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\subsection{Magnetic Force on a Moving Charge}
A charge that moves through a magnetic field experiences a force perpendicular to both its velocity and the field. This is the magnetic part of the Lorentz force. Unlike the electric force, the magnetic force acts only on moving charges and is always perpendicular to the direction of motion.
\dfn{Magnetic force on a point charge}{A particle of charge $q$ moving with velocity $\vec{v}$ through a magnetic field $\vec{B}$ experiences a magnetic force
\[
\vec{F}_B = q\,\vec{v}\times\vec{B}.
\]
The magnitude of this force is
\[
F_B = |q|\,v\,B\,\sin\theta,
\]
where $\theta$ is the angle between the vectors $\vec{v}$ and $\vec{B}$ measured in the plane they span. The direction of $\vec{F}_B$ is perpendicular to that plane and is determined by the right-hand rule, reversed for negative charge.}
\nt{The magnetic force vanishes when the charge is at rest ($\vec{v}=\vec{0}$), when $\vec{v}$ is parallel or antiparallel to $\vec{B}$ ($\theta=0^\circ$ or $180^\circ$), or when $B=0$. The force is maximal when $\vec{v}\perp\vec{B}$ ($\theta=90^\circ$).}
\thm{Lorentz magnetic force law}{Let $q$ be the charge of a particle, $\vec{v}$ its velocity vector, and $\vec{B}$ the magnetic field at the particle's position. Then the magnetic force on the particle is
\[
\vec{F}_B = q\,\vec{v}\times\vec{B}.
\]
\begin{itemize}
\item \textbf{Magnitude:} $F_B = |q|\,v\,B\,\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.
\item \textbf{Direction:} Point the fingers of your right hand along $\vec{v}$, then curl them toward $\vec{B}$. Your thumb points in the direction of $\vec{F}_B$ if $q>0$. If $q<0$, the force is opposite to your thumb.
\item \textbf{SI unit of $B$:} The tesla, $\mathrm{T} = \dfrac{\mathrm{N}}{\mathrm{C}\cdot\mathrm{m/s}} = \dfrac{\mathrm{N}}{\mathrm{A}\cdot\mathrm{m}} = \dfrac{\mathrm{kg}}{\mathrm{C}\cdot\mathrm{s}}$.
\end{itemize}}
\pf{Lorentz magnetic force law from cross-product geometry}{The vector cross product $\vec{v}\times\vec{B}$ is defined to have magnitude $vB\sin\theta$ and direction given by the right-hand rule. Multiplying by $q$ scales the magnitude by $|q|$ and reverses direction if $q<0$. Thus
\[
\vec{F}_B = q\,(\vec{v}\times\vec{B})
\]
has magnitude $|q|vB\sin\theta$ and the correct directional behaviour. This is the experimentally determined magnetic force law for a point charge.}
\cor{Charge at rest or parallel to field}{When $\vec{v}=\vec{0}$ or when $\vec{v}\parallel\vec{B}$, we have $\sin\theta=0$ and therefore $F_B=0$. The magnetic field exerts no force on a stationary charge or on a charge moving exactly along the field lines.}
\mprop{Magnetic force vs.\ electric force}{For the same charge $q$ placed in both an electric field $\vec{E}$ and a magnetic field $\vec{B}$, the total Lorentz force is
\[
\vec{F} = q\,\vec{E} + q\,\vec{v}\times\vec{B}.
\]
Key differences:
\begin{itemize}
\item $\vec{F}_E$ points parallel (or antiparallel) to $\vec{E}$, regardless of motion.
\item $\vec{F}_B$ is always perpendicular to $\vec{v}$, so it does zero work on the charge.
\item $\vec{F}_B$ vanishes when $\vec{v}=\vec{0}$; $\vec{F}_E$ does not.
\end{itemize}}
\thm{Magnetic force does no work}{Since $\vec{F}_B\perp\vec{v}$ at every instant, the instantaneous power delivered by the magnetic force is
\[
P = \vec{F}_B\cdot\vec{v} = q\,(\vec{v}\times\vec{B})\cdot\vec{v} = 0.
\]
The magnetic force can change the direction of a particle's velocity but never its kinetic energy. This is the mathematical expression of the scalar triple-product identity $(\vec{a}\times\vec{b})\cdot\vec{a}=0$.}
\ex{Illustrative example}{When a charged particle enters a uniform magnetic field perpendicularly, it follows a circular path. The magnetic force provides the centripetal force:
\[
|q|\,v\,B = \frac{m\,v^2}{R} \quad\Rightarrow\quad R = \frac{m\,v}{|q|\,B},
\]
where $R$ is the radius of the circular path. The period of revolution is
\[
T = \frac{2\pi R}{v} = \frac{2\pi m}{|q|\,B},
\]
which is independent of the particle's speed. This circular-motion analysis is developed fully in section~E12.2.}
\qs{Worked example}{A proton (charge $q=+1.60\times 10^{-19}\,\mathrm{C}$, mass $m=1.67\times 10^{-27}\,\mathrm{kg}$) moves with speed $v=3.0\times 10^6\,\mathrm{m/s}$ in the $+\hat{\imath}$ direction. It enters a region with a uniform magnetic field $\vec{B}=(0.50\,\mathrm{T})\,\hat{\jmath}$. The proton's velocity is perpendicular to the field.
Find:
\begin{enumerate}[label=(\alph*)]
\item the magnitude of the magnetic force on the proton,
\item the direction of the magnetic force as a unit vector, and
\item the initial acceleration vector of the proton.
\end{enumerate}}
\sol \textbf{(a) Force magnitude.} The magnetic force is $\vec{F}_B=q\,\vec{v}\times\vec{B}$. With $\vec{v}=v\,\hat{\imath}$ and $\vec{B}=B\,\hat{\jmath}$, the angle between them is $\theta=90^\circ$ and $\sin\theta=1$. The magnitude is
\[
F_B = |q|\,v\,B\,\sin 90^\circ = |q|\,v\,B.
\]
Substitute the given values:
\[
F_B = (1.60\times 10^{-19}\,\mathrm{C})(3.0\times 10^6\,\mathrm{m/s})(0.50\,\mathrm{T}).
\]
Compute step by step:
\[
(1.60\times 10^{-19})(3.0\times 10^6) = 4.8\times 10^{-13},
\]
\[
(4.8\times 10^{-13})(0.50) = 2.4\times 10^{-13}\,\mathrm{N}.
\]
Thus
\[
F_B = 2.4\times 10^{-13}\,\mathrm{N}.
\]
\textbf{(b) Force direction.} Use the cross product directly:
\[
\vec{F}_B = q\,(\vec{v}\times\vec{B}) = q\,(v\,\hat{\imath})\times(B\,\hat{\jmath}) = q\,v\,B\,(\hat{\imath}\times\hat{\jmath}).
\]
Since $\hat{\imath}\times\hat{\jmath}=\hat{k}$,
\[
\vec{F}_B = q\,v\,B\,\hat{k}.
\]
Because $q>0$ and $v,B>0$, the force points in the $+\hat{k}$ direction. In unit-vector form:
\[
\vec{F}_B = (2.4\times 10^{-13}\,\mathrm{N})\,\hat{k}.
\]
The right-hand rule confirms this: fingers along $+\hat{\imath}$, curl toward $+\hat{\jmath}$, thumb points along $+\hat{k}$.
\textbf{(c) Acceleration vector.} By Newton's second law,
\[
\vec{a} = \frac{\vec{F}_B}{m}.
\]
Substitute:
\[
\vec{a} = \frac{(2.4\times 10^{-13}\,\mathrm{N})\,\hat{k}}{1.67\times 10^{-27}\,\mathrm{kg}}.
\]
Compute the magnitude:
\[
\frac{2.4\times 10^{-13}}{1.67\times 10^{-27}} = \frac{2.4}{1.67}\times 10^{14} \approx 1.44\times 10^{14}\,\mathrm{m/s^2}.
