\\[0.2em]
\large\color{white} Electromagnetism \;•\; First Last
\end{tcolorbox}
\end{center}
\vspace{1em}
\begin{problem}{1}
A parallel-plate capacitor has plate area $A$ and separation $d$. Derive the capacitance in vacuum.
\end{problem}
\begin{solution}
Apply Gauss's law to one plate: $E = \sigma/\epsilon_0 = Q/(A\epsilon_0)$.
The potential difference is $V = Ed$. Capacitance: $C = Q/V = \epsilon_0 A / d$.
\end{solution}
\begin{problem}{2}
Compute the magnetic field inside an infinite solenoid with $n$ turns per unit length carrying current $I$.
\end{problem}
\begin{solution}
Applying Ampère's law to a rectangular loop of length $\ell$ half inside the solenoid:
$\oint \mathbf{B}\cdot d\boldsymbol\ell = \mu_0 n \ell I$, so $B = \mu_0 n I$ (axial, uniform inside).
\end{solution}
\begin{problem}{3}
A charge $+q$ is placed at the origin. Find the work required to bring a charge $+q$ from infinity to $(a, 0, 0)$.
\end{problem}
\begin{solution}
$W = k q^2 / a$ where $k = 1/(4\pi\epsilon_0)$.
\end{solution}
\begin{problem}{4}
State and explain Faraday's law of induction.
\end{problem}
\begin{solution}
The EMF induced in a closed loop equals the negative rate of change of magnetic flux through the loop:
\[ \mathcal{E} = -\frac{d\Phi_B}{dt}. \]
The minus sign encodes Lenz's law --- the induced current opposes the change in flux.
\end{solution}
\end{document}

PDF Preview
Create an account to compile and preview