我想让行之间的距离更大，如果这样做，换行符的大小也会增加，这是不想要的，因为目标是隔离行以明确哪些符号和文本是关联的。我在换行表环境中使用它，让文本位于左侧
\newpage\subsection{Current Loop}\label{CurrentLoop}
\begin{wraptable}{r}{3.9cm}\vspace{-0.52cm}
\footnotesize
\onehalfspacing
\caption{Annotation and their explanations for the current loop}\label{AnnotationsCurrent}
\begin{tabular}{p{4mm} p{2.8cm}}
\textbf{I} & is current, even throughout the section\\
\textbf{$\vv{J}$} & is the current density vector\\
\textbf{$\vv{dl}$} & is the line segment vector\\
\textbf{$\mu_{0}$} & is the vacuum permeability\\
\textbf{$\vv{B_n}$} & is the B-field at segment \textbf{n}\\
\textbf{$\vv{B_{\perp n}}$} & is the perpendicular B-field at segment \textbf{n}\\
\textbf{$\vv{B_{\parallel n}}$} & is the parallel B-field at segment \textbf{n}\\
\textbf{$\vv{R}$} & Radius from center to a segment\\
\textbf{d} & distance from center to probing point \textbf{$P_b$}\\
\textbf{$\psi$} & is the angle between \textbf{d} \& $\vv{r}$\\
\textbf{$\vv{r}$} & is the vector from segment \textbf{n} to $P_B$\\
\textbf{$\hat{\mathrm{r}}$} & is the unit vector\\
\\
\end{tabular}
\end{wraptable}
To theoretically analyze the behavior of a coil, a great model to analyze is a closed loop, where we consider it to be one turn of a coil. The model is imaginary and would not be realistic to use outside of the theory.\\
We consider a loop where the current is constant, and we divide this loop into 24 evenly-sized sections of wire for simplicity of the calculation. At last, we believe infinite sections are a more accurate version of a coil. The ring is made out of a conducting material.
To find the B-field of section 1, the below formula is used
\begin{equation}\label{BfieldSection1}
\vv{\mathrm{B}}_1=\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{\vv{\mathrm{dl}} \times \hat{\mathrm{r}}}{\mathrm{r}^2}
\end{equation}
The single-loop coil model is illustrated in figure xx.
The B-Field of the sections can be split into a parallel component and a perpendicular component. When this is done, we see the diametrically opposing sections, which allows us to see that they cancel each other out.
\begin{equation}\label{totalBfield24SegmentVersion}
\vv{\mathrm{B}}_{\text {tot }}=\vv{\mathrm{B}}_{\| 1}+\vv{\mathrm{B}}_{\| 2}+\cdots+\vv{\mathrm{B}}_{\| 24}
\end{equation}Knowing all of the sections are parallel and have the same direction, we can take the length and add them up
\begin{equation}
\left|\vv{\mathrm{B}}_{\text {tot }}\right|=\left|\vv{\mathrm{B}}_{\| 1}\right|+\left|\vv{\mathrm{B}}_{\| 2}\right|+\cdots+\left|\vv{\mathrm{B}}_{\| 24}\right|
\end{equation}
The first section of the looped current segment is calculated
\begin{equation}\label{FirstLoopedCurrentSegment}
\left|\vv{B}_{\| 1}\right|=\left|\vv{B}_1\right| \sin \psi=\left|\vv{B}_1\right| \frac{R}{\left(d^2+R^2\right)^{1 / 2}}
\end{equation}
$\vv{dl}$ and $R$ are perpendicular this lets us write \eqref{FirstLoopedCurrentSegment} into
\begin{equation}
\left|\vv{\mathrm{B}}_1\right|=\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{|\vv{\mathrm{dl}} \times \hat{\mathrm{r}}|}{\mathrm{r}^2}=\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{|\vv{\mathrm{dl}}|}{\mathrm{r}^2}=\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{|\vv{\mathrm{dl}}|}{\mathrm{d}^2+\mathrm{R}^2}
\end{equation}
答案1
这是我的尝试。
\documentclass{article}
\usepackage{amsmath,esvect,wrapfig}
\usepackage{tabularx}
\usepackage{caption}
\setlength{\intextsep}{3pt}
\begin{document}
\subsection{Current Loop}\label{CurrentLoop}
\begin{wraptable}{r}{3.9cm}
\renewcommand{\arraystretch}{1.3}
\footnotesize
\caption{Annotation and their explanations for the current loop}\label{AnnotationsCurrent}
\begin{tabularx}{\linewidth}{@{}l>{\raggedright\arraybackslash}X@{}}
$I$ & is current, even throughout the section\\
$\vv{J}$ & is the current density vector\\
$\vv{dl}$ & is the line segment vector\\
$\mu_{0}$ & is the vacuum permeability\\
$\vv{B_n}$ & is the $B$-field at segment $n$\\
$\vv{B_{\perp n}}$ & is the perpendicular $B$-field at segment $n$\\
$\vv{B_{\parallel n}}$ & is the parallel $B$-field at segment $n$\\
$\vv{R}$ & Radius from center to a segment\\
$d$ & distance from center to probing point $P_b$\\
$\psi$ & is the angle between $d$ and $\vv{r}$\\
$\vv{r}$ & is the vector from segment $n$ to $P_B$\\
$\hat{\mathrm{r}}$ & is the unit vector\\
\end{tabularx}
\bigskip
\end{wraptable}
To theoretically analyze the behavior of a coil, a great model to analyze
is a closed loop, where we consider it to be one turn of a coil. The model
is imaginary and would not be realistic to use outside of the theory.
