In short

A point charge q moving with velocity \vec{v} through a magnetic field \vec{B} experiences a force

\boxed{\;\vec{F}_\text{mag} \;=\; q\,\vec{v}\times\vec{B}\;}

Magnitude. F = qvB\sin\theta, where \theta is the angle between \vec{v} and \vec{B}. The force is zero when the charge moves parallel (or anti-parallel) to the field, and maximum (F = qvB) when it moves perpendicular to it.

Direction. By the right-hand rule for the cross product: point your fingers along \vec{v}, curl them toward \vec{B}, thumb gives \vec{v}\times\vec{B}. For a positive charge \vec{F} is along the thumb; for a negative charge the force is opposite.

When an electric field \vec{E} is also present, the total electromagnetic force on the charge is the Lorentz force:

\boxed{\;\vec{F} \;=\; q\vec{E} \;+\; q\,\vec{v}\times\vec{B}\;}

Five properties to remember.

  1. \vec{F}_\text{mag}\perp\vec{v} always. So the magnetic force does no work — it can bend a trajectory but cannot change speed.
  2. \vec{F}_\text{mag}\perp\vec{B} too. Field lines are never along the force.
  3. The force depends on velocity: a charge at rest in a pure magnetic field feels nothing.
  4. Reversing \vec{v} or reversing the sign of q reverses the force; reversing both leaves it unchanged.
  5. The unit of \vec{B}, the tesla, is defined so that a 1 C charge moving at 1 m/s perpendicular to a 1 T field feels 1 N of force: 1\ \text{T} = 1\ \text{N}/(\text{C}\cdot\text{m/s}).

The Lorentz force is the single rule that governs every charged particle in every electromagnetic field — auroras above Leh, the proton beam at BARC, the plasma in the Sun, the electrons drifting in the copper wire behind your phone charger.

A solar storm launches a stream of protons across 150 million kilometres of vacuum. They arrive at Earth travelling at about 500 km/s, aimed roughly at the magnetic north pole. The magnetic field of the Earth — a curving dipole field, strongest near the poles — grabs them by the scruff of the neck and hurls them sideways, spiralling along the field lines toward the upper atmosphere. When they crash into oxygen and nitrogen atoms 100 km above Leh, they paint the sky green and magenta. That is the aurora you have seen photographs of from Ladakh in winter.

None of this happens without a single vector equation. It is the equation every charged particle — whether it is a proton in a coronal mass ejection, an electron drifting through copper, or a lithium ion in a mass spectrometer — has to obey.

Building the equation from what you know about currents

You have already met the force on a current-carrying conductor in a magnetic field: a straight wire of length L carrying current I feels a force \vec{F} = I\vec{L}\times\vec{B}. A current is nothing but charges moving through the wire — so the force on the wire must be the sum of forces on individual charges. Extract one and see what you get.

Take a wire of cross-sectional area A carrying current I. Let n be the number density of mobile charges, each with charge q, each moving with drift velocity \vec{v}_d. In time dt every charge advances a distance v_d\,dt; all the charges in a cylinder of length v_d\,dt cross any given cross-section. The current is therefore

I \;=\; \frac{\text{charge crossing}}{dt} \;=\; \frac{(nAv_d\,dt)\,q}{dt} \;=\; nqAv_d.

Why: current is charge per unit time. The volume of the cylinder of drift is Av_d\,dt; it contains nAv_d\,dt charges; each carries charge q.

The wire's force law \vec{F} = I\vec{L}\times\vec{B}, for a segment of length L pointing along the current direction \hat{v}_d, becomes

\vec{F}_\text{wire} \;=\; I L\,\hat{v}_d\times\vec{B} \;=\; (nqAv_d)\,L\,\hat{v}_d\times\vec{B} \;=\; (nAL)\,q\,\vec{v}_d\times\vec{B}.

Why: substitute I = nqAv_d and write L\hat{v}_d = L\hat{v}_d; then group v_d\hat{v}_d = \vec{v}_d. The factor nAL is simply the total number of mobile charges in the wire segment.

