\subsection{Force on Current-Carrying Conductors and Loops}

A current-carrying wire in a magnetic field experiences a force because the moving charge carriers inside the wire each feel a magnetic force. When the current flows through a closed loop, the forces on individual segments can produce a net torque, causing the loop to rotate. This is the operating principle of electric motors.

\dfn{Force on a current-carrying wire}{A wire carrying current $I$ and placed in a magnetic field $\vec{B}$ experiences a magnetic force. For a straight wire segment of length $\ell$ carrying current $I$, with the vector $\vec{\ell}$ pointing in the direction of the current, the force is
\[
\vec{F} = I\,\vec{\ell}\times\vec{B}.
\]
The magnitude of this force is
\[
F = I\,\ell\,B\,\sin\theta,
\]
where $\theta$ is the angle between the current direction (the direction of $\vec{\ell}$) and the magnetic field $\vec{B}$. The direction is given by the right-hand rule for cross products, reversed for negative current carriers.}

\nt{The vector $\vec{\ell}$ has magnitude equal to the length of the wire segment and points along the wire in the direction of conventional current. In a curved wire, the total force is obtained by integrating the differential-force expression over the entire path: $\vec{F} = \displaystyle\oint I\,d\vec{\ell}\times\vec{B}$.}

\thm{Magnetic force on a current-carrying conductor}{Let a wire carry steady current $I$ through a magnetic field $\vec{B}$.
\begin{itemize}
\item \textbf{Straight wire:} For a straight wire segment of length $\ell$, with $\vec{\ell}$ pointing along the wire in the current direction,
\[
\vec{F} = I\,\vec{\ell}\times\vec{B}.
\]
The magnitude is $F = I\,\ell\,B\,\sin\theta$, where $\theta$ is the angle between $\vec{\ell}$ and $\vec{B}$.
\item \textbf{Differential element:} For an arbitrary wire path, the force on a differential element $d\vec{\ell}$ is
\[
d\vec{F} = I\,d\vec{\ell}\times\vec{B},
\]
and the total force is
\[
\vec{F} = \int_{\text{wire}} I\,d\vec{\ell}\times\vec{B}.
\]
\item \textbf{Right-hand rule:} Point your right hand's fingers along $\vec{\ell}$ (current direction), then curl them toward $\vec{B}$. Your thumb gives the direction of $\vec{F}$.
\end{itemize}}

\pf{Force on a current-carrying wire from the force on moving charges}{The force on a single charge $q$ moving with drift velocity $\vec{v}_d$ is $\vec{F}_q = q\,\vec{v}_d\times\vec{B}$. In a wire segment of length $\ell$ and cross-sectional area $A$, the number of charge carriers is $N = n\,A\,\ell$, where $n$ is the carrier number density. The total force is
\[
\vec{F} = N\,q\,\vec{v}_d\times\vec{B} = n\,A\,\ell\,q\,\vec{v}_d\times\vec{B}.
\]
The current is $I = n\,q\,v_d\,A$, and the direction of $\vec{v}_d$ for positive carriers is the current direction. Letting $\vec{\ell}$ point in that direction with magnitude $\ell$, we have $n\,q\,\vec{v}_d = (I/A)\,\hat{\ell}$, and so
\[
\vec{F} = I\,\vec{\ell}\times\vec{B},
\]
where we used $\vec{\ell} = \ell\,\hat{\ell}$. This derivation confirms that the macroscopic force on a current-carrying wire follows directly from the Lorentz force on individual charge carriers.}

\cor{Straight wire parallel or perpendicular to field}{When the wire is parallel or antiparallel to $\vec{B}$ ($\theta = 0^\circ$ or $180^\circ$), then $\sin\theta = 0$ and $F = 0$. When the wire is perpendicular to $\vec{B}$ ($\theta = 90^\circ$), the force is maximal: $F = I\,\ell\,B$.}

\cor{Closed loop in a uniform field}{When a closed current loop sits entirely in a uniform magnetic field, the net force is zero:
\[
\vec{F}_{\text{net}} = I\oint d\vec{\ell}\times\vec{B} = I\left(\oint d\vec{\ell}\right)\times\vec{B} = \vec{0}.
\]
The integral $\oint d\vec{\ell}$ around any closed loop is the zero vector. Thus, no net force acts on a closed loop in a uniform field, though individual segments still feel forces.}

