\subsection{Magnetic Flux}

Magnetic flux is a scalar quantity that measures the total magnetic field passing through a given surface. It is the magnetic analogue of electric flux and provides the natural framework for understanding Faraday's law of induction, which is the subject of the next section.

\dfn{Magnetic flux}{Let $\vec{B}$ denote the magnetic-field vector and let $S$ be an oriented surface. The \emph{magnetic flux} through $S$ is the surface integral
\[
\Phi_B = \iint_S \vec{B}\cdot d\vec{A},
\]
where $d\vec{A}$ is the vector area element: its magnitude is the area of the infinitesimal surface patch, and its direction is normal to the surface. For an open surface the orientation (and hence the direction of $d\vec{A}$) must be specified; for a closed surface the convention is that $d\vec{A}$ points outward.

The magnetic flux $\Phi_B$ is a scalar quantity. Its sign depends on the relative orientation of the surface normal and the field: positive when $\vec{B}$ has a component along the surface normal, negative when it opposes the normal, and zero when $\vec{B}$ is tangent to the surface.}

\nt{Think of magnetic flux as the ``number of magnetic field lines'' passing through a surface. A larger surface catches more lines; tilting the surface reduces the effective area and hence the flux; reversing the surface normal flips the sign of the flux. This picture is qualitative, but it is extremely useful for building intuition about how $\Phi_B$ changes when the surface moves or rotates.}

\thm{Magnetic flux through a flat surface in a uniform field}{When the magnetic field $\vec{B}$ is uniform (constant in magnitude and direction) and the surface is flat with area $A$ and unit normal $\hat{n}$, the surface integral simplifies to a single dot product:
\[
\Phi_B = \vec{B}\cdot\vec{A} = B\,A\,\cos\theta,
\]
where $\vec{A}=A\,\hat{n}$ is the area vector, $B=|\vec{B}|$, and $\theta$ is the angle between $\vec{B}$ and the area normal $\hat{n}$.}

\nt{The angle $\theta$ is always measured between $\vec{B}$ and the \emph{surface normal}, \emph{not} between $\vec{B}$ and the surface itself. When $\vec{B}$ is perpendicular to the surface, $\theta=0^\circ$ and $\cos\theta=1$, giving maximum flux $\Phi_B = BA$. When $\vec{B}$ is parallel to the surface, $\theta=90^\circ$ and $\cos\theta=0$, giving zero flux. These are the two extremes that frequently appear on exams.}

\mprop{Magnetic flux: key properties}{
\begin{itemize}
\item \textbf{SI unit:} The weber, $\mathrm{Wb}$, where $1\,\mathrm{Wb}=1\,\mathrm{T\!\cdot\!m^2}$.
\item \textbf{Scalar:} $\Phi_B$ is a scalar. It can be positive, negative, or zero, depending on the choice of surface orientation.
\item \textbf{Orientation dependence:} $\Phi_B = BA\cos\theta$, where $\theta$ is the angle between $\vec{B}$ and the surface normal. Flux is maximal when $\vec{B}\perp$ surface ($\theta=0^\circ$) and zero when $\vec{B}\parallel$ surface ($\theta=90^\circ$).
\item \textbf{Comparison with electric flux:} The electric flux through a surface is $\Phi_E = \iint_S \vec{E}\cdot d\vec{A}$. The mathematical structure is identical; the only difference is that $\Phi_E$ can be non-zero for closed surfaces enclosing net charge ($\Phi_E = Q_{\text{enc}}/\varepsilon_0$, Gauss's law), while $\Phi_B$ is always zero through a closed surface.
\item \textbf{Time dependence:} If any of $B$, $A$, or $\theta$ changes with time, the flux changes: $\Delta\Phi_B = \Phi_{B,f} - \Phi_{B,i}$. A changing magnetic flux is the fundamental ingredient behind electromagnetic induction (Faraday's law).
\end{itemize}}

\cor{Gauss's law for magnetism (qualitative)}{The net magnetic flux through any \emph{closed} surface is zero:
\[
\oiint_{\text{closed}} \vec{B}\cdot d\vec{A} = 0.
\]
This is Gauss's law for magnetism. It reflects the experimental fact that magnetic monopoles have never been observed: magnetic field lines always form closed loops, so every field line that enters a closed surface must also exit it. This stands in contrast to electric flux through a closed surface, which is proportional to the enclosed charge.}

\ex{Illustrative example}{A square loop of side $0.20\,\mathrm{m}$ sits in a uniform magnetic field $B=0.30\,\mathrm{T}$. The field makes an angle $\theta=35^\circ$ with the loop's normal. The area is $A=(0.20\,\mathrm{m})^2=0.040\,\mathrm{m^2}$, so the flux is
\[
\Phi_B = B\,A\,\cos\theta = (0.30\,\mathrm{T})(0.040\,\mathrm{m^2})\cos 35^\circ = 0.012 \times 0.8192 = 9.83\times 10^{-3}\,\mathrm{Wb}.
\]}

\qs{Worked example}{A rectangular loop of wire has width $w=12.0\,\mathrm{cm}=0.120\,\mathrm{m}$ and height $h=8.00\,\mathrm{cm}=0.0800\,\mathrm{m}$. The loop is situated in a uniform magnetic field of magnitude $B=0.450\,\mathrm{T}$. The field is constant in space. The loop has area $A=w\,h$.

