Motion in Combined Electric and Magnetic Fields

The starting point

• In both fields together the charge obeys the full Lorentz force $$\vec{F} = q\left[\vec{E} + \vec{v}\times\vec{B}\right]$$
• The electric part acts along $\vec{E}$ whatever the velocity — it can speed the particle up or slow it down. The magnetic part acts perpendicular to $\vec{v}$ and does no work. The motion is decided by how the two are arranged relative to $\vec{v}$.

Crossed fields: the velocity selector

• Arrange $\vec{E}$, $\vec{B}$ and $\vec{v}$ mutually perpendicular (this is what "crossed fields" means, and stating it is worth a mark).
• The electric force $qE$ and the magnetic force $qvB$ are then antiparallel. The particle goes straight through undeflected when they balance: $$qE = qvB \;\Rightarrow\; v = \dfrac{E}{B}$$
• Note what this condition does *not* contain: neither $q$ nor $m$. So out of a beam of particles of many speeds, only those with the single speed $v = E/B$ pass undeflected; all others are deflected out of the beam. That is the principle of the velocity selector, and it is the answer CBSE wants when a question asks how to "select charged particles of a particular speed".
• Reading it the other way round, to hold a particle of speed $v$ on a straight line in a field $B$, apply $$E = vB$$ with the electric force directed exactly opposite to $q\vec{v}\times\vec{B}$. For a positive charge whose magnetic force is along $-y$, the required $\vec{E}$ points along $+y$; for a negative charge, along $-y$.
• Units check: $E$ in $\text{V m}^{-1}$ divided by $B$ in T gives $\text{m s}^{-1}$.

Parallel fields, particle released from rest

• Let $\vec{E}$ and $\vec{B}$ both point along the same axis and let the charge start at rest.
• At $t=0$, $v = 0$, so the magnetic force $q(\vec{v}\times\vec{B})$ is zero. Only the electric force acts, and it accelerates the charge along the common axis.
• As soon as the charge has a velocity, that velocity is parallel (or antiparallel) to $\vec{B}$, so $\vec{v}\times\vec{B} = 0$ at every later instant too.
• The particle therefore moves in a straight line along the common field direction, accelerating continuously. It does not circle and does not spiral.

Velocity with a component along $\vec{B}$

• If, in a magnetic field, the velocity has a component along $\vec{B}$, that component is force-free and gives uniform straight-line motion, while the perpendicular component circles — the resultant is a helix with its axis along $\vec{B}$.
• Adding an electric field along $\vec{B}$ turns this into a helix of increasing (or decreasing) pitch, since the axial motion is now accelerated. For CBSE, saying "helical path" with the reason above is enough.

The standard three-mark question, worked

• A positive charge $q$ moving with speed $v$ along $-x$ enters a uniform field $B$ into the plane of the paper (along $-z$), with $x$ horizontal and $y$ vertical.
• Path: $\vec{v}\perp\vec{B}$, so the force $q(\vec{v}\times\vec{B})$ is perpendicular to $\vec{v}$ and constant in magnitude — it acts as a centripetal force and the charge moves in a circle. The circle lies in the plane perpendicular to $\vec{B}$, i.e. in the plane of the paper (the $x$–$y$ plane).
• If the velocity acquires a component parallel to $\vec{B}$, that component feels $F = qvB\sin 0^\circ = 0$ and travels in a straight line, while the perpendicular component keeps circling; the combined motion is a helix.
• To keep the charge on its original straight line, add an electric field whose force cancels the magnetic force. Here the magnetic force on the positive charge is along $-y$, so the electric force must be along $+y$, and since $q$ is positive, $\vec{E}$ itself must be along $+y$. Its magnitude follows from $qE = qvB$, i.e. $E = vB$.

Diagram in words

• Draw a horizontal beam entering from the left. Above it draw the plates of a parallel-plate capacitor, giving $\vec{E}$ vertically downwards between them; mark $\vec{B}$ as crosses (into the page) in the same region.
• Show three trajectories leaving the region: one straight through (the selected speed $v = E/B$), one deflected upwards (slower — magnetic force wins) and one deflected downwards (faster). Label the undeflected one $v = E/B$.

Results to carry away

• Undeflected motion in crossed fields requires $\vec{E}$, $\vec{B}$, $\vec{v}$ mutually perpendicular and $qE = qvB$, giving $v = E/B$, independent of $q$ and $m$.
• To balance a magnetic deflection, $E = vB$ with $\vec{E}$ opposing $q\vec{v}\times\vec{B}$.
• Parallel $\vec{E}$ and $\vec{B}$ with the charge starting from rest: straight-line accelerated motion.
• Velocity partly along $\vec{B}$: helical path.