\dfn{Magnetic dipole moment of a current loop}{A planar loop carrying current $I$ with enclosed area $A$ has a \emph{magnetic dipole moment}
\[
\vec{\mu} = N\,I\,A\,\hat{n},
\]
where $N$ is the number of turns in the loop, $A$ is the area enclosed by one turn, and $\hat{n}$ is a unit vector perpendicular to the plane of the loop. The direction of $\hat{n}$ is given by the right-hand rule: curl the fingers of your right hand around the loop in the direction of the current, and your thumb points along $\hat{n}$.}

\mprop{Torque and potential energy of a current loop in a uniform field}{Let a planar current loop with magnetic dipole moment $\vec{\mu} = N I A\,\hat{n}$ be placed in a uniform magnetic field $\vec{B}$. Then:
\begin{enumerate}
\item The torque on the loop is
\[
\vec{\tau} = \vec{\mu}\times\vec{B}.
\]
The magnitude is
\[
\tau = \mu\,B\,\sin\phi = N\,I\,A\,B\,\sin\phi,
\]
where $\phi$ is the angle between $\hat{n}$ and $\vec{B}$.
\item The potential energy of the loop is
\[
U = -\vec{\mu}\cdot\vec{B} = -\mu\,B\,\cos\phi = -N\,I\,A\,B\,\cos\phi.
\]
\end{enumerate}
Equilibrium occurs when $\phi = 0^\circ$ (stable, $\hat{n}\parallel\vec{B}$, torque zero, energy minimum) or $\phi = 180^\circ$ (unstable, $\hat{n}$ antiparallel to $\vec{B}$, torque zero, energy maximum).}

\pf{Torque on a current loop in a uniform field}{Consider a rectangular loop of width $a$ (in the $x$-direction) and height $b$ (in the $y$-direction), carrying current $I$ in a uniform field $\vec{B}=B\,\hat{k}$ (out of the plane). Let the normal $\hat{n}$ to the loop make angle $\phi$ with $\hat{k}$. The loop rotates about an axis through its center, perpendicular to $\vec{B}$.

The forces on the top and bottom segments (length $a$) are equal and opposite and collinear, so they cancel. The forces on the two side segments (length $b$) are
\[
\vec{F}_1 = I\,\vec{b}_1\times\vec{B}
\quad\text{and}\quad
\vec{F}_2 = I\,\vec{b}_2\times\vec{B},
\]
with $\vec{b}_1 = -\vec{b}_2$. These forces have magnitude $F = I\,b\,B$ and act at perpendicular distances $(a/2)\,\sin\phi$ from the axis. Each produces a torque of magnitude
\[
\tau_{\text{side}} = F\cdot\frac{a}{2}\,\sin\phi = I\,b\,B\cdot\frac{a}{2}\,\sin\phi.
\]
Both sides contribute in the same rotational sense, so
\[
\tau = 2\cdot I\,b\,B\cdot\frac{a}{2}\,\sin\phi = I\,a\,b\,B\,\sin\phi.
\]
Since $A = a\,b$ and $\vec{\mu}$ has magnitude $\mu = I A$ (for $N=1$),
\[
\tau = \mu\,B\,\sin\phi.
\]
The vector form is $\vec{\tau} = \vec{\mu}\times\vec{B}$. For $N$ turns, multiply by $N$.

For the potential energy, the torque tends to align $\hat{n}$ with $\vec{B}$. The work done by an external agent rotating the loop from angle $\phi_0$ to $\phi$ equals the change in potential energy:
\[
\Delta U = -\int_{\phi_0}^{\phi} \tau_{\text{ext}}\,d\phi'
= -\int_{\phi_0}^{\phi} \mu B\,\sin\phi'\,d\phi'
= -\mu B\left(\cos\phi_0 - \cos\phi\right).
\]
Choosing the reference $U(\phi_0 = 90^\circ) = 0$, we find
\[
U = -\mu B\,\cos\phi = -\vec{\mu}\cdot\vec{B}.
\]

\nt{The torque tries to align $\vec{\mu}$ with $\vec{B}$. The stable equilibrium orientation has $\hat{n}\parallel\vec{B}$, i.e., the plane of the loop is perpendicular to the field. The unstable equilibrium has $\hat{n}$ antiparallel to $\vec{B}$. A DC motor exploits this by periodically reversing the current so the loop continues rotating.}}

