\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}}

The magnetic force on a single charge (this section) generalizes to the magnetic force on a current-carrying wire (Section 12.3) by summing over all charge carriers. The Biot-Savart law (Section 12.4) gives the reverse: how currents produce the $\vec{B}$ field that exerts these forces.

\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.}

\wc{A stationary charge produces zero magnetic field}{A point charge at rest produces only an \emph{electric} field. A magnetic field is produced only by \emph{moving} charges (currents) or changing electric fields (displacement current). A single stationary charge $q$ has $\vec{B}=\vec{0}$ everywhere.}

\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 $(\vec{a}\times\vec{b})\cdot\vec{a}=0$.}

\wc{Magnetic force does zero work --- always}{Because $\vec{F}_B=q\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$, the power $P=\vec{F}_B\cdot\vec{v}=0$. The magnetic force can change a particle's direction but never its speed or kinetic energy. Even in complex field configurations, the magnetic force contributes zero to the work integral $\int\vec{F}\cdot d\vec{r}$.}

\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}