There are N = nAL charges in that segment. Dividing the total force by N gives the force on one charge:

\vec{F}_\text{single charge} \;=\; \frac{\vec{F}_\text{wire}}{N} \;=\; q\,\vec{v}_d\times\vec{B}.

Why: the total force is the sum of forces on identical charges, each moving with the same drift velocity. Dividing by the count isolates one.

Renaming \vec{v}_d \to \vec{v} (the argument works for any velocity, not just drift):

\boxed{\;\vec{F}_\text{mag} \;=\; q\,\vec{v}\times\vec{B}\;}

This is the magnetic force on a moving point charge. It is the microscopic law from which the macroscopic \vec{F} = I\vec{L}\times\vec{B} is built, not the other way around.

Reading the cross product — direction and magnitude

The cross product \vec{v}\times\vec{B} is a vector perpendicular to both \vec{v} and \vec{B}, with magnitude vB\sin\theta where \theta is the angle between them.

Magnitude.

F \;=\; qvB\sin\theta.

Direction. Use the right-hand rule for the cross product:

  1. Point the fingers of your right hand along \vec{v}.
  2. Curl them through the angle \theta toward \vec{B}.
  3. Your thumb now points along \vec{v}\times\vec{B}.

If q > 0 the force \vec{F} = q\,\vec{v}\times\vec{B} is along the thumb. If q < 0 (an electron), the force is opposite to the thumb — reverse it.

Right-hand rule for the magnetic force on a positive charge A positive charge moves to the right with velocity v. The magnetic field points into the page, shown as crossed circles. The force on the charge points upward, perpendicular to both v and B. Reversing the sign of q flips the force direction. B into page ×××××××××× ×××××××××× ×××××××××× ×××××××××× ×××××××××× +q v F = qv × B
A positive charge moves rightward through a field that points into the page ($\vec{B}$ shown as $\times$ symbols — tails of arrows going away from you). Fingers along $\vec{v}$, curl into the page along $\vec{B}$, thumb points up. For a negative charge the force would point down.

A useful mnemonic: "v cross B" is the direction the palm pushes when you orient your right hand so fingers point along \vec{v} and can fold along \vec{B}. Either the finger-curl or the palm-push version works; pick whichever sticks.

Why the magnetic force does no work

This is the single most surprising property of the magnetic force. No matter how hard a magnetic field pushes a charge, it cannot change the charge's kinetic energy. Here is why, derived cleanly.

Step 1. The rate at which any force does work on a moving body is

P \;=\; \vec{F}\cdot\vec{v}.

Why: work is force times displacement along the direction of motion. Power — work per unit time — is force dotted with velocity.

Step 2. For the magnetic force, \vec{F} = q\,\vec{v}\times\vec{B}. Substitute:

P \;=\; (q\,\vec{v}\times\vec{B})\cdot\vec{v}.

Step 3. Any vector crossed with itself or containing itself in a dot-plus-cross is zero. Specifically, \vec{v}\times\vec{B} is perpendicular to \vec{v} by construction (the cross product of \vec{v} with anything is perpendicular to \vec{v}):

(\vec{v}\times\vec{B})\cdot\vec{v} \;=\; 0.

Why: the cross product \vec{v}\times\vec{B} is defined to be perpendicular to \vec{v}. A vector dotted with a perpendicular vector gives zero. This is geometry, not physics — it is a property of the cross product.

Step 4. Therefore

P_\text{mag} \;=\; q\,(\vec{v}\times\vec{B})\cdot\vec{v} \;=\; 0.

Why: the magnetic force is always perpendicular to the velocity. A force perpendicular to motion does no work — exactly like gravity on a ball moving horizontally, or the tension in the string of a whirling stone.

The consequence is profound: a magnetic field can change the direction of a charged particle but never its speed. A proton entering a magnetic field at 500 km/s leaves at exactly 500 km/s, no matter how complicated the field was inside. The particle's kinetic energy \tfrac{1}{2}mv^2 is conserved.

