23 May 2025

Newtonian Mechanics and Charged Particles in Uniform Fields

Newtonian motion in uniform electric, magnetic, and crossed fields, including gyroradius, gyrofrequency, and drift.

bsc semester-v classical-mechanics mj-10 unit-i charged-particle

Newtonian mechanics: a concise review

In an inertial frame, a particle of constant mass obeys

\[\mathbf p=m\mathbf v, \qquad \frac{d\mathbf p}{dt}=\mathbf F, \qquad \mathbf v=\dot{\mathbf r}.\]

Thus Newton’s second law is $m\ddot{\mathbf r}=\mathbf F$. The first law is the zero-force case, and the third law makes the internal forces of an isolated system cancel in pairs. Consequently,

\[\frac{d\mathbf P}{dt}=\mathbf F_{\mathrm{ext}}, \qquad \frac{d\mathbf L_O}{dt}=\boldsymbol\tau_{O,\mathrm{ext}},\]

where $\mathbf P$ is total linear momentum and $\mathbf L_O$ is angular momentum about $O$. For one particle,

\[\frac{dT}{dt}=\mathbf F\cdot\mathbf v, \qquad T=\frac12mv^2.\]

If $\mathbf F=-\nabla V$ and $V$ has no explicit time dependence, $T+V$ is conserved.

Lorentz-force equation

For a particle of charge $q$ in prescribed fields,

\[\boxed{m\ddot{\mathbf r}=q(\mathbf E+\dot{\mathbf r}\times\mathbf B)}.\]

In SI units, $qE$ and $qvB$ are forces in newtons. Taking the scalar product with $\mathbf v$ gives

\[\frac{dT}{dt}=q\mathbf E\cdot\mathbf v,\]

so a magnetic field changes the direction of motion but does no work.

Uniform electric field

With constant $\mathbf E$ and $\mathbf B=0$, direct integration gives

\[\boxed{\mathbf v(t)=\mathbf v_0+\frac{q\mathbf E}{m}t}, \qquad \boxed{\mathbf r(t)=\mathbf r_0+\mathbf v_0t+\frac{q\mathbf E}{2m}t^2}.\]

The acceleration $q\mathbf E/m$ has units $\mathrm{m\,s^{-2}}$. The velocity perpendicular to $\mathbf E$ remains constant, while the parallel component changes uniformly. For example, take $\mathbf E=E\hat{\mathbf y}$, $\mathbf r_0=0$, and $\mathbf v_0=v_0\hat{\mathbf x}$. Eliminating $t$ from

\[x=v_0t, \qquad y=\frac{qE}{2m}t^2\]

gives the parabola

\[\boxed{y=\frac{qE}{2mv_0^2}x^2}.\]

The opening reverses when the sign of $qE$ reverses.

Uniform magnetic field

Let $\mathbf B=B\hat{\mathbf z}$ with $B>0$ and $\mathbf E=0$. The component equations are

\[\dot v_x=\omega_c v_y, \qquad \dot v_y=-\omega_c v_x, \qquad \dot v_z=0, \qquad \omega_c=\frac{qB}{m}.\]

Here $\omega_c$ is signed: its sign fixes the sense of rotation. For $v_x(0)=v_\perp$ and $v_y(0)=0$,

\[v_x=v_\perp\cos(\omega_ct), \qquad v_y=-v_\perp\sin(\omega_ct).\]

Integrating once more,

\[x=x_0+\frac{v_\perp}{\omega_c}\sin(\omega_ct), \qquad y=y_0+\frac{v_\perp}{\omega_c}\bigl[\cos(\omega_ct)-1\bigr].\]

With $x_c=x_0$ and $y_c=y_0-v_\perp/\omega_c$,

\[(x-x_c)^2+(y-y_c)^2=\left(\frac{v_\perp}{\omega_c}\right)^2.\]

Therefore the gyroradius, gyrofrequency magnitude, and gyroperiod are

\[\boxed{r_g=\frac{mv_\perp}{\lvert q\rvert B}}, \qquad \boxed{\Omega_g=\frac{\lvert q\rvert B}{m}}, \qquad \boxed{T_g=\frac{2\pi}{\Omega_g}}.\]

$r_g$ is measured in metres, $\Omega_g$ in $\mathrm{s^{-1}}$, and $T_g$ in seconds. A constant component $v_\parallel$ along $\mathbf B$ turns the circle into a helix with pitch $2\pi v_\parallel/\Omega_g$.