\nt{The SI unit of magnetic dipole moment $\vec{\mu}$ is ampere-square meter, $\mathrm{A\cdot m^2}$. This is equivalent to joules per tesla, $\mathrm{J/T}$.}

\ex{Illustrative example}{A coil with $N=50$ turns, each of area $A = 8.0\,\mathrm{cm^2} = 8.0\times 10^{-4}\,\mathrm{m^2}$, carries current $I = 2.0\,\mathrm{A}$ in a field $B = 0.40\,\mathrm{T}$. The maximum torque occurs at $\phi = 90^\circ$:}
\[
\tau_{\text{max}} = N\,I\,A\,B = (50)(2.0\,\mathrm{A})(8.0\times 10^{-4}\,\mathrm{m^2})(0.40\,\mathrm{T}) = 3.2\times 10^{-2}\,\mathrm{N\cdot m}.
\]

\qs{Worked example}{A rectangular wire loop has width $a=0.10\,\mathrm{m}$ (horizontal side) and height $b=0.050\,\mathrm{m}$ (vertical side). The loop has $N=100$ turns and carries current $I=2.0\,\mathrm{A}$ in the direction shown: clockwise when viewed from the front (from the $+x$-axis toward the $yz$-plane). The loop sits in a uniform magnetic field $\vec{B} = (0.60\,\mathrm{T})\,\hat{\imath}$ pointing to the right. The normal vector $\hat{n}$ points along the $-\hat{k}$ direction (into the page) by the right-hand rule for clockwise current.

The loop lies in the $xy$-plane, centered at the origin, with its sides parallel to the $x$- and $y$-axes.

Find:
\begin{enumerate}[label=(\alph*)]
\item the net magnetic force on the loop,
\item the net magnetic torque on the loop (magnitude and direction), and
\item the magnetic potential energy of the loop in this orientation, and determine whether the loop will rotate clockwise or counterclockwise when released.
\end{enumerate}}

\textbf{Given quantities:}
\begin{itemize}
\item Width (horizontal): $a = 0.10\,\mathrm{m}$
\item Height (vertical): $b = 0.050\,\mathrm{m}$
\item Number of turns: $N = 100$
\item Current: $I = 2.0\,\mathrm{A}$ (clockwise in the $xy$-plane)
\item Magnetic field: $\vec{B} = (0.60\,\mathrm{T})\,\hat{\imath}$
\item Normal vector: $\hat{n} = -\hat{k}$
\end{itemize}

\sol \textbf{(a) Net force.} The magnetic field is uniform, and the loop is a closed current loop. From the corollary, the net force on a closed loop in a uniform magnetic field is zero. We can verify this segment by segment.

The loop has four segments:
\begin{itemize}
\item \textit{Bottom segment} (length $a$, current to the right): $\vec{\ell}_1 = a\,\hat{\imath}$, so $\vec{F}_1 = I\,(a\,\hat{\imath})\times(B\,\hat{\imath}) = \vec{0}$ (parallel to $\vec{B}$).
\item \textit{Top segment} (length $a$, current to the left): $\vec{\ell}_3 = -a\,\hat{\imath}$, so $\vec{F}_3 = I\,(-a\,\hat{\imath})\times(B\,\hat{\imath}) = \vec{0}$.
\item \textit{Right segment} (length $b$, current downward): $\vec{\ell}_2 = -b\,\hat{\jmath}$, so
\[
\vec{F}_2 = I\,(-b\,\hat{\jmath})\times(B\,\hat{\imath}) = I\,b\,B\,(-\hat{\jmath}\times\hat{\imath}) = I\,b\,B\,\hat{k}.
\]
\item \textit{Left segment} (length $b$, current upward): $\vec{\ell}_4 = b\,\hat{\jmath}$, so
\[
\vec{F}_4 = I\,(b\,\hat{\jmath})\times(B\,\hat{\imath}) = I\,b\,B\,(\hat{\jmath}\times\hat{\imath}) = -I\,b\,B\,\hat{k}.
\]
\end{itemize}