We consider a loop where the current is constant, and we divide this loop
into 24 evenly-sized sections of wire for simplicity of the calculation.
At last, we believe infinite sections are a more accurate version of a coil.
The ring is made out of a conducting material.
To find the $B$-field of section 1, the below formula is used
\begin{equation}\label{BfieldSection1}
\vv{\mathrm{B}}_1=
\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{\vv{\mathrm{dl}} \times \hat{\mathrm{r}}}{\mathrm{r}^2}
\end{equation}
The single-loop coil model is illustrated in figure~xx.
The $B$-Field of the sections can be split into a parallel component and
a perpendicular component. When this is done, we see the diametrically
opposing sections, which allows us to see that they cancel each other out.
\begin{equation}\label{totalBfield24SegmentVersion}
\vv{\mathrm{B}}_{\mathrm{tot}}=
\vv{\mathrm{B}}_{\| 1}+\vv{\mathrm{B}}_{\| 2}+\cdots+\vv{\mathrm{B}}_{\| 24}
\end{equation}
Knowing all of the sections are parallel and have the same direction,
we can take the length and add them up
\begin{equation}
|\vv{\mathrm{B}}_{\mathrm{tot}}|=
|\vv{\mathrm{B}}_{\| 1}|+|\vv{\mathrm{B}}_{\| 2}|+\cdots+|\vv{\mathrm{B}}_{\| 24}|
\end{equation}
The first section of the looped current segment is calculated
\begin{equation}\label{FirstLoopedCurrentSegment}
|\vv{B}_{\| 1}|=|\vv{B}_1| \sin \psi=
|\vv{B}_1| \frac{R}{(d^2+R^2)^{1 / 2}}
\end{equation}
$\vv{dl}$ and $R$ are perpendicular this lets us write \eqref{FirstLoopedCurrentSegment} into
\begin{equation}
|\vv{\mathrm{B}}_1|=
\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{|\vv{\mathrm{dl}} \times \hat{\mathrm{r}}|}{\mathrm{r}^2}=
\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{|\vv{\mathrm{dl}}|}{\mathrm{r}^2}=
\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{|\vv{\mathrm{dl}}|}{\mathrm{d}^2+\mathrm{R}^2}
\end{equation}
\end{document}
\onehalfspacing如果为文档添加，请务必\singlelinespacing为执行wraptable。
您还应该修复符号中的各种不一致之处：有时您会得到$r$，而其他时候 的情况\mathrm{r}类似；在表格中dl也有，但在文本中：箭头总是覆盖下标或从不覆盖下标（并且不同于，尽管它们通常打印相同）。请注意，在表格的第一列中什么也不做（除了产生其他不一致之外）。\vv{B_{\parallel n}}\vv{B}_{\| 1}\|\parallel\textbf
我删除了所有\left和\right命令，因为您可以看到它们会产生过大的分隔符。您可能想要\bigl|和\bigr|（在我看来不需要），但使用\left和时，\right分隔符确实太长了。
答案2
- 你应该提供一个 MWE（最小工作示例），这样我们就能知道你在文档序言中加载了哪些与你的问题相关的包和定义
- 用于增加表格行之间的垂直空间（通常）使用
\renewcommand\arraystretch{1.5}，如@DavidCarlisle在他的评论中提到的那样，或者使用能够定义这一点的包（如cellspace或tabularray - 使用该
tabularray包的一个可能的解决方案是：
\documentclass{article}
%---------------- show page layout. don't use in a real document!