\]
Thus
\[
\vec{a} = (1.44\times 10^{14}\,\mathrm{m/s^2})\,\hat{k}.
\]
\bigskip
\textbf{Final answers:}
\begin{enumerate}[label=(\alph*)]
\item $F_B = 2.4\times 10^{-13}\,\mathrm{N}$
\item $\vec{F}_B = (2.4\times 10^{-13}\,\mathrm{N})\,\hat{k}$
\item $\vec{a} = (1.44\times 10^{14}\,\mathrm{m/s^2})\,\hat{k}$
\end{enumerate}

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\subsection{Circular and Helical Motion in a Uniform Magnetic Field}
When a charged particle moves through a uniform magnetic field, the magnetic force acts as a centripetal force, bending the particle's path. The resulting motion depends on the angle between the velocity $\vec{v}$ and the field $\vec{B}$: perpendicular entry produces circular motion, while oblique entry produces helical motion.
\dfn{Circular motion of a charge in a uniform B-field}{Let a particle of mass $m$ and charge $q$ move with speed $v$ through a uniform magnetic field $\vec{B}$, with $\vec{v}\perp\vec{B}$. The magnetic force has constant magnitude $|q|vB$ and is always perpendicular to $\vec{v}$, so it provides the centripetal force for uniform circular motion:
\[
|q|\,v\,B=\frac{m\,v^{2}}{R}.
\]
Solving for the radius,
\[
R=\frac{m\,v}{|q|\,B}=\frac{m\,v_{\perp}}{|q|\,B},
\]
where $v_{\perp}=v$ is the perpendicular speed. The period of revolution is
\[
T=\frac{2\pi R}{v}=\frac{2\pi m}{|q|\,B}.
\]
The frequency of revolution (cyclotron frequency) is
\[
f=\frac{1}{T}=\frac{|q|\,B}{2\pi m},
\]
and the angular frequency is
\[
\omega=2\pi f=\frac{|q|\,B}{m}.
\]
Note that $T$, $f$, and $\omega$ are independent of the particle's speed.}
\nt{Because the magnetic force does no work (it is always perpendicular to $\vec{v}$), the particle's speed and kinetic energy remain constant throughout the circular motion. The magnetic field only changes the direction of the velocity, not its magnitude. This is why the radius depends on $v$ but the period does not.}
\thm{Cyclotron motion}{Let $m$ be the mass and $q$ the charge of a particle entering a uniform magnetic field $\vec{B}$ with velocity component $v_{\perp}$ perpendicular to $\vec{B}$. The particle undergoes uniform circular motion in the plane perpendicular to $\vec{B}$ with:
\begin{itemize}
\item \textbf{Radius:} $R=\dfrac{m\,v_{\perp}}{|q|\,B}$
\item \textbf{Period:} $T=\dfrac{2\pi m}{|q|\,B}$
\item \textbf{Cyclotron frequency:} $f=\dfrac{|q|\,B}{2\pi m}$
\item \textbf{Angular frequency:} $\omega=\dfrac{|q|\,B}{m}$
\end{itemize}
The sense of rotation is counter-clockwise for $q>0$ and clockwise for $q<0$ when viewing along the direction of $\vec{B}$.}
\pf{Cyclotron motion from Newton's second law}{The magnetic force on the particle is $\vec{F}_B=q\,\vec{v}\times\vec{B}$. When $\vec{v}\perp\vec{B}$, the force magnitude is $F_B=|q|vB$ and its direction is always perpendicular to $\vec{v}$ (toward the center of curvature). For uniform circular motion, Newton's second law requires
\[
F_{\text{net}}=\frac{m\,v^{2}}{R}.
\]
Equating the magnetic force to the required centripetal force:
\[
|q|\,v\,B=\frac{m\,v^{2}}{R}.
\]
Solving for $R$:
\[
R=\frac{m\,v^{2}}{|q|\,v\,B}=\frac{m\,v}{|q|\,B}.
\]
The period is the circumference divided by the speed:
\[
T=\frac{2\pi R}{v}=\frac{2\pi}{v}\cdot\frac{m\,v}{|q|\,B}=\frac{2\pi m}{|q|\,B}.
\]
Thus $T$ is independent of $v$ and $R$. The angular frequency is $\omega=2\pi/T=|q|B/m$, known as the cyclotron angular frequency.}
\dfn{Helical motion of a charge in a uniform B-field}{Let a particle of mass $m$ and charge $q$ move with speed $v$ through a uniform magnetic field $\vec{B}$, with the velocity making an angle $\alpha$ with $\vec{B}$ (where $0^\circ<\alpha<90^\circ$). Decompose the velocity into components parallel and perpendicular to $\vec{B}$:
\[
v_{\parallel}=v\cos\alpha,\qquad v_{\perp}=v\sin\alpha.
\]
The perpendicular component produces circular motion with radius
\[
R=\frac{m\,v_{\perp}}{|q|\,B}=\frac{m\,v\,\sin\alpha}{|q|\,B}.
\]
The parallel component is unaffected by the magnetic force and produces uniform linear motion along $\vec{B}$. The combination is helical motion.
The \emph{pitch} $p$ of the helix is the distance advanced along $\vec{B}$ during one full revolution:
\[
p=v_{\parallel}\,T=(v\cos\alpha)\cdot\frac{2\pi m}{|q|\,B}=\frac{2\pi m\,v\,\cos\alpha}{|q|\,B}.
\]
The sense of the circular rotation follows the same rule as cyclotron motion: counter-clockwise for $q>0$ and clockwise for $q<0$, when viewing along $\vec{B}$.}
\nt{If $\alpha=0^\circ$, then $v_{\perp}=0$ and the particle travels in a straight line along $\vec{B}$ (no magnetic force). If $\alpha=90^\circ$, then $v_{\parallel}=0$ and the particle undergoes pure circular motion (no drift along $\vec{B}$). Helical motion interpolates between these two extremes. The pitch increases as $\alpha\to 0^\circ$ and approaches zero as $\alpha\to 90^\circ$.}
\mprop{Cyclotron and helical motion parameters}{For a particle of mass $m$ and charge $q$ in a uniform magnetic field $\vec{B}$, with velocity $\vec{v}$ at angle $\alpha$ to $\vec{B}$:
\begin{align}
R&=\frac{m\,v\,\sin\alpha}{|q|\,B} && \text{(helix radius)} \\
T&=\frac{2\pi m}{|q|\,B} && \text{(period of revolution, independent of }v\text{ and }\alpha)\\
f&=\frac{|q|\,B}{2\pi m} && \text{(cyclotron frequency)} \\
p&=\frac{2\pi m\,v\,\cos\alpha}{|q|\,B} && \text{(helix pitch)}
\end{align}}
\mprop{Magnetic force does no work}{The magnetic force $\vec{F}_B=q\,\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$, so
\[
P=\vec{F}_B\cdot\vec{v}=0\qquad\text{and}\qquad\Delta K=0.
\]
The kinetic energy and speed of the particle remain constant. This holds for both pure circular motion and helical motion.}
\ex{Illustrative example}{An electron and a proton, each with the same speed $v$, enter perpendicular to the same uniform magnetic field. The proton's radius is larger by the mass ratio $m_p/m_e\approx 1836$, but both complete one revolution in the same time $T=2\pi m/|q|B$, because the proton's period is 1836 times longer due to its mass but it travels a proportionally longer path (radius 1836 times larger), so the time cancels out.}
\qs{Worked example}{An electron (mass $m_e=9.11\times 10^{-31}\,\mathrm{kg}$, charge $q=-1.60\times 10^{-19}\,\mathrm{C}$) enters a region with a uniform magnetic field $\vec{B}=(0.040\,\mathrm{T})\,\hat{\jmath}$. At the moment it enters, its velocity is
\[
\vec{v}=(4.0\times 10^{5}\,\mathrm{m/s})\,\hat{\imath}+(2.0\times 10^{5}\,\mathrm{m/s})\,\hat{k}.
