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.