Lorentz force

In the CBSE Class 12 unit Moving Charges and Magnetism, the syllabus first establishes the magnetic force on a moving charge, $\vec F = q\vec v\times\vec B$, and then explicitly asks students to combine it with the familiar electric force $q\vec E$ into the single "Lorentz force" expression — this combined force is squarely in scope and is a standard short-answer and numerical topic. It sits as a bridge: it reuses electrostatics ($q\vec E$) from Class 11, reuses the freshly learned magnetic force on a moving charge, and feeds forward into force on a current-carrying conductor (where the same $q\vec v\times\vec B$ idea is summed over a whole wire of moving charges). The velocity-selector application, where the two forces are set to cancel, is the syllabus's standard worked scenario for this topic and is examined regularly; a full cyclotron treatment is not required here and is only ever quoted qualitatively elsewhere in the chapter, never derived from the Lorentz force in this article.

Picture a charge $q$ sitting inside a region carrying both a uniform electric field $\vec E$ and a uniform magnetic field $\vec B$, and moving with velocity $\vec v$ at the instant we look at it. Two independent forces act on it at once, and the total is their vector sum:

where: $q$ is the charge (in coulombs, with its own sign), $\vec E$ is the electric field (volts per metre, or equivalently newtons per coulomb), $\vec v$ is the instantaneous velocity of the charge (metres per second), $\vec B$ is the magnetic field (tesla), and $\vec F$ is the net force (newtons). Read term by term: the first piece, $q\vec E$, is exactly the electrostatic force from Class 11 — it exists for a charge at rest just as much as for a moving one, and it points along $\vec E$ (or opposite to it, if $q$ is negative). The second piece, $q\,\vec v\times\vec B$, is new to this chapter and behaves completely differently — it vanishes the instant $\vec v = 0$, and its direction is fixed by the cross product, always perpendicular to the plane containing $\vec v$ and $\vec B$, with the right-hand rule (or the corkscrew rule) giving the sense for a positive charge, and the opposite sense for a negative one. The physical reading of "perpendicular to $\vec v$" is the single most important fact about the magnetic piece: a force that is always perpendicular to velocity can never have a component along the direction of motion, so it can never speed the charge up or slow it down — it does zero work, however strong $B$ is or however long the charge stays in the field. Only the electric piece $q\vec E$ can add or remove kinetic energy from the charge. This split — "electric force can do work, magnetic force cannot" — is worth memorising as a standalone fact, independent of any numerical problem.

A key application built directly on this splitting is the velocity selector: arrange $\vec E$, $\vec B$ and $\vec v$ mutually perpendicular, and orient the fields so that the electric force $q\vec E$ and the magnetic force $q\vec v\times\vec B$ point in exactly opposite directions. Because both forces are proportional to $q$, the charge itself cancels out of the balance condition — the selector works the same way for any charge, positive or negative, light or heavy, as long as it is moving at the right speed. Only one speed satisfies the balance, and every charge moving at that speed sails straight through; every other speed leaves a net sideways force that deflects the charge out of the beam.

Board level - undeflected speed through a velocity selector

An electron moving horizontally enters a region where a uniform electric field $E = 3\times10^4\ \text{V/m}$ points vertically downward and a uniform magnetic field $B = 0.02\ \text{T}$ points into the page, both perpendicular to the electron's velocity. Find the speed at which the electron passes through undeflected.

Step 1. Recognise the setup as a velocity selector: for the electron to travel straight, the net Lorentz force must be zero, which requires the electric and magnetic forces to be equal in magnitude and opposite in direction.

Step 2. Apply the balance condition derived above: $qE = qvB$, so $v = \dfrac{E}{B}$. The electron's charge cancels out of this relation, so the result holds regardless of the sign of the moving charge, provided the fields are oriented so the two forces oppose.

Step 3. Substitute the given values: $v = \dfrac{3\times10^4}{0.02} = 1.5\times10^6\ \text{m/s}$.

Step 4. Sanity-check the answer. A speed of $1.5\times10^6\ \text{m/s}$ is about $0.5\%$ of the speed of light, safely in the non-relativistic regime assumed throughout this syllabus, so the classical Lorentz-force treatment used here is valid. The electron passes through undeflected at $v = 1.5\times10^6\ \text{m/s}$; electrons moving faster or slower are deflected off this straight path.