\]
Find:
\begin{enumerate}[label=(\alph*)]
\item the radius of the helical path,
\item the period of revolution,
\item the pitch of the helix, and
\item the sense of rotation (clockwise or counter-clockwise when viewing along $+\vec{B}$).
\end{enumerate}}
\sol \textbf{(a) Radius of the helix.} Decompose the velocity into components parallel and perpendicular to $\vec{B}$ (which points along $\hat{\jmath}$):
\[
v_{\parallel}=v_{k}=2.0\times 10^{5}\,\mathrm{m/s},\qquad v_{\perp}=\sqrt{v_{i}^{2}+v_{j}^{2}}=\sqrt{(4.0\times 10^{5})^{2}+0^{2}}=4.0\times 10^{5}\,\mathrm{m/s}.
\]
Note that $v_j=0$, so the entire $x$-component is perpendicular to $\vec{B}$.
The radius of the helical path is
\[
R=\frac{m\,v_{\perp}}{|q|\,B}.
\]
Substitute the values:
\[
R=\frac{(9.11\times 10^{-31}\,\mathrm{kg})(4.0\times 10^{5}\,\mathrm{m/s})}{(1.60\times 10^{-19}\,\mathrm{C})(0.040\,\mathrm{T})}.
\]
Compute the numerator:
\[
(9.11\times 10^{-31})(4.0\times 10^{5})=3.644\times 10^{-25}\,\mathrm{kg{\cdot}m/s}.
\]
Compute the denominator:
\[
(1.60\times 10^{-19})(0.040)=6.4\times 10^{-21}\,\mathrm{C{\cdot}T}.
\]
Thus
\[
R=\frac{3.644\times 10^{-25}}{6.4\times 10^{-21}}\,\mathrm{m}=5.69\times 10^{-5}\,\mathrm{m}.
\]
So
\[
R=5.69\times 10^{-5}\,\mathrm{m}=56.9\,\mathrm{\mu m}.
\]
\textbf{(b) Period of revolution.} The period depends only on the particle's properties and the field strength:
\[
T=\frac{2\pi m}{|q|\,B}.
\]
Substitute:
\[
T=\frac{2\pi(9.11\times 10^{-31}\,\mathrm{kg})}{(1.60\times 10^{-19}\,\mathrm{C})(0.040\,\mathrm{T})}.
\]
The denominator is $6.4\times 10^{-21}\,\mathrm{C{\cdot}T}$ as computed above. The numerator is
\[
2\pi(9.11\times 10^{-31})=5.724\times 10^{-30}\,\mathrm{kg}.
\]
Thus
\[
T=\frac{5.724\times 10^{-30}}{6.4\times 10^{-21}}\,\mathrm{s}=8.94\times 10^{-10}\,\mathrm{s}.
\]
So
\[
T=8.94\times 10^{-10}\,\mathrm{s}=0.894\,\mathrm{ns}.
\]
\textbf{(c) Pitch of the helix.} The pitch is the distance advanced along $\vec{B}$ in one period:
\[
p=v_{\parallel}\,T.
\]
Substitute:
\[
p=(2.0\times 10^{5}\,\mathrm{m/s})(8.94\times 10^{-10}\,\mathrm{s}).
\]
Compute:
\[
p=1.788\times 10^{-4}\,\mathrm{m}.
\]
So
\[
p=1.79\times 10^{-4}\,\mathrm{m}=179\,\mathrm{\mu m}.
\]
\textbf{(d) Sense of rotation.} The magnetic field points in the $+\hat{\jmath}$ direction. To determine the sense of rotation, note the initial force on the electron at the entry point. The velocity has a $+x$ component, so at $t=0$ the perpendicular velocity is $\vec{v}_{\perp}=(4.0\times 10^{5}\,\mathrm{m/s})\,\hat{\imath}$. The magnetic force is
\[
\vec{F}=q\,(\vec{v}\times\vec{B})=(-1.60\times 10^{-19})\,[(4.0\times 10^{5}\,\hat{\imath})\times(0.040\,\hat{\jmath})].
\]
Since $\hat{\imath}\times\hat{\jmath}=\hat{k}$,
\[
\vec{F}=(-1.60\times 10^{-19})(4.0\times 10^{5})(0.040)\,\hat{k}=-(2.56\times 10^{-15}\,\mathrm{N})\,\hat{k}.
\]
The initial force points in the $-z$ direction, meaning the electron curves toward $-z$ from its initial $+x$ direction. Viewing along $+\hat{\jmath}$ (the direction of $\vec{B}$), the $+x$ axis is to the right and the $+z$ axis points toward you. Starting at $+x$ and curving toward $-z$, the electron moves clockwise. This is consistent with the general rule: for negative charge, the rotation is clockwise when viewing along $\vec{B}$.
\bigskip
\textbf{Final answers:}
\begin{enumerate}[label=(\alph*)]
\item $R=5.69\times 10^{-5}\,\mathrm{m}=56.9\,\mathrm{\mu m}$
\item $T=8.94\times 10^{-10}\,\mathrm{s}=0.894\,\mathrm{ns}$
\item $p=1.79\times 10^{-4}\,\mathrm{m}=179\,\mathrm{\mu m}$
\item Clockwise (viewing along $+\hat{\jmath}$)
\end{enumerate}

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\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}

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\subsection{The Biot--Savart Law}
Currents produce magnetic fields. The Biot--Savart law gives the magnetic field at a point in space due to a steady current distribution. It is the magnetic analogue of Coulomb's law in electrostatics: just as Coulomb's law tells you the electric field of a charge distribution by integrating over point charges, the Biot--Savart law tells you the magnetic field of a current distribution by integrating over current elements.
\dfn{Biot--Savart law (differential form)}{Let a steady current $I$ flow through a thin wire. Consider a differential element of the wire of length $d\ell$ carrying current $I$, represented by the vector $d\vec{\ell}$ pointing in the direction of the current. Let $P$ be an observation point, and let $\vec{r}$ be the displacement vector from the current element to $P$, with magnitude $r = |\vec{r}|$ and unit vector $\hat{r} = \vec{r}/r$. The differential magnetic field at $P$ due to this current element is
\[
d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{\ell}\times\hat{r}}{r^2}
= \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{\ell}\times\vec{r}}{r^3}.
\]
Here $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\cdot m/A}$ is the permeability of free space (the magnetic constant).}
\nt{The Biot--Savart law is fundamentally a \emph{superposition principle}: the total field is the vector sum (integral) of all the differential contributions from every current element. Because each contribution involves a cross product, $d\vec{B}$ is always perpendicular to both the current element $d\vec{\ell}$ and the displacement $\vec{r}$. The $1/r^2$ dependence mirrors Coulomb's law, making the Biot--Savart law a Green's-function solution to the static Maxwell equations for $\vec{B}$.}
\thm{Biot--Savart law}{For a steady current $I$ flowing along a wire path $C$, the total magnetic field at a point $P$ is
\[
\vec{B} = \int_C \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{\ell}\times\hat{r}}{r^2}.
\]
\begin{itemize}
\item \textbf{Constants:} $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\cdot m/A}$ (permeability of free space). In SI units, $d\vec{B}$ has units of tesla ($\mathrm{T}$).
\item \textbf{Geometry:} $d\vec{\ell}$ points along the wire in the direction of conventional current. The vector $\vec{r}$ points \emph{from} the current element \emph{to} the observation point. The unit vector $\hat{r} = \vec{r}/r$ is the normalized version of this displacement.