Summing:
\[
\vec{F}_{\text{net}} = \vec{F}_1+\vec{F}_2+\vec{F}_3+\vec{F}_4 = \vec{0} + I\,b\,B\,\hat{k} + \vec{0} - I\,b\,B\,\hat{k} = \vec{0}.
\]
For $N$ turns, each segment's force is multiplied by $N$, but they still cancel:
\[
\boxed{\vec{F}_{\text{net}} = \vec{0}}.
\]

\textbf{(b) Net torque.} The magnetic dipole moment has magnitude
\[
\mu = N\,I\,A = N\,I\,(a\,b).
\]
Substitute the values:
\[
\mu = (100)(2.0\,\mathrm{A})(0.10\,\mathrm{m})(0.050\,\mathrm{m}) = (100)(2.0)(0.0050)\,\mathrm{A\cdot m^2} = 1.0\,\mathrm{A\cdot m^2}.
\]
The dipole moment vector is $\vec{\mu} = \mu\,\hat{n} = -(1.0\,\mathrm{A\cdot m^2})\,\hat{k}$.

The torque is
\[
\vec{\tau} = \vec{\mu}\times\vec{B} = [-(1.0)\,\hat{k}]\times[(0.60)\,\hat{\imath}] = -(1.0)(0.60)\,(\hat{k}\times\hat{\imath}).
\]
Since $\hat{k}\times\hat{\imath}=\hat{\jmath}$,
\[
\vec{\tau} = -(0.60)\,\hat{\jmath}\,\mathrm{N\cdot m}.
\]
The magnitude is
\[
\tau = 0.60\,\mathrm{N\cdot m}.
\]
To interpret the direction, $-\hat{\jmath}$ points downward. Using the right-hand rule for torque, the loop tends to rotate such that $\vec{\mu}$ aligns with $\vec{B}$ — that is, $\hat{n}$ rotates from $-\hat{k}$ toward $+\hat{\imath}$. This corresponds to a rotation about the $y$-axis.

More physically: the right side of the loop (at $+x/2$) feels force $\vec{F}_2 = N\,I\,b\,B\,\hat{k} = 100(2.0)(0.050)(0.60)\,\hat{k} = 6.0\,\mathrm{N}$ upward, while the left side (at $-x/2$) feels $6.0\,\mathrm{N}$ downward. This pair of forces creates a torque that rotates the right side up and the left side down — a rotation about the $y$-axis.

\[
\boxed{\tau = 0.60\,\mathrm{N\cdot m},\quad \text{rotation about the }-\hat{\jmath}\text{ axis (right side up, left side down)}}.
\]

\textbf{(c) Potential energy and rotational tendency.} The potential energy is
\[
U = -\vec{\mu}\cdot\vec{B} = -[-(1.0)\,\hat{k}]\cdot[(0.60)\,\hat{\imath}].
\]
Since $\hat{k}\perp\hat{\imath}$, their dot product is $0$, so
\[
\boxed{U = 0}.
\]

This corresponds to $\phi = 90^\circ$ between $\vec{\mu}$ (pointing in $-\hat{k}$) and $\vec{B}$ (pointing in $+\hat{\imath}$), since $\cos 90^\circ = 0$.

The loop will rotate toward the stable equilibrium orientation where $\vec{\mu}\parallel\vec{B}$. Currently $\vec{\mu}$ points into the page ($-\hat{k}$) while $\vec{B}$ points right ($+\hat{\imath}$). The stable orientation has $\hat{n}$ aligned with $+\hat{\imath}$, meaning the loop's plane is perpendicular to the field (normal pointing along $\vec{B}$). The torque computed in part~(b) drives this rotation: the right side is pushed upward and the left side downward, rotating the loop about the $y$-axis. Viewed from the $+y$ direction (looking down from above), this rotation appears counterclockwise.

\[
\boxed{\text{Rotates about the }y\text{-axis toward stable equilibrium (right side up, left side down).}}
\]

\bigskip
\textbf{Summary of results:}
\begin{enumerate}[label=(\alph*)]
\item $\vec{F}_{\text{net}} = \vec{0}$
\item $\tau = 0.60\,\mathrm{N\cdot m}$, rotation about the $-\hat{\jmath}$ axis
\item $U = 0$, loop rotates toward stable equilibrium where $\hat{n}\parallel\vec{B}$
\end{enumerate}