Motion in Combined Electric and Magnetic Fields

For an undeflected charged particle (velocity selector principle), the Lorentz force must be zero: $q(\vec{E} + \vec{v} \times \vec{B}) = 0 \implies \vec{E} = -(\vec{v} \times \vec{B})$. Thus $\vec{E}$ must be perpendicular to $\vec{B}$ and $\vec{v}$. The velocity is $v = \frac{c}{100} = \frac{3 \times 10^8}{100} = 3 \times 10^6\text{ m/s}$. The required electric field magnitude is $E = vB = (3 \times 10^6\text{ m/s})(9 \times 10^{-4}\text{ T}) = 27 \times 10^2\text{ V m}^{-1}$.

Since the electron's velocity is along the direction of $\vec B$, the magnetic force ($q\vec v\times\vec B$) on it is zero. The electric force on the (negatively charged) electron, moving along the direction of $\vec E$, decelerates it, so its speed will decrease.

Radius of the path in a magnetic field: $r=\frac{p}{qB}$ For the same momentum and field, $r\propto\frac{1}{q}$. $\frac{r_H}{r_\alpha}=\frac{q_\alpha}{q_H}=\frac{2e}{e}=\frac{2}{1}$ $r_H:r_\alpha=2:1$

$R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$
$R_\alpha = R_p$: $\frac{2(4m_p)K_\alpha}{(2e)^2B^2} = \frac{2m_pK_p}{e^2B^2}$
$K_\alpha = K_p = 1$ MeV

$R = \frac{mv}{qB} = \frac{\sqrt{2mE}}{qB}$
For equal radii: $\frac{\sqrt{2m_pE_p}}{q_pB} = \frac{\sqrt{2m_\alpha E_\alpha}}{q_\alpha B}$
$E_\alpha = \left(\frac{q_\alpha}{q_p}\right)^2\left(\frac{m_p}{m_\alpha}\right)E_p$, with $\frac{m_p}{m_\alpha} = \frac{1}{4}$ and $\frac{q_\alpha}{q_p} = 2$
$E_\alpha = (2)^2\times\frac{1}{4}\times E_p = E_p$
$E_\alpha = 1$ MeV

Magnetic force: $\vec{F}_m = q(\vec{v}\times\vec{B})$. The velocity is parallel to $\vec{B}$, so $\vec{v}\times\vec{B} = 0$ and there is no magnetic force; the electron does not turn.
Electric force: $\vec{F}_e = q\vec{E} = -e\vec{E}$. Since the electron has negative charge, this force is opposite to $\vec{E}$.
The velocity is along $\vec{E}$, so the electric force is opposite to the velocity.
The electron is therefore retarded along its line of motion.
Hence its speed will decrease.

Force on the particle: $\vec{F} = q\vec{E} + q(\vec{v}\times\vec{B})$. The particle keeps its direction if there is no net force perpendicular to its velocity.
(b): if $\vec{B}$ is along $\vec{v}$, the magnetic force $q(\vec{v}\times\vec{B})$ is zero; with $\vec{E}$ also along $\vec{v}$ the electric force only changes the speed, not the direction. So (b) is possible.
(c): if $\vec{E}$, $\vec{B}$ and $\vec{v}$ are mutually perpendicular, the electric force $q\vec{E}$ and the magnetic force $q(\vec{v}\times\vec{B})$ act along the same line and can be made equal and opposite ($v = \frac{E}{B}$, the velocity selector), so the particle goes undeflected. So (c) is possible.
(a): with $\vec{B}$ perpendicular to $\vec{v}$ there is an unbalanced magnetic force perpendicular to the velocity (the electric force along $\vec{v}$ cannot cancel it), so the path bends.
(d): with $\vec{E}$ perpendicular to $\vec{v}$ and no magnetic force, the electric force deflects the particle.
Hence the possible pair is (b) and (c).

The beam is undeflected when the electric and magnetic forces balance: $eE = evB$, so $v = \frac{E}{B}$
The electrons gain their speed from the accelerating potential V: $eV = \frac{1}{2}mv^2$
$\frac{e}{m} = \frac{v^2}{2V}$
Substituting $v = \frac{E}{B}$: $\frac{e}{m} = \frac{E^2}{2VB^2}$

The magnetic force on a moving charge is $\vec{F} = Q(\vec{v} \times \vec{B})$.
This force is always perpendicular to both the velocity $\vec{v}$ and the magnetic field $\vec{B}$.
Power delivered by the force: $P = \vec{F}\cdot\vec{v} = 0$, because $\vec{F} \perp \vec{v}$.
Equivalently, the work done in a small displacement is $dW = \vec{F}\cdot d\vec{r} = 0$, because $\vec{F} \perp d\vec{r}$.
A force that does no work cannot change the kinetic energy or the speed of the particle; it only changes the direction of motion (the particle moves on a circular path).
Hence after 3 seconds (or after any time) the kinetic energy is still $T$.

The beam passes undeflected, so the electric force balances the magnetic force: $eE = evB$.
This is possible only when the velocity of the electrons is perpendicular to both the electric and the magnetic field (velocity selector), with $v = \dfrac{E}{B}$.
When the electric field is switched off, only the magnetic force $F = evB$ acts on the electrons.
This force is always perpendicular to the velocity, so it does no work and the speed stays constant; it only changes the direction of motion.
A force of constant magnitude that is always perpendicular to the velocity acts as a centripetal force: $evB = \dfrac{mv^2}{r}$, giving $r = \dfrac{mv}{eB}$.
Since the electrons travel in a magnetic field perpendicular to their velocity, they move in a circular orbit.

In a mass spectrometer, when the ions are accelerated through a potential $V$: $\dfrac{1}{2}mv^2 = qV$ ...(i)
The magnetic field curves the path of the ions into a semicircular orbit, the magnetic force providing the centripetal force:
$Bqv = \dfrac{mv^2}{R} \Rightarrow v = \dfrac{BqR}{m}$ ...(ii)
Substituting (ii) in (i): $\dfrac{1}{2}m\left(\dfrac{BqR}{m}\right)^2 = qV$
$\dfrac{B^2 q^2 R^2}{2m} = qV$
$\dfrac{q}{m} = \dfrac{2V}{B^2 R^2}$
Since $V$ and $B$ are constants, $\dfrac{q}{m} \propto \dfrac{1}{R^2}$

The magnetic force provides the centripetal force: $qvB = \dfrac{mv^2}{R}$
$\Rightarrow R = \dfrac{mv}{qB}$
Time period: $T = \dfrac{2\pi R}{v} = \dfrac{2\pi}{v} \times \dfrac{mv}{qB}$
$T = \dfrac{2\pi m}{qB}$
The speed $v$ cancels out: a faster particle moves in a proportionally bigger circle, so the time for one revolution is the same.
The time period depends only on the mass, the charge and the magnetic field, so it is independent of both $R$ and $v$.

Only the velocity component perpendicular to the field causes circular motion: $v_\perp = v\sin 30^\circ$.

$r = \dfrac{mv_\perp}{qB} = \dfrac{m}{q} \cdot \dfrac{v\sin 30^\circ}{B}$
$r = \dfrac{3 \times 10^5 \times 0.5}{10^8 \times 0.3} = \dfrac{3 \times 10^5}{10^8 \times 0.3 \times 2}$
$r = 0.5 \times 10^{-2}$ m $= 0.5$ cm
(The particle moves along a helix of this radius.)