Charged Particles in Fields

The electric field between parallel plates points from the positive plate to the negative plate. A positive charge feels a force in the direction of the field, so it is pushed from the positive plate toward the negative plate.

The magnetic force is $F = qvB\sin\theta$. It is zero when the charge is stationary ($v = 0$) or moving parallel to the field ($\theta = 0$). A force arises only when the charge moves with a component across the field.

The magnetic force is always perpendicular to the velocity, so it does no work ($W = Fd\cos 90^\circ = 0$). It changes the direction of motion but never the speed, which is why charges travel in circles at constant speed in a magnetic field.

Between parallel plates the field is uniform: $E = \dfrac{V}{d} = \dfrac{200}{0.050} = 4000\text{ V/m}$. (The gap must be in metres: $5.0\text{ cm} = 0.050\text{ m}$.)

$F = qE = 3.0\times10^{-6} \times 4000 = 0.012\text{ N}$, in the direction of the field for a positive charge.

$W = qV = 2.0\times10^{-6} \times 500 = 1.0\times10^{-3}\text{ J} = 1.0\text{ mJ}$. Potential difference is the energy transferred per coulomb of charge.

All the electrical work becomes kinetic energy: $qV = \tfrac{1}{2}mv^2$, so $v = \sqrt{\dfrac{2qV}{m}} = \sqrt{\dfrac{2(1.6\times10^{-19})(100)}{9.1\times10^{-31}}} \approx 5.9\times10^{6}\text{ m/s}$.

$F = qvB = (1.6\times10^{-19})(2.0\times10^{6})(0.50) = 1.6\times10^{-13}\text{ N}$, directed perpendicular to both the velocity and the field.

The magnetic force provides the centripetal force: $qvB = \dfrac{mv^2}{r}$, so $r = \dfrac{mv}{qB} = \dfrac{(1.67\times10^{-27})(2.0\times10^{6})}{(1.6\times10^{-19})(0.50)} \approx 0.042\text{ m} = 4.2\text{ cm}$.

The period is $T = \dfrac{2\pi m}{qB}$, which does not contain $v$. A faster particle travels a larger circle ($r \propto v$) but covers the extra distance in the same time, so the period is unchanged. (This constant period is the key to the cyclotron.)

The particle goes straight only when the electric and magnetic forces balance: $qE = qvB$, so $v = \dfrac{E}{B} = \dfrac{3.0\times10^{4}}{0.10} = 3.0\times10^{5}\text{ m/s}$. This speed is independent of the charge.

It has a constant velocity along its original direction and a constant acceleration perpendicular to it, exactly like a projectile in gravity. This combination gives a parabolic path. (In a magnetic field the path would instead be circular.)

$a = \dfrac{F}{m} = \dfrac{qE}{m} = \dfrac{(1.6\times10^{-19})(4000)}{9.1\times10^{-31}} \approx 7.0\times10^{14}\text{ m/s}^2$. Because the electron is so light, even a modest field gives an enormous acceleration.

Just like a projectile: the time in the field is $t = \dfrac{L}{v} = \dfrac{0.10}{1.0\times10^{7}} = 1.0\times10^{-8}\text{ s}$, and the sideways deflection is $y = \tfrac{1}{2}at^2 = \tfrac{1}{2}(7.0\times10^{14})(1.0\times10^{-8})^2 = 0.035\text{ m} = 3.5\text{ cm}$.

The electric force $F = qE$ depends only on the field, not the speed, so it is constant. The magnetic force $F = qvB$ is proportional to the speed, so it grows. This speed dependence is exactly what a velocity selector exploits.

From $r = \dfrac{mv}{qB}$, rearrange for the charge-to-mass ratio: $\dfrac{q}{m} = \dfrac{v}{rB} = \dfrac{3.0\times10^{6}}{(0.10)(0.20)} = 1.5\times10^{8}\text{ C/kg}$.

Using the right-hand rule ($\vec{F} = q\vec{v}\times\vec{B}$) for a positive charge: point the fingers along $\vec{v}$ (right) and curl them into $\vec{B}$ (into the page); the thumb points upward. So the force deflects the charge upward.

Since $r = \dfrac{mv}{qB}$, with the same $m$, $q$ and $B$, the radius is proportional to the speed: $r \propto v$. Doubling the speed doubles the radius, giving a ratio of $2 : 1$.

First the selector fixes the speed: $v = \dfrac{E}{B} = \dfrac{3.0\times10^{4}}{0.10} = 3.0\times10^{5}\text{ m/s}$. Then from the circular arc, $m = \dfrac{qBr}{v} = \dfrac{(1.6\times10^{-19})(0.20)(0.15)}{3.0\times10^{5}} \approx 1.6\times10^{-26}\text{ kg}$.

$r = \dfrac{mv}{qB}$, so with the same $v$, $q$ and $B$, $r \propto m$. The heavier ion A follows a circle of twice the radius: $2 : 1$. This mass-dependent separation is how a mass spectrometer distinguishes isotopes.

First the speed from $qV = \tfrac{1}{2}mv^2$: $v = \sqrt{\dfrac{2qV}{m}} \approx 1.3\times10^{7}\text{ m/s}$. Then $r = \dfrac{mv}{qB} = \dfrac{(9.1\times10^{-31})(1.3\times10^{7})}{(1.6\times10^{-19})(0.010)} \approx 7.5\times10^{-3}\text{ m} = 7.5\text{ mm}$.

At the maximum radius, $r = \dfrac{mv}{qB}$ rearranges to $v = \dfrac{qBr}{m} = \dfrac{(1.6\times10^{-19})(1.0)(0.50)}{1.67\times10^{-27}} \approx 4.8\times10^{7}\text{ m/s}$. A bigger magnet and stronger field give faster, more energetic particles.

The selected speed is $v = \dfrac{E}{B}$. With $B$ fixed, increasing $E$ increases $v$: a larger electric force needs a larger magnetic force ($qvB$) to balance it, which requires a higher speed.

$r = \dfrac{mv}{qB}$. The two have the same charge magnitude and speed, so $r \propto m$. The proton is about $1836$ times more massive than the electron, so its circular path has about $1836$ times the radius.

Combine acceleration ($qV = \tfrac{1}{2}mv^2$) with the circular path ($r = \frac{mv}{qB}$) to eliminate the speed: $m = \dfrac{qB^2 r^2}{2V} = \dfrac{(1.6\times10^{-19})(0.30)^2(0.050)^2}{2(1000)} \approx 1.8\times10^{-26}\text{ kg}$ (about $11$ atomic mass units).