\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}