\usepackage{showframe}
\renewcommand\ShowFrameLinethickness{0.15pt}
\renewcommand*\ShowFrameColor{\color{red}}
%---------------------------------------------------------------%
\usepackage{amsmath}
\usepackage{esvect}
\usepackage{tabularray}
\usepackage{wrapfig}
\usepackage{setspace}
\onehalfspacing
\begin{document}
\subsection{Current Loop}\label{CurrentLoop}
\begin{wraptable}{r}{3.9cm}
\vspace{-1.7\baselineskip}
\caption{Annotation and their explanations for the current loop}\label{AnnotationsCurrent}
\footnotesize
\begin{tblr}{colspec={@{} Q[l, mode=math]
Q[l, wd=28mm, font=\linespread{1}\selectfont] @{}}
}
\textbf{I} & is current, even throughout the section\\
\vv{J} & is the current density vector\\
\vv{dl} & is the line segment vector\\
\mu_{0} & is the vacuum permeability\\
\vv{B}_n & is the B-field at segment \textbf{n}\\
\vv{B}_{\perp n} & is the perpendicular B-field at segment \textbf{n}\\
\vv{B}_{\parallel n} & is the parallel B-field at segment \textbf{n}\\
\vv{R} & Radius from center to a segment\\
\mathbf{d} & distance from center to probing point $P_b$\\
\psi & is the angle between \textbf{d} \& $\vv{r}$\\
\vv{r} & is the vector from segment \textbf{n} to $P_B$\\
\hat{\mathrm{r}} & is the unit vector
\end{tblr}
\end{wraptable}
To theoretically analyze the behavior of a coil, a great model to analyze is a closed loop, where we consider it to be one turn of a coil. The model is imaginary and would not be realistic to use outside of the theory.\\
We consider a loop where the current is constant, and we divide this loop into 24 evenly-sized sections of wire for simplicity of the calculation. At last, we believe infinite sections are a more accurate version of a coil. The ring is made out of a conducting material.
To find the B-field of section 1, the below formula is used
\begin{equation}\label{BfieldSection1}
\vv{\mathrm{B}}_1=\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{\vv{\mathrm{dl}} \times \hat{\mathrm{r}}}{\mathrm{r}^2}
\end{equation}
The single-loop coil model is illustrated in figure xx.
The B-Field of the sections can be split into a parallel component and a perpendicular component. When this is done, we see the diametrically opposing sections, which allows us to see that they cancel each other out.
\begin{equation}\label{totalBfield24SegmentVersion}
\vv{\mathrm{B}}_{\text {tot }}=\vv{\mathrm{B}}_{\| 1}+\vv{\mathrm{B}}_{\| 2}+\cdots+\vv{\mathrm{B}}_{\| 24}
\end{equation}Knowing all of the sections are parallel and have the same direction, we can take the length and add them up
\begin{equation}
\left|\vv{\mathrm{B}}_{\text {tot }}\right|=\left|\vv{\mathrm{B}}_{\| 1}\right|+\left|\vv{\mathrm{B}}_{\| 2}\right|+\cdots+\left|\vv{\mathrm{B}}_{\| 24}\right|
\end{equation}
The first section of the looped current segment is calculated
\begin{equation}\label{FirstLoopedCurrentSegment}
\left|\vv{B}_{\| 1}\right|=\left|\vv{B}_1\right| \sin \psi=\left|\vv{B}_1\right| \frac{R}{\left(d^2+R^2\right)^{1 / 2}}
\end{equation}
$\vv{dl}$ and $R$ are perpendicular this lets us write \eqref{FirstLoopedCurrentSegment} into
\begin{equation}
\left|\vv{\mathrm{B}}_1\right|=\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{|\vv{\mathrm{dl}} \times \hat{\mathrm{r}}|}{\mathrm{r}^2}=\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{|\vv{\mathrm{dl}}|}{\mathrm{r}^2}=\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{|\vv{\mathrm{dl}}|}{\mathrm{d}^2+\mathrm{R}^2}
\end{equation}
\end{document}
这是（接近）您所寻找的吗？