Motion of charged particle in magnetic field

Fire an electron into a magnetic field at right angles to it and something strange happens: the field pushes on the electron constantly, yet the electron never speeds up or slows down. It simply turns — forever. That single fact, that a magnetic force can steer a charge without ever touching its speed, is the seed of some of the biggest machines in physics: the particle accelerators that smash protons together at CERN, the magnetic bottles that trap plasma in fusion reactors, and the old cathode-ray tubes that once bent electron beams to paint a television picture line by line. The reason is geometry, not new physics. The magnetic force on a moving charge, $\vec F = q\vec v\times\vec B$ , is built from a cross product, and a cross product is always perpendicular to both vectors that make it. A force that is permanently perpendicular to velocity can never do work — it can only bend the path, never stretch or shrink the speed along it. Perpendicular force of constant size acting on a particle of constant speed is precisely the recipe for uniform circular motion, the same mechanics you met when a stone was whirled on a string, except now the "string" is invisible and made of field lines. This article takes the magnetic force law you already know and turns the crank: equate it to the centripetal force requirement, solve for the radius of the circle, and read off the time for one full loop. Along the way a genuinely surprising result falls out — that time does not depend on how fast the particle is going. We will also see what happens when the charge does not enter exactly perpendicular to the field, and work two full numbers-in numbers-out problems, one for an electron and one for a proton, extending the Pass-1 results with a fresh look at frequency and the helical case.

Key takeaways

A charge $q$ moving perpendicular to a uniform field $\vec B$ feels a force $F = qvB$ that is always perpendicular to its velocity, so the force does no work and the speed $v$ never changes.
A constant-magnitude force always perpendicular to velocity is exactly the condition for uniform circular motion — the magnetic field bends the path into a circle.
Equating the magnetic force to the centripetal requirement $qvB = mv^2/r$ gives the radius $r = \dfrac{mv}{qB}$ .
The period $T = \dfrac{2\pi m}{qB}$ and cyclotron frequency $f = \dfrac{qB}{2\pi m}$ depend only on the charge-to-mass ratio and the field — never on the particle's speed or radius.
If the velocity has a component along $\vec B$ , that component is untouched by the force and the particle traces a helix — a circle in the plane perpendicular to $\vec B$ combined with steady straight-line drift along $\vec B$ .
Positive and negative charges circulate in opposite senses in the same field, because the force $q\vec v\times\vec B$ flips direction when the sign of $q$ flips.
Key idea: because the magnetic force $q\vec v\times\vec B$ is always perpendicular to velocity it does no work, so a charge entering a uniform field at right angles moves on a circle of radius $r = mv/(qB)$ whose period $T = 2\pi m/(qB)$ is independent of speed.

The core insight

A charged particle crossing a uniform magnetic field perpendicular to it moves on a circle because the magnetic force, always perpendicular to velocity, supplies exactly the centripetal force needed, giving radius $r = mv/(qB)$ and a period $T = 2\pi m/(qB)$ that depends only on charge, mass and field — never on speed.

Where it fits in the syllabus

This topic sits inside the CBSE Class 12 unit Moving Charges and Magnetism, immediately after the magnetic force law $\vec F = q\vec v\times\vec B$ and the force on a current-carrying conductor. The board syllabus expects a student to derive the radius, period and frequency of circular motion for a charge entering a uniform field perpendicular to it, using nothing beyond the centripetal-force idea from mechanics. The circular-motion result also quietly underlies how a cyclotron confines particles, but the CBSE 2025-26 prescription keeps the cyclotron itself at most a one-line qualitative mention — its full quantitative treatment as a device is explicitly out of scope here, so this article stays with the algebra of $r$ , $T$ and $f$ and does not build an accelerator around them.