Find the magnetic flux $\Phi_B$ through the loop for each of the following orientations:
\begin{enumerate}[label=(\alph*)]
\item The loop is in the $xy$-plane and the magnetic field points in the $+\hat{k}$ direction (perpendicular to the loop, with the area normal also taken as $+\hat{k}$).
\item The loop is in the $xz$-plane and the magnetic field points in the $+\hat{k}$ direction (the field is parallel to the loop).
\item The loop is in the $xy$-plane and the magnetic field lies in the $xz$-plane, making an angle $\theta=30.0^\circ$ below the $x$-axis. Take the area normal as $+\hat{k}$.
\item The loop is in the $yz$-plane and the magnetic field points in the $+\hat{k}$ direction (the field is perpendicular to the loop, with the area normal taken as $+\hat{k}$).
\item The loop is in the $xy$-plane. The magnetic field is $\vec{B}=(0.450\,\mathrm{T})\,\hat{\imath}+(0.300\,\mathrm{T})\,\hat{\jmath}$. Take the area normal as $+\hat{k}$.
\end{enumerate}}

\sol First compute the loop's area:
\[
A = w\,h = (0.120\,\mathrm{m})(0.0800\,\mathrm{m}) = 9.60\times 10^{-3}\,\mathrm{m^2}.
\]
The area vector is $\vec{A} = A\,\hat{n}$, where $\hat{n}$ is the unit normal determined by the loop's orientation. Since $\vec{B}$ is uniform, we use $\Phi_B = \vec{B}\cdot\vec{A} = B\,A\,\cos\theta$, where $\theta$ is the angle between $\vec{B}$ and $\hat{n}$.

\textbf{(a)} The loop is in the $xy$-plane with normal $\hat{n}=+\hat{k}$, and $\vec{B} = B\,\hat{k}$. The angle between $\vec{B}$ and $\hat{n}$ is $\theta = 0^\circ$:
\[
\Phi_B = B\,A\,\cos 0^\circ = (0.450\,\mathrm{T})(9.60\times 10^{-3}\,\mathrm{m^2})(1).
\]
Compute:
\[
\Phi_B = 0.450 \times 9.60\times 10^{-3}\,\mathrm{Wb} = 4.32\times 10^{-3}\,\mathrm{Wb}.
\]

\textbf{(b)} The loop is in the $xz$-plane with normal $\hat{n}=+\hat{j}$ (by the right-hand rule, or equivalently, the area vector points along $y$). The field is $\vec{B}=B\,\hat{k}$. The angle between $\vec{B}$ and $\hat{n}$ is $\theta = 90^\circ$:
\[
\Phi_B = B\,A\,\cos 90^\circ = 0.
\]
No field lines pass through the loop; the field runs parallel to the plane of the loop.

\textbf{(c)} The loop is in the $xy$-plane with normal $\hat{n}=+\hat{k}$. The field is $\vec{B}$ lying in the $xz$-plane at $30.0^\circ$ below the $x$-axis. We need the angle between $\vec{B}$ and $\hat{k}$. Since $\vec{B}$ is in the $xz$-plane and makes $30.0^\circ$ with the $x$-axis, the angle with the $z$-axis ($\hat{k}$) is $\theta = 90^\circ - 30.0^\circ = 60.0^\circ$:
\[
\Phi_B = B\,A\,\cos 60.0^\circ = (0.450\,\mathrm{T})(9.60\times 10^{-3}\,\mathrm{m^2})\,(0.500).
\]
Compute:
\[
\Phi_B = 0.450 \times 9.60\times 10^{-3} \times 0.500\,\mathrm{Wb} = 2.16\times 10^{-3}\,\mathrm{Wb}.
\]

\textbf{(d)} The loop is in the $yz$-plane with normal $\hat{n}=+\hat{k}$, and the field is $\vec{B}=B\,\hat{k}$. Again $\theta = 0^\circ$:
\[
\Phi_B = B\,A\,\cos 0^\circ = (0.450\,\mathrm{T})(9.60\times 10^{-3}\,\mathrm{m^2})(1).
\]
\[
\Phi_B = 4.32\times 10^{-3}\,\mathrm{Wb}.
\]
The orientation of the loop in space does not matter; only the relative angle between $\vec{B}$ and the area normal matters. Since $\vec{B}$ is again perpendicular to the loop and aligned with the normal, the flux is the same as in part (a).

\textbf{(e)} The loop is in the $xy$-plane with normal $\hat{n}=+\hat{k}$. The field is $\vec{B}=(0.450\,\mathrm{T})\,\hat{\imath}+(0.300\,\mathrm{T})\,\hat{\jmath}$. The area vector is $\vec{A} = A\,\hat{k}$. The flux is the dot product:
\[
\Phi_B = \vec{B}\cdot\vec{A} = \bigl[(0.450)\,\hat{\imath}+(0.300)\,\hat{\jmath}\bigr]\cdot\bigl[(9.60\times 10^{-3})\,\hat{k}\bigr].
\]
Since $\hat{\imath}\cdot\hat{k}=0$ and $\hat{\jmath}\cdot\hat{k}=0$:
\[
\Phi_B = 0.
\]
The field lies entirely in the $xy$-plane, parallel to the loop, so no flux passes through.

\bigskip
\textbf{Final answers:}
\begin{enumerate}[label=(\alph*)]
\item $\Phi_B = 4.32\times 10^{-3}\,\mathrm{Wb} = 4.32\,\mathrm{mWb}$
\item $\Phi_B = 0$
\item $\Phi_B = 2.16\times 10^{-3}\,\mathrm{Wb} = 2.16\,\mathrm{mWb}$
\item $\Phi_B = 4.32\times 10^{-3}\,\mathrm{Wb} = 4.32\,\mathrm{mWb}$
\item $\Phi_B = 0$
\end{enumerate}