This is why cyclotrons (chapter 188) need an electric field to accelerate particles — a pure magnetic field could bend them in circles forever without ever speeding them up. The magnetic field is the bouncer, not the engine.

What changes when the field is non-uniform? The magnitude of \vec{v} is still unchanged at every instant, because at every instant \vec{F}\perp\vec{v}. Speed is conserved even through arbitrary \vec{B}(\vec{r}). What does change is the particle's direction of motion — curving trajectories are the rule, not the exception.

The full Lorentz force — electric and magnetic together

In the real world charges feel both electric and magnetic forces simultaneously. An electron in a cathode-ray tube (back when television sets had one), a proton in the ISRO plasma experiment Aditya-L1, a lithium ion in a mass spectrometer at BARC Mumbai — all of these move under the combined force

\boxed{\;\vec{F} \;=\; q\vec{E} \;+\; q\,\vec{v}\times\vec{B}\;}

This is the Lorentz force. It is the cleanest possible statement of the force on a charged particle in classical electromagnetism — one line, two terms, nothing else needed. Maxwell's four equations tell you what \vec{E} and \vec{B} look like; the Lorentz force tells you what those fields do to a charge.

A few sanity checks on the expression:

The electric term is parallel (or anti-parallel) to \vec{E} and does not depend on velocity. The magnetic term is perpendicular to \vec{v} and does depend on it. These two terms have completely different characters — which is why separating them is useful.

Work–energy theorem for the Lorentz force

Combine the power of each term. The electric part contributes q\vec{E}\cdot\vec{v}; the magnetic part contributes zero. So

P_\text{Lorentz} \;=\; q\vec{E}\cdot\vec{v}.

Only the electric field can change a charge's kinetic energy. This is the rule that lets you separate "electrostatic energy" from "magnetic steering" in every device you will meet — the electric field is the accelerator, the magnetic field is the steering wheel.

Worked examples

Example 1: A proton in a coronal mass ejection above Leh

A proton from a solar wind stream travels at v = 5\times 10^5 m/s due north above Leh (magnetic latitude ≈ 24° N; take the local geomagnetic field to be |\vec{B}| = 4.5\times 10^{-5} T, pointing mostly downward into the Earth at this latitude — assume straight down for a clean estimate). Compute the magnitude and direction of the magnetic force on the proton at the instant it enters this region.

Proton moving north through a downward geomagnetic field A proton moves north (into the page) while the Earth's magnetic field points downward. The cross product v times B points to the east (right side of the page), which is where a positive charge is pushed by the Lorentz force. viewed from above; north into page +p v (north, into page) B (down) F (east) E W
Looking down at Leh from directly above. The proton moves north (into the page). The geomagnetic field points down. The Lorentz force $q\vec{v}\times\vec{B}$ pushes the proton eastward.

Step 1. Identify the inputs.

q = +1.6\times 10^{-19} C, v = 5\times 10^5 m/s, B = 4.5\times 10^{-5} T, angle between \vec{v} and \vec{B} is \theta = 90^\circ (north is horizontal; field is vertical).

Why: the proton's charge is elementary. The angle is 90° because a horizontal velocity is perpendicular to a vertical field — maximum magnitude of the cross product.

Step 2. Compute the magnitude.

F \;=\; qvB\sin\theta \;=\; (1.6\times 10^{-19})(5\times 10^5)(4.5\times 10^{-5})(1)
F \;=\; 1.6\times 5\times 4.5\times 10^{-19+5-5}\ \text{N} \;=\; 36\times 10^{-19}\ \text{N} \;=\; 3.6\times 10^{-18}\ \text{N}.

Why: straight substitution into qvB\sin\theta with \sin 90^\circ = 1. Collect the exponents: -19 + 5 - 5 = -19.

Step 3. Find the direction.

Fingers of the right hand point north (along \vec{v}). Curl them downward toward \vec{B}. The thumb points east. Since the proton is positive, \vec{F} is along the thumb — east.

Why: this is the geometric part of the cross product. The sign of the charge is positive, so no reversal is needed.