\item \textbf{Direction:} The direction of $d\vec{B}$ is given by the right-hand rule for the cross product $d\vec{\ell}\times\hat{r}$. $d\vec{B}$ is perpendicular to the plane containing $d\vec{\ell}$ and $\vec{r}$.
\item \textbf{Magnitude:} The magnitude of the differential field is
\[
dB = \frac{\mu_0}{4\pi}\,\frac{I\,d\ell\,\sin\theta}{r^2},
\]
where $\theta$ is the angle between $d\vec{\ell}$ and $\vec{r}$.
\item \textbf{Symmetry considerations:} In many symmetric geometries (long straight wires, circular loops, solenoids), symmetry allows you to argue that certain components of $\vec{B}$ cancel upon integration, dramatically simplifying the calculation.
\end{itemize}}
\pf{Biot--Savart law from the macroscopic field of a long wire and superposition}{The field of a long straight wire carrying current $I$ is known experimentally (from Amp\`ere's law or direct measurement) to be
\[
B = \frac{\mu_0 I}{2\pi s},
\]
where $s$ is the perpendicular distance from the wire. The Biot--Savart law is the differential statement that, when integrated for an infinite straight wire, reproduces this result.
Consider an infinite straight wire along the $z$-axis carrying current $I$ in the $+\hat{k}$ direction. The observation point is in the $xy$-plane at distance $s$ from the wire. A current element at position $z$ has
\[
d\vec{\ell} = dz\,\hat{k}, \qquad
\vec{r} = s\,\hat{s} - z\,\hat{k}, \qquad
r = \sqrt{s^2+z^2}.
\]
The cross product is
\[
d\vec{\ell}\times\vec{r} = dz\,\hat{k}\times(s\,\hat{s}-z\,\hat{k}) = s\,dz\,(\hat{k}\times\hat{s}) = s\,dz\,\hat{\phi},
\]
which points in the azimuthal direction. The magnitude of the field contribution is
\[
dB = \frac{\mu_0}{4\pi}\,\frac{I\,s\,dz}{(s^2+z^2)^{3/2}}.
\]
Integrating from $z=-\infty$ to $z=+\infty$ using $z = s\tan\theta$, $dz = s\,\sec^2\theta\,d\theta$:
\[
B = \frac{\mu_0 I}{4\pi}\int_{-\infty}^{\infty}\frac{s\,dz}{(s^2+z^2)^{3/2}}
= \frac{\mu_0 I}{4\pi}\int_{-\pi/2}^{\pi/2}\frac{s^2\,\sec^2\theta}{s^3\,\sec^3\theta}\,d\theta
= \frac{\mu_0 I}{4\pi s}\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta
= \frac{\mu_0 I}{4\pi s}\Bigl[\sin\theta\Bigr]_{-\pi/2}^{\pi/2}
= \frac{\mu_0 I}{2\pi s}.
\]
This matches the known result for the field of an infinite wire. By linearity, the Biot--Savart law applied to any current distribution gives the correct total field via superposition.
\qed}
\mprop{Field of a long straight wire}{An infinitely long straight wire carrying steady current $I$ produces a magnetic field at perpendicular distance $s$ given by
\[
B = \frac{\mu_0 I}{2\pi s}.
\]
The field lines are concentric circles around the wire, with direction given by the right-hand rule: point your right thumb along the current, and your fingers curl in the direction of $\vec{B}$. In cylindrical coordinates $(s,\phi,z)$ with the wire along the $z$-axis,
\[
\vec{B} = \frac{\mu_0 I}{2\pi s}\,\hat{\phi}.\]
}
\mprop{Field at the center of a circular current loop}{A circular loop of radius $R$ carrying steady current $I$ produces a magnetic field at its center given by
\[
B = \frac{\mu_0 I}{2R}.
\]
The direction is perpendicular to the plane of the loop, given by the right-hand rule: curl your fingers in the direction of the current, and your thumb points along $\vec{B}$. For $N$ tightly wound turns, multiply by $N$:
\[
B = \frac{\mu_0 N I}{2R}.\]
}
\mprop{Field on the axis of a circular current loop}{A circular loop of radius $R$ carrying steady current $I$ produces a magnetic field on its symmetry axis at distance $z$ from the center given by
\[
B(z) = \frac{\mu_0 I\,R^2}{2\,(R^2+z^2)^{3/2}}.
\]
The field points along the axis in the direction given by the right-hand rule. At the center ($z=0$), this reduces to $B = \mu_0 I/(2R)$. Far from the loop ($z\gg R$), the field falls off as
\[
B \approx \frac{\mu_0 I\,R^2}{2\,z^3} = \frac{\mu_0}{4\pi}\,\frac{2\pi R^2 I}{z^3} = \frac{\mu_0}{4\pi}\,\frac{2\mu}{z^3},
\]
which is the dipole-field form, where $\mu = \pi R^2 I$ is the magnetic dipole moment of the loop.
}
\cor{Far-field (dipole) limit of a current loop}{Far from any compact current loop, the magnetic field has the universal dipole form
\[
\vec{B}_{\text{dipole}}(\vec{r}) = \frac{\mu_0}{4\pi}\left[\frac{3(\vec{\mu}\cdot\hat{r})\hat{r}-\vec{\mu}}{r^3}\right],
\]
where $\vec{\mu} = I\,A\,\hat{n}$ is the magnetic dipole moment ($A$ is the loop area, $\hat{n}$ its normal). This is the analogue of the electric dipole field in electrostatics.}
\cor{Straight wire of finite length}{For a finite straight wire segment of length $L$ carrying current $I$, the field at a point perpendicular to the midpoint of the wire at distance $s$ is
\[
B = \frac{\mu_0 I}{2\pi s}\,\frac{L/2}{\sqrt{s^2+(L/2)^2}}.
\]
In the limit $L\to\infty$, the fraction approaches $1$, recovering the infinite-wire result $B=\mu_0 I/(2\pi s)$.}
\ex{Illustrative example}{A square loop of side $a$ carries current $I$. Each of the four sides contributes equally. From the finite-wire formula with $L=a$ and the perpendicular distance from the midpoint to the center being $s=a/2$:
\[
B_{\text{one side}} = \frac{\mu_0 I}{2\pi(a/2)}\cdot\frac{a/2}{\sqrt{(a/2)^2+(a/2)^2}}
= \frac{\mu_0 I}{\pi a}\cdot\frac{a/2}{a/\sqrt{2}}
= \frac{\mu_0 I}{\pi a}\cdot\frac{\sqrt{2}}{2}.
\]
Four sides contribute in the same direction (perpendicular to the square's plane), so
\[
B_{\text{center}} = 4\cdot\frac{\mu_0 I\sqrt{2}}{2\pi a}
= \frac{2\sqrt{2}\,\mu_0 I}{\pi a}.\]
\qed}
\nt{Key symmetry principles for Biot--Savart calculations:}
\begin{itemize}
\item \textbf{Straight wire segments aimed directly at (or away from) the observation point} contribute \emph{zero} field: when $d\vec{\ell}\parallel\vec{r}$, the cross product $d\vec{\ell}\times\hat{r}=\vec{0}$. This is a very useful shortcut.
\item \textbf{Circular arcs} centered on the observation point contribute field proportional to the arc angle. For an arc of angle $\theta$ (in radians) and radius $R$, $B = (\mu_0 I/4\pi R)\cdot\theta$.
\item \textbf{Perpendicular geometry} maximises the contribution: when $d\vec{\ell}\perp\vec{r}$ at every point (as on a circular arc centered at $P$), the magnitude is $dB = (\mu_0/4\pi)\,I\,d\ell/R^2$ with no angle factor.
\end{itemize}
\qs{Worked example}{A wire bent into the shape shown carries a steady current $I$ in the direction indicated. The wire consists of three segments:
\begin{enumerate}[label=(\roman*)]
\item A straight horizontal segment running from $x=-\infty$ to $x=-R$ along the line $y=0$, approaching the origin.