Competitive level - full Lorentz force from perpendicular E and B on a moving charge

A charge $q = 2\times10^{-6}\ \text{C}$ moves with velocity $v = 5\times10^4\ \text{m/s}$ along the $+x$-axis. It is in a region with $\vec E = 400\ \text{N/C}$ along $+y$ and $\vec B = 0.1\ \text{T}$ along $+z$. Find the magnitude and direction of the total Lorentz force on the charge.

Step 1. Compute the electric part first, since it needs no cross product: $F_E = qE = (2\times10^{-6})(400) = 8\times10^{-4}\ \text{N}$, directed along $+y$ (the same direction as $\vec E$, since $q$ is positive).

Step 2. Compute the magnetic part. Since $\vec v$ (along $+x$) is perpendicular to $\vec B$ (along $+z$), the angle between them is $90°$, so $F_B = qvB\sin90° = qvB = (2\times10^{-6})(5\times10^4)(0.1) = 1\times10^{-5}\ \text{N}$. Using $\hat x\times\hat z = -\hat y$, this magnetic force points along $-y$ — directly opposite the electric force.

Step 3. Combine the two forces as vectors. Because both lie along the same $y$-axis, they subtract rather than add at an angle: $F_{net} = F_E - F_B = 8\times10^{-4} - 1\times10^{-5} = 7.9\times10^{-4}\ \text{N}$, directed along $+y$ (the electric force wins since $F_E \gg F_B$ here).

Step 4. State the answer clearly, distinguishing it from the earlier velocity-selector example: here the two forces do not cancel — this charge is not moving at the special selector speed for these fields, so it experiences a net sideways-along-$y$ push and would be deflected. The Lorentz force has magnitude $7.9\times10^{-4}\ \text{N}$, directed along $+y$. Timing note: this is a two-minute problem if you keep the electric and magnetic pieces separate and only combine them at the very last step. The recurring trap is trying to add $F_E$ and $F_B$ as plain numbers before checking their directions from the cross product — always work out the direction of $\vec v\times\vec B$ first (right-hand rule), then decide whether the two forces add, subtract, or need full vector addition at an angle.

• Adding the electric and magnetic force magnitudes as plain numbers, $F = qE + qvB$, without first checking whether the two force vectors point the same way, opposite ways, or at an angle — they must be combined as vectors, using the actual geometry of $\vec E$, $\vec v$ and $\vec B$ in the problem.
• Believing the Lorentz force always does zero work on a charge — only the magnetic piece $q\vec v\times\vec B$ is guaranteed to do zero work; the electric piece $q\vec E$ routinely changes the charge's kinetic energy, and many exam questions hinge on exactly this distinction.
• Assuming the velocity-selector condition $v = E/B$ depends on the sign or size of the charge — it does not, since $q$ cancels out of $qE = qvB$; the same fields select the same speed for an electron, a proton, or an alpha particle.
• Getting the cross-product order backwards and writing $\vec B\times\vec v$ instead of $\vec v\times\vec B$ — the two differ by a sign, so a flipped order silently reverses the predicted direction of the magnetic force.
• Forgetting that for a negative charge, both the electric force ($q\vec E$) and the magnetic force ($q\,\vec v\times\vec B$) reverse direction compared with a positive charge in the identical fields — the sign of $q$ must be carried through every term, not just the electric one.

• Write the full Lorentz force expression $\vec F = q\vec E + q\,\vec v\times\vec B$ and correctly identify which term is electric and which is magnetic.
• Explain why only the electric part of the Lorentz force can do work on a moving charge, and why the magnetic part never can.
• Derive the velocity-selector balance condition $v = E/B$ from first principles, starting from the full Lorentz force expression.
• Combine electric and magnetic force vectors correctly using the right-hand rule for $\vec v\times\vec B$, including cases where they are not simply parallel or antiparallel.
• Track the sign of a negative charge through both the electric and magnetic terms consistently, rather than only through one.
• Distinguish a pure magnetic-force problem (only $\vec v\times\vec B$ present) from a full Lorentz-force problem where both $\vec E$ and $\vec B$ act together.

Ask the tutor: Want to see how the same velocity-selector balance condition feeds directly into finding a charge's mass in a simple mass-spectrometer setup — ask the AI tutor to walk through that next step and test you with a non-perpendicular field arrangement.