Step 4. Sanity check with acceleration.

a \;=\; \frac{F}{m_p} \;=\; \frac{3.6\times 10^{-18}}{1.67\times 10^{-27}}\ \text{m/s}^2 \;\approx\; 2.2\times 10^{9}\ \text{m/s}^2.

Why: at two billion metres per second squared, the proton bends noticeably within a tiny fraction of a second — which is exactly how auroral protons get guided down the field lines into the upper atmosphere.

Result. The proton feels a force of magnitude 3.6\times 10^{-18} N directed eastward. This tiny-looking force, acting on a particle with tiny mass, produces enormous acceleration — enough to curl the proton into a tight spiral around the field line as it descends toward 100 km altitude and lights up the aurora.

Example 2: A velocity selector at BARC — crossed E and B fields

A lithium ion (q = +1.6\times 10^{-19} C, mass \sim 7 u, irrelevant here) is sent into a region of crossed fields used as a velocity selector in the mass-spectrometer line at BARC Mumbai. The electric field is \vec{E} = 2.0\times 10^4\ \hat{j} N/C (pointing up, in the +y direction). The magnetic field is \vec{B} = 0.050\ \hat{k} T (pointing out of the page, in the +z direction). The ion enters along +x with speed v_x. Find the speed at which the ion passes through undeflected.

Velocity selector with crossed electric and magnetic fields An ion moves along the positive x direction through a region where E points up and B points out of the page. The electric force on a positive charge is upward. The magnetic force is downward. At one particular speed the two forces cancel. B out of page (shown ⊙) ⊙⊙⊙⊙⊙⊙ ⊙⊙⊙⊙⊙⊙ + v qE (up) qv×B (down)
Crossed fields: $\vec{E}$ pushes the positive ion up; $\vec{v}\times\vec{B}$ pushes it down. For one particular $v_x$ the two forces are equal and opposite, and the ion sails through undeflected.

Step 1. Write each force on the ion in vector form.

\vec{F}_E \;=\; q\vec{E} \;=\; qE\,\hat{j}.
\vec{F}_B \;=\; q\vec{v}\times\vec{B} \;=\; q(v_x\,\hat{i})\times(B\,\hat{k}) \;=\; qv_xB\,(\hat{i}\times\hat{k}).

Why: the ion's velocity is purely along \hat{i} and the field is purely along \hat{k}, so the cross product collapses to a single term.

Step 2. Evaluate the unit-vector cross product. Using the right-hand rule (or the cyclic rule \hat{i}\times\hat{j}=\hat{k}, \hat{j}\times\hat{k}=\hat{i}, \hat{k}\times\hat{i}=\hat{j}, and the antisymmetry \hat{i}\times\hat{k} = -\hat{j}):

\hat{i}\times\hat{k} \;=\; -\hat{j}.

Why: the cyclic order is \hat{i}\to\hat{j}\to\hat{k}\to\hat{i}. Going against the cycle (\hat{i}\to\hat{k}) introduces a minus sign.

Therefore

\vec{F}_B \;=\; -qv_xB\,\hat{j}.

Step 3. Undeflected means zero net force in the \hat{j} direction: \vec{F}_E + \vec{F}_B = 0.

qE\,\hat{j} + (-qv_xB)\,\hat{j} \;=\; 0
qE \;=\; qv_xB
\boxed{\;v_x \;=\; \frac{E}{B}\;}.

Why: the q cancels — so the selector picks out a single speed independent of the charge or mass of the particle. Any ion travelling at exactly v_x = E/B sails through; anything faster gets deflected up, anything slower gets deflected down.

Step 4. Plug in numbers.

v_x \;=\; \frac{2.0\times 10^4}{0.050} \;=\; 4.0\times 10^{5}\ \text{m/s}.

Why: 2\times 10^4 / 5\times 10^{-2} = (2/5)\times 10^{4-(-2)} = 0.4\times 10^6 = 4\times 10^5. A lithium ion entering at this speed passes through the selector undeflected and enters the next stage of the spectrometer.