\item A circular arc of radius $R$ centered at the origin, extending from the point $(R,0)$ counterclockwise through the upper half-plane to the point $(-R,0)$ (a semicircle).
\item A straight horizontal segment running from $x=-R$ along the line $y=0$ to $x=+\infty$, extending to the right.
\end{enumerate}
Find the magnetic field $\vec{B}$ at the origin $O$ (the center of the circular arc).
Assume the wire lies entirely in the $xy$-plane and the current $I$ flows to the right on the incoming straight segment, then counterclockwise along the arc, then to the right on the outgoing straight segment. Use $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\cdot m/A}$.}
\textbf{Given quantities:}
\begin{itemize}
\item Current: $I$
\item Arc radius: $R$
\item Arc: semicircle in the upper half-plane ($y\geq 0$), counterclockwise
\item Observation point: origin $O$
\end{itemize}
\sol \textbf{Strategy.} By the superposition principle, $\vec{B}_{\text{total}} = \vec{B}_{\text{straight, incoming}} + \vec{B}_{\text{arc}} + \vec{B}_{\text{straight, outgoing}}$. We evaluate each contribution separately.
\medskip
\noindent\textbf{(i) Incoming straight wire (segment from $x=-\infty$ to $x=-R$, along $y=0$).}
The current flows along the $x$-axis (the line $y=0$) toward the origin. The observation point (the origin) lies \emph{on the line} of the wire. For every current element $d\vec{\ell}$ on this segment, the displacement vector $\vec{r}$ from the element to the origin points along the $+x$ direction, which is \emph{parallel} to $d\vec{\ell}$ (current flows in the $+x$ direction). Therefore
\[
d\vec{\ell}\times\hat{r} = 0
\]
for all elements of this segment, and
\[
\vec{B}_{\text{incoming}} = \vec{0}.
\]
The same reasoning applies to the outgoing wire.
\medskip
\noindent\textbf{(ii) Outgoing straight wire (segment from $x=-R$ to $x=+\infty$, along $y=0$).}
Again, the observation point lies on the line of the wire. The displacement $\vec{r}$ from every current element to the origin is collinear with $d\vec{\ell}$, so $d\vec{\ell}\times\hat{r}=\vec{0}$. Thus
\[
\vec{B}_{\text{outgoing}} = \vec{0}.
\]
\medskip
\noindent\textbf{(iii) Semicircular arc (radius $R$, upper half-plane, counterclockwise).}
For the circular arc, every current element is at distance $R$ from the origin, and every $d\vec{\ell}$ is tangent to the circle. The displacement vector from each element to the origin points radially inward (toward the center). The angle between $d\vec{\ell}$ (tangential) and $\vec{r}$ (radial) is $90^\circ$, so $\sin 90^\circ = 1$ everywhere.
The magnitude of each differential contribution is
\[
dB = \frac{\mu_0}{4\pi}\,\frac{I\,d\ell}{R^2}.
\]
The direction: by the right-hand rule, $d\vec{\ell}\times\hat{r}$ for counterclockwise current on the upper semicircle points in the $+\hat{k}$ direction (out of the page) everywhere along the arc.
The total magnitude is
\[
B_{\text{arc}} = \int_{\text{arc}} \frac{\mu_0 I}{4\pi R^2}\,d\ell
= \frac{\mu_0 I}{4\pi R^2}\int_{\text{arc}} d\ell
= \frac{\mu_0 I}{4\pi R^2}\cdot(\pi R)
= \frac{\mu_0 I}{4 R}.
\]
Here we used that the arc length is $\pi R$ (a semicircle).
In vector form, with $\hat{k}$ pointing out of the $xy$-plane:
\[
\vec{B}_{\text{arc}} = \frac{\mu_0 I}{4 R}\,\hat{k}.
\]
\medskip
\noindent\textbf{(iv) Total field.}
Summing the three contributions:
\[
\vec{B}_{\text{total}} = \vec{0} + \frac{\mu_0 I}{4 R}\,\hat{k} + \vec{0}
= \frac{\mu_0 I}{4 R}\,\hat{k}.
\]
\bigskip
\textbf{Final answer:}
\[
\boxed{\vec{B} = \frac{\mu_0 I}{4 R}\,\hat{k}}
\]
The field points out of the page (perpendicular to the wire plane, in the $+\hat{k}$ direction by the right-hand rule for the counterclockwise arc current).
\bigskip
\textbf{Check.} If the arc were a full loop, we would recover $B = \mu_0 I/(2R)$ (the result from the centre-of-loop formula). Since we have a semicircle (half a loop), the field should be half of that: $B = \mu_0 I/(4R)$. This matches our result, confirming consistency.

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\subsection{Amp\`ere's Law and Symmetry Reduction}
This subsection states Amp\`ere's law and shows how symmetry can reduce a difficult line integral to simple algebra when the current distribution is highly symmetric. It is the magnetic analogue of Gauss's law.
\dfn{Amperian loop and enclosed current}{Let $C$ be any closed curve in space (called an \emph{Amperian loop}). The \emph{enclosed current} $I_{\mathrm{enc}}$ is the algebraic sum of all steady currents passing through any open surface $S$ bounded by $C$. The sign of each current is determined by the right-hand rule: curl the fingers of your right hand along the direction of integration around $C$; if your thumb points in the direction of the current, that current counts as positive. Currents in the opposite direction count as negative.}
\nt{Just as with Gauss's law, Amp\`ere's law is always true, but it is not always useful for finding $\vec{B}$. In a general asymmetric current distribution, knowing only the total enclosed current does not tell you the field at each point on the loop. The main strategy is therefore: first identify strong symmetry, then choose an Amperian loop matched to that symmetry so that $B=|\vec{B}|$ is constant on the field-contributing parts of the loop and the angle between $\vec{B}$ and $d\vec{\ell}$ is everywhere $0^\circ$, $180^\circ$, or $90^\circ$.}
\thm{Amp\`ere's law and when symmetry makes it useful}{Let $C$ be any closed curve and $I_{\mathrm{enc}}$ the net steady current passing through any surface bounded by $C$. Then Amp\`ere's law states
\[
\oint_C \vec{B}\cdot d\vec{\ell}=\mu_0 I_{\mathrm{enc}}.
\]
This law is always true. It becomes a practical method for solving for the magnetic field when the current distribution has enough symmetry that one can choose an Amperian loop for which the magnitude $B=|\vec{B}|$ is constant on each field-contributing part of the loop and the angle between $\vec{B}$ and $d\vec{\ell}$ is everywhere $0^\circ$, $180^\circ$, or $90^\circ$. Then the line integral reduces to algebraic terms such as $B\ell$, $-B\ell$, or $0$. Common useful cases are cylindrical symmetry (long straight wires), planar symmetry (infinite current sheets), and solenoidal symmetry (ideal solenoids). The direction of $\vec{B}$ follows the right-hand rule relative to the enclosed current: if the thumb of your right hand points in the direction of the current, your fingers curl in the direction of the magnetic field circulation.}
\nt{Amp\`ere's law is the magnetic analogue of Gauss's law. Gauss's law relates the electric field flux through a closed surface to the enclosed charge, $\oint\vec{E}\cdot d\vec{A}=q_{\mathrm{enc}}/\varepsilon_0$. Amp\`ere's law relates the magnetic field circulation around a closed loop to the enclosed current, $\oint\vec{B}\cdot d\vec{\ell}=\mu_0 I_{\mathrm{enc}}$. Both are universally valid but are practically useful for finding fields only when the source distribution has high symmetry. The matching of symmetry to geometry is parallel: spherical symmetry $\to$ spherical Gaussian surface, cylindrical symmetry $\to$ circular Amperian loop, planar symmetry $\to$ rectangular Amperian loop.}
\pf{How symmetry reduces the line integral}{Let a long straight wire carry current $I$ along the $+z$ axis. By cylindrical symmetry, the magnetic field circulates around the wire in concentric circles in planes perpendicular to the wire, and its magnitude $B(r)$ depends only on the radial distance $r$ from the wire axis. Choose a circular Amperian loop of radius $r$ centred on the wire. Along this loop, $\vec{B}$ is everywhere tangent to $d\vec{\ell}$, so $\vec{B}\cdot d\vec{\ell}=B(r)\,d\ell$, and $B(r)$ is constant everywhere on the loop. Therefore,
\[
\oint_C \vec{B}\cdot d\vec{\ell}=B(r)\oint_C d\ell=B(r)(2\pi r).