Crossed electric and magnetic fields

Suppose $\mathbf E\cdot\mathbf B=0$ and both fields are uniform. Define

\[\boxed{\mathbf v_D=\frac{\mathbf E\times\mathbf B}{B^2}}, \qquad \mathbf v=\mathbf u+\mathbf v_D.\]

The vector identity

\[(\mathbf E\times\mathbf B)\times\mathbf B =-B^2\mathbf E\]

shows that $\mathbf v_D\times\mathbf B=-\mathbf E$. The equation of motion therefore becomes

\[m\dot{\mathbf u}=q\mathbf u\times\mathbf B.\]

The motion is a gyration superposed on a uniform guiding-centre drift. The drift speed has units $E/B=\mathrm{m\,s^{-1}}$ and is independent of both $m$ and $q$.

For the special case $q>0$, $\mathbf E=E\hat{\mathbf x}$, $\mathbf B=B\hat{\mathbf z}$, and $\mathbf r(0)=\mathbf v(0)=0$, put $a=E/B$ and $\Omega=qB/m$. The exact trajectory is

\[x(t)=\frac{a}{\Omega}\bigl[1-\cos(\Omega t)\bigr], \qquad y(t)=-\frac{a}{\Omega}\bigl[\Omega t-\sin(\Omega t)\bigr], \qquad z(t)=0.\]

It is a cycloid whose mean velocity is $-(E/B)\hat{\mathbf y}=\mathbf v_D$. For a negative charge the gyration reverses, but the $\mathbf E\times\mathbf B$ drift does not.

Parabolic electric-field motion, circular magnetic gyration, and the exact crossed-field cycloid described above.

Maxima verification: uniform-field equations and crossed-field trajectory residuals.

Solved Problems

1. Helical motion of an electron

An electron enters a uniform field $\mathbf B=0.20\,\hat{\mathbf z}\ \mathrm T$ with

\[\mathbf v_0=(3.0\times10^6\,\hat{\mathbf x} +1.2\times10^6\,\hat{\mathbf z})\ \mathrm{m\,s^{-1}}.\]

Find its gyroradius, gyroperiod, pitch, and sense of rotation. Use $e=1.602\,176\,634\times10^{-19}\ \mathrm C$ and $m_e=9.109\,383\,7139\times10^{-31}\ \mathrm{kg}$.

The magnetic field separates the velocity into

\[v_\perp=3.0\times10^6\ \mathrm{m\,s^{-1}}, \qquad v_\parallel=1.2\times10^6\ \mathrm{m\,s^{-1}}.\]

The positive gyrofrequency magnitude is

\[\Omega_g=\frac{eB}{m_e} =\frac{(1.602\,176\,634\times10^{-19})(0.20)} {9.109\,383\,7139\times10^{-31}} =3.5188\times10^{10}\ \mathrm{s^{-1}}.\]

Hence

\[r_g=\frac{v_\perp}{\Omega_g} =8.528\times10^{-5}\ \mathrm m,\] \[T_g=\frac{2\pi}{\Omega_g} =1.786\times10^{-10}\ \mathrm s,\]

and the advance along $\mathbf B$ in one turn is

\[p=v_\parallel T_g =2.143\times10^{-4}\ \mathrm m.\]

At entry, $\mathbf v_\perp\times\mathbf B=-v_\perp B\hat{\mathbf y}$. Since the electron has $q=-e$, its force is initially along $+\hat{\mathbf y}$; viewed from the $+\hat{\mathbf z}$ side, its velocity therefore rotates counterclockwise. Reversing the charge reverses this sense but leaves $r_g$, $T_g$, and the pitch magnitude unchanged.

2. Gyration relative to the crossed-field drift

A proton moves in $\mathbf E=3.0\times10^4\hat{\mathbf x}\ \mathrm{V\,m^{-1}}$ and $\mathbf B=0.15\hat{\mathbf z}\ \mathrm T$. Its initial velocity is $4.0\times10^5\hat{\mathbf y}\ \mathrm{m\,s^{-1}}$. Find the guiding-centre velocity and the gyroradius about that centre.