Step 5. Sanity-check a faster ion.

At v_x = 5\times 10^5 m/s the magnetic force is qv_xB = q\times 2.5\times 10^4 N/C-equivalent — larger than the electric force qE = q\times 2.0\times 10^4. The net force is downward, deflecting the ion out of the beam. Slower ions (v_x < 4\times 10^5 m/s) are deflected upward for the mirror-image reason.

Result. Only ions with v_x = 4.0\times 10^5 m/s survive. This is the principle behind every velocity selector, in labs from BARC and IUAC Delhi to CERN: two crossed fields, the electric force up, the magnetic force down, and the single speed v = E/B walks a knife edge between them.

A sim you can watch — an electron in crossed E and B fields

The next figure shows what happens when a charged particle moves through crossed electric and magnetic fields, starting from rest. The electric field accelerates it; the magnetic force bends the accelerating velocity; the combined motion traces a famous curve called a cycloid — the same curve drawn by a point on the rim of a rolling wheel. Watch how the particle drifts in the +x direction overall, even though neither \vec{E} nor \vec{B} points along +x.

Animated: cycloid trajectory from crossed E and B fields A positive charge released from rest in a region where E points in the y direction and B points out of the page. The particle traces a cycloid — it arches, returns to rest, arches again, drifting in the x direction. The drift speed is E over B. x (m) y (m) E = 1 V/m (↑), B = 1 T (out of page), drift = E/B = 1 m/s (→)
A positive charge released from rest with $\vec{E}=E\hat{j}$ and $\vec{B}=B\hat{k}$, in units where $qE/m = 1$ and $qB/m = 1$. The trajectory is $x(t) = t - \sin t$, $y(t) = 1 - \cos t$ — a textbook cycloid. The particle arches upward, returns to $y=0$ momentarily at rest, then arches again, drifting rightward at average speed $E/B$.

The drift speed E/B is exactly the speed the velocity selector (Example 2) was tuned to. That is not a coincidence — the selector singled out the one speed at which the electric and magnetic forces cancel, which is precisely the guiding-centre drift velocity of an otherwise-circular magnetic orbit displaced by the electric field. This drift is how charged particles migrate across magnetic field lines in plasmas — in the solar corona, in tokamak reactors, in the Van Allen belts around Earth.

Common confusions

If you came here to learn the Lorentz force law, apply the right-hand rule, and see why magnetic force does no work, you have what you need. What follows is for readers who want the deeper structure — why this single equation contains the whole story of electromagnetism, and how the E/B drift generalises.

The Lorentz force in component form

In Cartesian components, \vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k} and \vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}. The cross product expands as a determinant:

\vec{v}\times\vec{B} \;=\; \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ v_x & v_y & v_z \\ B_x & B_y & B_z\end{vmatrix}
\;=\; (v_yB_z - v_zB_y)\hat{i} \;+\; (v_zB_x - v_xB_z)\hat{j} \;+\; (v_xB_y - v_yB_x)\hat{k}.

The three components of the equation of motion m\,d\vec{v}/dt = q\vec{E} + q\vec{v}\times\vec{B} are then

m\,\dot{v}_x = qE_x + q(v_yB_z - v_zB_y),
m\,\dot{v}_y = qE_y + q(v_zB_x - v_xB_z),
m\,\dot{v}_z = qE_z + q(v_xB_y - v_yB_x).

This is a system of three coupled first-order ODEs. For a uniform \vec{B} = B\hat{k} and uniform \vec{E} = E\hat{j}, the z-equation decouples: m\,\dot{v}_z = 0, so v_z is constant. The x- and y-equations couple through B:

m\,\dot{v}_x = qBv_y,
m\,\dot{v}_y = qE - qBv_x.

Define \omega_c \equiv qB/m — the cyclotron frequency. Define the drift velocity v_d \equiv E/B. Then \dot{v}_x = \omega_c v_y and \dot{v}_y = \omega_c(v_d - v_x). Let u \equiv v_x - v_d; then \dot{u} = \dot{v}_x = \omega_c v_y and \dot{v}_y = -\omega_c u.