\]
If the enclosed current is $I_{\mathrm{enc}}$, Amp\`ere's law gives
\[
B(r)(2\pi r)=\mu_0 I_{\mathrm{enc}}.
\]
The law itself is general, but the symmetry is what allowed $B(r)$ to be pulled outside the integral.}
\mprop{Magnetic field of a long straight wire}{Let $\mu_0=4\pi\times 10^{-7}\,\mathrm{T\cdot m/A}$ be the permeability of free space. Consider a long straight wire of radius $R$ carrying total steady current $I$ with uniform current density $J$ across its cross section. The magnetic field magnitude is
\[
B(r)=\begin{cases}
\dfrac{\mu_0 I\,r}{2\pi R^2} & \text{for }r<R\;\text{(inside the wire)},\\[1.2ex]
\dfrac{\mu_0 I}{2\pi r} & \text{for }r>R\;\text{(outside the wire)}.
\end{cases}
\]
Inside the wire the field grows linearly with $r$; outside it falls as $1/r$. In both regions, $\vec{B}$ is tangent to circles centred on the wire axis, in the direction given by the right-hand rule.}
\cor{Continuity of $B$ at the wire surface}{At the surface of the wire ($r=R$), both the inside and outside formulas give the same result:
\[
B(R)=\frac{\mu_0 I\,R}{2\pi R^2}=\frac{\mu_0 I}{2\pi R}.
\]
The magnetic field is continuous at the boundary even though the functional form changes. The maximum field occurs at the surface.}
\mprop{Magnetic field of an infinite current sheet}{An infinite planar sheet carrying uniform surface current density $K$ (current per unit width, in units of $\mathrm{A/m}$). Choose a rectangular Amperian loop of length $L$ straddling the sheet, with two long sides parallel to the field and two short sides perpendicular. By symmetry, the field has constant magnitude on each side of the sheet and is parallel to the sheet but perpendicular to the current direction. Amp\`ere's law gives
\[
2BL=\mu_0(KL)\quad\Rightarrow\quad B=\frac{\mu_0 K}{2}.
\]
The field reverses direction on opposite sides of the sheet. The field outside is independent of distance from the sheet.}
\ex{Illustrative example}{A square loop of side $0.1\,\mathrm{m}$ carries a current $I=2\,\mathrm{A}$ in a uniform magnetic field $B=0.5\,\mathrm{T}$ perpendicular to the plane of the loop. The magnetic flux through the loop is
\[
\Phi_B = BA = (0.5\,\mathrm{T})(0.1\,\mathrm{m})^2 = 5.0\times 10^{-3}\,\mathrm{T\cdot m^2}.
\]}
\qs{Worked example}{A long cylindrical wire of radius $R=2.0\times 10^{-3}\,\mathrm{m}=2.0\,\mathrm{mm}$ carries a steady current $I=10\,\mathrm{A}$ uniformly distributed across its cross section. The current flows in the $+\hat{k}$ direction.
Find the magnetic field magnitude and direction at:
\begin{enumerate}[label=(\alph*)]
\item $r_1=1.0\times 10^{-3}\,\mathrm{m}=1.0\,\mathrm{mm}$ (inside the wire), and
\item $r_2=5.0\times 10^{-3}\,\mathrm{m}=5.0\,\mathrm{mm}$ (outside the wire).
\end{enumerate}}
\sol Let the wire lie along the $z$ axis with current flowing in the $+\hat{k}$ direction. By cylindrical symmetry, the magnetic field circulates around the wire in concentric circles in planes perpendicular to the wire. The field magnitude depends only on the radial distance $r$ from the wire axis.
Choose a circular Amperian loop of radius $r$ centred on the wire axis. Along this loop, $\vec{B}$ is everywhere tangent to $d\vec{\ell}$, so $\vec{B}\cdot d\vec{\ell}=B(r)\,d\ell$. The line integral becomes
\[
\oint_C \vec{B}\cdot d\vec{\ell}=B(r)\oint_C d\ell=B(r)(2\pi r).
\]
The enclosed current depends on whether the loop is inside or outside the wire. The uniform current density is
\[
J=\frac{I}{\pi R^2}.
\]
\textbf{(a) Inside the wire ($r_1<R$).} The enclosed current is the fraction of the total current passing through the area inside the Amperian loop:
\[
I_{\mathrm{enc}}=J(\pi r_1^2)=\frac{I}{\pi R^2}(\pi r_1^2)=I\,\frac{r_1^2}{R^2}.
\]
Substitute the values:
\[
I_{\mathrm{enc}}=(10\,\mathrm{A})\,\frac{(1.0\times 10^{-3}\,\mathrm{m})^2}{(2.0\times 10^{-3}\,\mathrm{m})^2}=(10\,\mathrm{A})\,\frac{1.0\times 10^{-6}}{4.0\times 10^{-6}}.
\]
Thus
\[
I_{\mathrm{enc}}=(10\,\mathrm{A})\times\frac{1}{4}=2.5\,\mathrm{A}.
\]
Apply Amp\`ere's law:
\[
B(r_1)(2\pi r_1)=\mu_0 I_{\mathrm{enc}}.
\]
Solve for $B(r_1)$:
\[
B(r_1)=\frac{\mu_0 I_{\mathrm{enc}}}{2\pi r_1}=\frac{(4\pi\times 10^{-7}\,\mathrm{T\cdot m/A})(2.5\,\mathrm{A})}{2\pi(1.0\times 10^{-3}\,\mathrm{m})}.
\]
Compute step by step:
\[
(4\pi\times 10^{-7})(2.5)=10\pi\times 10^{-7},
\]
\[
2\pi(1.0\times 10^{-3})=2\pi\times 10^{-3},
\]
\[
B(r_1)=\frac{10\pi\times 10^{-7}}{2\pi\times 10^{-3}}\,\mathrm{T}=5.0\times 10^{-4}\,\mathrm{T}.
\]
The direction follows the right-hand rule: thumb along $+\hat{k}$ (current direction), fingers curl counterclockwise when viewed from above.
\textbf{(b) Outside the wire ($r_2>R$).} The Amperian loop encloses the entire wire, so
\[
I_{\mathrm{enc}}=I=10\,\mathrm{A}.
\]
Apply Amp\`ere's law:
\[
B(r_2)(2\pi r_2)=\mu_0 I.
\]
Solve for $B(r_2)$:
\[
B(r_2)=\frac{\mu_0 I}{2\pi r_2}=\frac{(4\pi\times 10^{-7}\,\mathrm{T\cdot m/A})(10\,\mathrm{A})}{2\pi(5.0\times 10^{-3}\,\mathrm{m})}.
\]
Compute step by step:
\[
(4\pi\times 10^{-7})(10)=40\pi\times 10^{-7},
\]
\[
2\pi(5.0\times 10^{-3})=10\pi\times 10^{-3},
\]
\[
B(r_2)=\frac{40\pi\times 10^{-7}}{10\pi\times 10^{-3}}\,\mathrm{T}=4.0\times 10^{-4}\,\mathrm{T}.
\]
The direction is again given by the right-hand rule: counterclockwise when viewed from above.