First,

\[\mathbf v_D=\frac{\mathbf E\times\mathbf B}{B^2} =-\frac EB\hat{\mathbf y} =-2.00\times10^5\hat{\mathbf y}\ \mathrm{m\,s^{-1}}.\]

The velocity that gyrates is not $\mathbf v_0$ but

\[\mathbf u_0=\mathbf v_0-\mathbf v_D =6.00\times10^5\hat{\mathbf y}\ \mathrm{m\,s^{-1}}.\]

With $m_p=1.672\,621\,923\,69\times10^{-27}\ \mathrm{kg}$,

\[\Omega_p=\frac{eB}{m_p}=1.437\times10^7\ \mathrm{s^{-1}},\]

so

\[r_g=\frac{u_0}{\Omega_p} =\frac{m_pu_0}{eB} =4.176\times10^{-2}\ \mathrm m.\]

The proton’s mean motion is therefore $-2.00\times10^5\hat{\mathbf y}\ \mathrm{m\,s^{-1}}$, with a $4.176\ \mathrm{cm}$ gyration superposed. The result separates continuously into pure magnetic gyration when $E\to0$ and pure drift when $\mathbf u_0\to0$.

Descriptive Questions

1. Derive the signed component equations for a charged particle in $\mathbf B=B\hat{\mathbf z}$ and explain how the sign of the charge fixes the sense of gyration.
2. Explain why a uniform magnetic field changes momentum but not kinetic energy, whereas a uniform electric field can change both.
3. Starting from the Lorentz equation, obtain the charge-independent $\mathbf E\times\mathbf B$ guiding-centre velocity for mutually perpendicular uniform fields.
4. Distinguish gyroradius, gyrofrequency, gyroperiod, and helical pitch, stating the dependence of each on $m$, $q$, $B$, $v_\perp$, and $v_\parallel$.

Numerical Problems

1. A proton starts from rest in $\mathbf E=2.5\times10^4\hat{\mathbf x}\ \mathrm{V\,m^{-1}}$. Find its acceleration, speed, and displacement after $2.0\ \mu\mathrm s$. Answer: $2.395\times10^{12}\ \mathrm{m\,s^{-2}}$, $4.789\times10^6\ \mathrm{m\,s^{-1}}$, $4.789\ \mathrm m$.
2. An electron moves perpendicular to a $0.080\ \mathrm T$ field at $2.0\times10^6\ \mathrm{m\,s^{-1}}$. Find its orbit radius and ordinary cyclotron frequency $f=\Omega_g/(2\pi)$. Answer: $1.421\times10^{-4}\ \mathrm m$, $2.239\times10^9\ \mathrm{Hz}$.
3. An alpha particle has $v_\perp=5.0\times10^5\ \mathrm{m\,s^{-1}}$ and $v_\parallel=8.0\times10^5\ \mathrm{m\,s^{-1}}$ in $B=0.30\ \mathrm T$. Take $q=2e$ and $m_\alpha=6.6447\times10^{-27}\ \mathrm{kg}$. Find its gyroradius and pitch. Answer: $3.456\times10^{-2}\ \mathrm m$, $3.474\times10^{-1}\ \mathrm m$.
4. In $\mathbf E=1.8\times10^4\hat{\mathbf x}\ \mathrm{V\,m^{-1}}$ and $\mathbf B=0.12\hat{\mathbf z}\ \mathrm T$, a proton is launched at $-3.0\times10^5\hat{\mathbf y}\ \mathrm{m\,s^{-1}}$. Find the guiding-centre velocity and its gyroradius relative to that centre. Answer: $\mathbf v_D=-1.50\times10^5\hat{\mathbf y}\ \mathrm{m\,s^{-1}}$, $r_g=1.305\times10^{-2}\ \mathrm m$.

Maxima verification: all solved-problem and numerical values.

References

1. Lorentz force — Wikipedia
2. H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Chapter 1, Pearson (2002).
3. D. J. Griffiths, Introduction to Electrodynamics, 4th ed., Chapter 5, Pearson (2013).
4. J. D. Jackson, Classical Electrodynamics, 3rd ed., Chapter 12, Wiley (1998).