Differentiating once more: \ddot{u} = \omega_c\dot{v}_y = -\omega_c^2 u. The solution is simple harmonic motion in u:

u(t) = A\cos(\omega_c t + \phi), \qquad v_y(t) = -A\sin(\omega_c t + \phi).

Going back to v_x = u + v_d, you see explicitly that the motion is circular motion at frequency \omega_c superposed on a uniform drift at speed v_d = E/B. If the particle starts at rest (A = v_d, \phi = 0), you recover the cycloid from the earlier animation: v_x(t) = v_d(1 - \cos\omega_c t), v_y(t) = -v_d\sin\omega_c t, which integrates to x(t) = v_d t - (v_d/\omega_c)\sin\omega_c t and y(t) = (v_d/\omega_c)(\cos\omega_c t - 1).

Why this matters: the same algebra that gives you the velocity selector (Example 2) gives you the cyclotron frequency, the cycloid trajectory, and the general \vec{E}\times\vec{B}/B^2 drift. Three different-looking results, one coupled pair of ODEs.

The general \vec{E}\times\vec{B}/B^2 drift

The drift v_d = E/B is a special case of a vector result. For arbitrary constant \vec{E} and \vec{B} (with \vec{E}\perp\vec{B}), the guiding-centre drift velocity is

\vec{v}_d \;=\; \frac{\vec{E}\times\vec{B}}{B^2}.

This drift has two remarkable properties:

  1. It is the same for all charges, regardless of sign or magnitude. Positive ions and electrons drift together, carrying no net current.
  2. It is the velocity at which the electric and magnetic forces cancel. In the frame moving with \vec{v}_d, only the magnetic force remains, and the particle executes pure cyclotron motion.

The second property is the clue to why the velocity selector works: v_d = E/B is the velocity at which a particle going along +x (perpendicular to both \vec{E} and \vec{B}) feels zero net Lorentz force.

Why \vec{F} = q\vec{E} + q\vec{v}\times\vec{B} is forced by relativity

The most remarkable fact about the Lorentz force law is that you can derive it from Coulomb's law plus special relativity. An electric charge in one frame has only electric field around it; in a frame where the charge is moving, the field looks partially magnetic. The transformation rules for \vec{E} and \vec{B} between frames are uniquely determined by the Lorentz transformations of space and time, and they force the force law to take exactly the form \vec{F} = q\vec{E} + q\vec{v}\times\vec{B}.

Said differently: magnetism is what electricity looks like when you change reference frames. The factor qv\times B is not an independent force — it is the velocity-dependent piece of the electric force as viewed from a different frame. This is why the magnetic constant \mu_0 is related to the electric constant \varepsilon_0 by c^2 = 1/(\mu_0\varepsilon_0): the speed of light ties them together, because relativity ties them together.

You will not need this for JEE, but it is the deepest statement of what the Lorentz force is: the unique force law consistent with Coulomb's law and the symmetry between inertial frames.

Aditya-L1 and ISRO's space-plasma measurements

ISRO's Aditya-L1 satellite, parked at the Sun–Earth L1 Lagrange point 1.5 million km from Earth, measures the solar wind's plasma properties in real time. Its payload includes a suite of detectors that exploit exactly the Lorentz force: incoming charged particles (electrons at keV energies, protons at hundreds of keV, heavier ions at MeV) enter a region of known \vec{E} and \vec{B}, get deflected by angles that depend on their mass, charge, and energy, and land on position-sensitive detectors. From the landing positions, the full particle spectrum of the solar wind can be reconstructed.

Every such measurement — whether in Aditya-L1, in BARC's tandem accelerator, in the Indus-2 synchrotron at RRCAT Indore, or in the upcoming Indian Neutrino Observatory's charged-particle veto — is a direct application of \vec{F} = q\vec{E} + q\vec{v}\times\vec{B}. The single line of vector algebra is what makes the observations possible.

Where this leads next