\bigskip
\textbf{Final answers:}
\begin{enumerate}[label=(\alph*)]
\item $B(r_1)=5.0\times 10^{-4}\,\mathrm{T}=0.50\,\mathrm{mT}$, directed counterclockwise around the wire.
\item $B(r_2)=4.0\times 10^{-4}\,\mathrm{T}=0.40\,\mathrm{mT}$, directed counterclockwise around the wire.
\end{enumerate}

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\subsection{Solenoids, Parallel Currents, and Magnetic Dipoles}
This subsection covers three closely related topics: the magnetic field inside a solenoid and a toroid (derived from Amp\`ere's law), the magnetic force per unit length between two parallel current-carrying wires, and the magnetic dipole moment of a current loop. Together these describe how steady currents in geometrically ordered configurations produce well-defined magnetic fields and forces.
\dfn{Ideal solenoid}{An \emph{ideal solenoid} is a long, tightly wound helical coil of wire. When the winding is close and the length $L$ is much greater than the radius $R$, the magnetic field inside is uniform, axial, and of magnitude
\[
B = \mu_0\,n\,I,
\]
where $I$ is the current, $n = N/L$ is the number of turns per unit length ($N$ is the total number of turns, $L$ is the length of the solenoid), and $\mu_0 = 4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A}$ is the permeability of free space. Outside the ideal solenoid the field is approximately zero. The field direction follows the right-hand rule: curl the fingers of your right hand in the direction of the current around the coil, and your thumb points along $\vec{B}$ inside the solenoid.}
\nt{The field inside a solenoid depends only on the turn density $n$ and the current $I$; it is independent of the radius of the coil. This is analogous to the electric field inside a uniformly charged infinite sheet, which depends only on the surface charge density and not on the lateral dimensions.}
\thm{Magnetic field of a long solenoid}{For an ideal solenoid of length $L \gg R$ with $N$ turns carrying current $I$, the magnetic field inside is uniform and directed along the axis. Its magnitude is
\[
B = \mu_0\,n\,I
\qquad\text{where}\qquad
n = \frac{N}{L}.
\]
Outside the solenoid $B \approx 0$. The field lines are straight, parallel lines inside and loop back around outside (forming closed loops).}
\dfn{Toroid}{A \emph{toroid} is a solenoid bent into a circle (a doughnut shape). It consists of $N$ turns wound around a circular core of mean radius $r$. By applying Amp\`ere's law to a circular Amperian loop of radius $r$ inside the toroid, the magnetic field inside the torus is
\[
B = \frac{\mu_0\,N\,I}{2\pi\,r}.
\]
Outside the toroid ($r$ is not within the winding region), $B \approx 0$. The field lines are circles concentric with the toroid axis.}
\nt{The toroid field formula $B = \mu_0 N I/(2\pi r)$ shows a $1/r$ dependence, unlike the uniform solenoid. For a toroid with a large mean radius and small cross-section (so $r$ varies little across the windings), the field is nearly uniform and approximately $B \approx \mu_0 n I$ where $n = N/(2\pi r_{\text{avg}})$.}
\mprop{Solenoid and toroid magnetic fields}{The magnetic field produced by steady currents in these common device geometries is:
\begin{itemize}
\item \textbf{Long solenoid (inside):} $B = \mu_0\,n\,I$, where $n = N/L$ is the turn density. The field is uniform and axial.
\item \textbf{Toroid (inside the windings):} $B = \dfrac{\mu_0\,N\,I}{2\pi\,r}$, where $r$ is the radial distance from the toroid centre. The field circulates along circular field lines and decreases as $1/r$.
\item \textbf{Outside both devices:} $B \approx 0$ (ideal case).
\end{itemize}}
\pf{Solenoid field from Amp\`ere's law (outline)}{Consider an ideal solenoid with $n$ turns per unit length. Choose a rectangular Amperian loop with one long side (length $\ell$) inside the solenoid parallel to the axis, the opposite side outside, and the two short sides perpendicular to the axis. The line integral of $\vec{B}\cdot d\vec{\ell}$ around the loop receives contributions only from the inside segment, because $\vec{B}\approx 0$ outside and $\vec{B}\perp d\vec{\ell}$ on the perpendicular segments. Thus
\[
\oint \vec{B}\cdot d\vec{\ell} = B\,\ell.
\]
The enclosed current is $I_{\text{enc}} = n\,\ell\,I$ (each of the $n\ell$ turns inside the loop carries current $I$). By Amp\`ere's law, $B\,\ell = \mu_0\,n\,\ell\,I$, giving $B = \mu_0 n I$.}
\pf{Toroid field from Amp\`ere's law}{Choose a circular Amperian loop of radius $r$ inside the toroid, concentric with the toroid axis. By symmetry, $\vec{B}$ is tangent to the circle and has constant magnitude $B$ at fixed $r$. The line integral is
\[
\oint \vec{B}\cdot d\vec{\ell} = B\,(2\pi r).
\]
The loop encloses $N$ turns each carrying current $I$, so $I_{\text{enc}} = N I$. Amp\`ere's law gives $B(2\pi r) = \mu_0 N I$, yielding $B = \mu_0 N I/(2\pi r)$.}
\nt{Both derivations rely on Amp\`ere's law $\oint \vec{B}\cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$ and the presence of sufficient symmetry to pull $B$ out of the integral. The solenoid requires the infinite-length approximation; the toroid requires circular symmetry. These are the two magnetostatic configurations in the AP Physics C curriculum where Amp\`ere's law gives a clean result.}
\dfn{Magnetic force between two parallel current-carrying wires}{Two long, straight, parallel wires separated by distance $d$, carrying steady currents $I_1$ and $I_2$, exert magnetic forces on each other. Wire 1 produces a magnetic field $B_1 = \dfrac{\mu_0 I_1}{2\pi d}$ at the location of wire 2. The force per unit length on wire 2 due to wire 1 is
\[
\frac{F_{12}}{L} = \frac{\mu_0\,I_1\,I_2}{2\pi\,d}.
\]
The force is \emph{attractive} when the currents flow in the \emph{same} direction and \emph{repulsive} when they flow in \emph{opposite} directions. By Newton's third law, the force per unit length on wire 1 due to wire 2 has the same magnitude and opposite direction.}
\nt{The rule for attraction and repulsion is the \emph{opposite} of what happens with electric charges. Parallel currents (same direction) \emph{attract}, while like electric charges \emph{repel}. A useful mnemonic: ``like currents attract, unlike repel'' -- but remember this refers to current \emph{directions}, not charge types.}
\thm{Force per unit length between two parallel wires}{Two long straight parallel wires separated by distance $d$ carry currents $I_1$ and $I_2$. The magnitude of the magnetic force per unit length is
\[
\frac{F}{L} = \frac{\mu_0\,I_1\,I_2}{2\pi\,d}.
\]
\begin{itemize}
\item Currents in the \textbf{same direction} $\;\rightarrow\;$ \textbf{attractive} force.
\item Currents in \textbf{opposite directions} $\;\rightarrow\;$ \textbf{repulsive} force.
\end{itemize}
The direction of the force on either wire is perpendicular to the wire and toward (or away from) the other wire, along the line connecting the wires.}
\pf{Force between parallel wires from the Lorentz force}{Wire 1 (carrying $I_1$) produces a magnetic field at the position of wire 2. By the right-hand rule for a long straight wire, $B_1$ circles wire 1 and has magnitude $B_1 = \mu_0 I_1/(2\pi d)$. The field at wire 2 is perpendicular to wire 2. The Lorentz force on a length $L$ of wire 2 is $\vec{F}_{12} = I_2\,\vec{L}\times\vec{B}_1$. Since $\vec{L}\perp\vec{B}_1$, the magnitude is
\[
F_{12} = I_2\,L\,B_1 = I_2\,L\,\frac{\mu_0 I_1}{2\pi d}.
\]
Dividing by $L$ gives $F_{12}/L = \mu_0 I_1 I_2/(2\pi d)$. The direction follows from the cross product: if both currents point upward, $\vec{B}_1$ at wire 2 points into the page, and $\vec{L}_2\times\vec{B}_1$ points toward wire 1 (attractive).}
\thm{Magnetic dipole moment of a current loop}{A planar current loop carrying current $I$ and enclosing area $A$ has a \emph{magnetic dipole moment}
\[
\vec{\mu} = I\,A\,\hat{n},
\]
where $\hat{n}$ is the unit normal to the plane of the loop, determined by the right-hand rule: curl the fingers of your right hand in the direction of the current, and your thumb points along $\hat{n}$. The SI unit of $\mu$ is the ampere-square metre ($\mathrm{A\!\cdot\!m^2}$), equivalent to joule per tesla ($\mathrm{J/T}$). For a coil with $N$ turns, $\vec{\mu} = N I A\,\hat{n}$.}
\nt{The magnetic dipole moment characterises the torque a current loop experiences in a uniform external field: $\vec{\tau} = \vec{\mu}\times\vec{B}$. It is also the quantity that determines the far-field of the loop -- at distances much greater than the loop size, a current loop produces the same magnetic field as a magnetic dipole. This is the quantum-mechanical basis of atomic magnetism (electron orbital and spin angular momenta give rise to magnetic dipole moments).}
\mprop{Magnetic dipole properties}{For a planar current loop (or coil of $N$ turns) with area $A$ and current $I$:
\begin{itemize}
\item \textbf{Dipole moment:} $\mu = N I A$. The direction is given by the right-hand rule.
\item \textbf{Torque in a uniform field:} $\vec{\tau} = \vec{\mu}\times\vec{B}$, with magnitude $\tau = \mu B \sin\theta$, where $\theta$ is the angle between $\vec{\mu}$ and $\vec{B}$.
\item \textbf{Potential energy:} $U = -\vec{\mu}\cdot\vec{B} = -\mu B \cos\theta$. The minimum energy (stable equilibrium) occurs when $\vec{\mu}$ aligns with $\vec{B}$ ($\theta = 0$).
\end{itemize}}
\cor{Rectangular and circular loops}{For a rectangular loop of sides $a$ and $b$, $A = a\,b$. For a circular loop of radius $R$, $A = \pi R^2$. In both cases $\mu = I A$ for a single turn, and the dipole moment points along the axis of symmetry.}
\ex{Illustrative example}{Two parallel wires are separated by $d = 8.0\,\mathrm{cm}$ and carry currents $I_1 = 3.0\,\mathrm{A}$ and $I_2 = 5.0\,\mathrm{A}$ in the same direction. The force per unit length is
\[
\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A})(3.0\,\mathrm{A})(5.0\,\mathrm{A})}{2\pi(0.080\,\mathrm{m})} = \frac{6\times 10^{-6}}{0.080}\,\mathrm{N/m} = 7.5\times 10^{-5}\,\mathrm{N/m},
\]
attractive. A circular loop of radius $5.0\,\mathrm{cm}$ carrying $2.0\,\mathrm{A}$ has dipole moment $\mu = I(\pi R^2) = (2.0)(\pi)(0.050)^2 = 1.57\times 10^{-2}\,\mathrm{A\!\cdot\!m^2}$.}
\qs{Worked example}{A solenoid is $40.0\,\mathrm{cm}$ long, has $N = 600$ turns uniformly distributed along its length, and a circular cross-section of diameter $3.0\,\mathrm{cm}$. A steady current $I = 4.0\,\mathrm{A}$ flows through the wire.
Find:
\begin{enumerate}[label=(\alph*)]
\item the turn density $n$ of the solenoid,
\item the magnitude of the magnetic field inside the solenoid,
\item the direction of the magnetic field if the current, viewed from the left end, flows counterclockwise,
\item the magnetic dipole moment of the solenoid, and
\item the torque on the solenoid if it is placed in a uniform external magnetic field $B_{\text{ext}} = 0.15\,\mathrm{T}$ with its axis at $30^\circ$ to the field.
\end{enumerate}}
\sol \textbf{Part (a).} The turn density is the number of turns divided by the length:
\[
n = \frac{N}{L} = \frac{600}{0.400\,\mathrm{m}} = 1500\,\mathrm{turns/m}.
\]
\textbf{Part (b).} The magnetic field inside the solenoid is
\[
B = \mu_0\,n\,I.
\]
Substitute the values:
\[
B = \bigl(4\pi\times 10^{-7}\,\mathrm{T\!\cdot\!m/A}\bigr)\bigl(1500\,\mathrm{m^{-1}}\bigr)\bigl(4.0\,\mathrm{A}\bigr).
\]
Compute step by step:
\[
1500\times 4.0 = 6000,
\]
\[
B = 4\pi\times 10^{-7}\times 6000 = 4\pi\times 6.0\times 10^{-4} = 24\pi\times 10^{-4}\,\mathrm{T}.
\]
Using $\pi \approx 3.1416$:
\[
B = 24\times 3.1416\times 10^{-4}\,\mathrm{T} = 7.54\times 10^{-3}\,\mathrm{T} = 7.54\,\mathrm{mT}.
\]
\textbf{Part (c).} The current, viewed from the left end, flows counterclockwise. By the right-hand rule: curl the fingers of your right hand counterclockwise (as seen from the left), and your thumb points to the right. Thus the magnetic field inside the solenoid points to the \textbf{right} (along the solenoid axis, away from the viewer of the left end).
\textbf{Part (d).} The magnetic dipole moment of a solenoid is $\mu = N I A$, where $A$ is the cross-sectional area of one turn. The radius of the solenoid is
\[
R = \frac{3.0\,\mathrm{cm}}{2} = 1.5\,\mathrm{cm} = 0.015\,\mathrm{m}.
\]
The area is
\[
A = \pi R^2 = \pi (0.015\,\mathrm{m})^2 = \pi \times 2.25\times 10^{-4}\,\mathrm{m^2} = 7.07\times 10^{-4}\,\mathrm{m^2}.
\]
The dipole moment magnitude is
\[
\mu = N I A = (600)(4.0\,\mathrm{A})(7.07\times 10^{-4}\,\mathrm{m^2}).
\]
Compute:
\[
600\times 4.0 = 2400,
\]
\[
\mu = 2400\times 7.07\times 10^{-4}\,\mathrm{A\!\cdot\!m^2} = 1.70\,\mathrm{A\!\cdot\!m^2}.
\]
The direction of $\vec{\mu}$ is along the axis to the right (same as $\vec{B}$ inside).
\textbf{Part (e).} The torque on a magnetic dipole in a uniform field is $\vec{\tau} = \vec{\mu}\times\vec{B}_{\text{ext}}$. Its magnitude is
\[
\tau = \mu\,B_{\text{ext}}\,\sin\theta,
\]
where $\theta = 30^\circ$. Substituting:
\[
\tau = (1.70\,\mathrm{A\!\cdot\!m^2})(0.15\,\mathrm{T})\,\sin 30^\circ.
\]
Since $\sin 30^\circ = 0.5$:
\[
\tau = 1.70\times 0.15\times 0.5 = 0.1275\,\mathrm{N\!\cdot\!m} \approx 0.13\,\mathrm{N\!\cdot\!m}.
\]
\bigskip
\textbf{Final answers:}
\begin{enumerate}[label=(\alph*)]
\item $n = 1500\,\mathrm{turns/m}$
\item $B = 7.54\,\mathrm{mT}$ (or $7.5\times 10^{-3}\,\mathrm{T}$)
\item To the right (along the solenoid axis)
\item $\mu = 1.70\,\mathrm{A\!\cdot\!m^2}$
\item $\tau = 0.13\,\mathrm{N\!\cdot\!m}$
\end{enumerate}