The physical picture

Picture a charge $q$ of mass $m$ shot into a region of uniform field $\vec B$ with its velocity $\vec v$ exactly perpendicular to $\vec B$ . At every instant the magnetic force has magnitude $F = qvB$ and points perpendicular to $\vec v$ — by the geometry of the cross product it can only ever push the particle sideways, never forward or backward along its own direction of travel. Because a force perpendicular to displacement does zero work, the particle's kinetic energy, and hence its speed $v$ , stays fixed for the entire motion; only the direction keeps changing, smoothly and continuously, which is the defining signature of uniform circular motion. To find the size of that circle, set the magnetic force equal to the centripetal force a body of mass $m$ needs to move at constant speed $v$ on a circle of radius $r$ :

qvB = \dfrac{mv^2}{r}

where: $q$ is the magnitude of the charge (coulombs), $v$ is its constant speed (m/s), $B$ is the magnetic field strength (tesla), $m$ is the particle's mass (kg), and $r$ is the radius of the circular path it settles into (m). The physical reading is simple bookkeeping: the left side is how hard the field pushes sideways, the right side is how hard a circle of radius $r$ demands to be pushed to keep a mass $m$ turning at speed $v$ — setting them equal fixes the one radius at which the two match.

r = \dfrac{mv}{qB}

A heavier or faster particle needs a bigger circle to turn the same corner; a stronger field or a bigger charge can bend it into a tighter one. From this the period follows from circumference over speed, $T = 2\pi r/v$ , which simplifies neatly because the $v$ cancels:

T = \dfrac{2\pi m}{qB}, \qquad f = \dfrac{1}{T} = \dfrac{qB}{2\pi m}, \qquad \omega = 2\pi f = \dfrac{qB}{m}

where: $T$ is the time for one complete loop (s), $f$ is the cyclotron frequency (Hz), and $\omega$ is the angular frequency (rad/s). The physical reading of the cancelled $v$ is the whole point of this topic: however fast the particle moves, it simply traces a proportionally larger circle at that higher speed, and the two effects cancel exactly, so the time per lap stays the same for every particle sharing the same charge-to-mass ratio $q/m$ in the same field.

Figure

A charged particle entering a uniform magnetic field region perpendicular to the field lines and curving into a circular arc, with the velocity vector, the magnetic force vector pointing toward the centre, and the radius marked

The physical picture continued: the helical case

Real beams rarely enter a field exactly at right angles. If $\vec v$ has a component along $\vec B$ as well as a component perpendicular to it, split it into the two pieces: $v_\parallel$ , parallel to the field, and $v_\perp$ , perpendicular to it. The parallel piece feels no force at all, because $\vec v_\parallel \times \vec B = 0$ when the two vectors point along the same line — so that component drifts on at constant velocity, untouched. Only the perpendicular piece $v_\perp$ feels the sideways push and circles as before, with radius $r = mv_\perp/(qB)$ . Combine steady drift along the field with circular motion around it and the path traced is a helix, a corkscrew spiralling forward along $\vec B$ — the shape that guides charged particles along magnetic field lines in the aurora and in laboratory plasma-confinement devices.

Derivation ARadius of the circular path of a charged particle in a magnetic field

The goal of this derivation is to obtain the radius of the circular path traced by a charge $q$ of mass $m$ moving with speed $v$ perpendicular to a uniform magnetic field $B$ , using only the magnetic force law $\vec F = q\vec v\times\vec B$ already established for a moving charge, together with the ordinary mechanics of uniform circular motion — no new physical law is introduced, only the recognition that a perpendicular, constant-magnitude force is a centripetal force.

Step 1. Start from the magnetic force on the moving charge, $\vec F = q\vec v\times\vec B$ . Because $\vec v$ is perpendicular to $\vec B$ by the setup of this problem, the magnitude of the cross product is simply $F = qvB\sin 90° = qvB$ , and its direction, by the right-hand rule, is perpendicular to $\vec v$ at that instant.

Step 2. Note that this perpendicularity never changes as the particle moves. At every point on the path, whichever way $\vec v$ happens to be pointing at that moment, the force $q\vec v\times\vec B$ is perpendicular to that instantaneous $\vec v$ and has the same magnitude $qvB$ , since $v$ and $B$ do not change. A force that stays perpendicular to velocity does no work on the particle ( $\vec F\cdot\vec v = 0$ always), so the kinetic energy $\tfrac12 mv^2$ is constant and the speed $v$ never changes throughout the motion.

Step 3. Recognise the mechanical condition this satisfies: a force of constant magnitude, always directed perpendicular to a constant-speed velocity, is precisely the condition for uniform circular motion, with the force playing the role of the centripetal force pointing toward the centre of the circle. This is the same requirement met by a stone whirled on a string or a satellite in a circular orbit, except that here the mechanism supplying the inward pull is entirely magnetic.

Step 4. Equate the magnetic force to the centripetal force required to keep a mass $m$ moving at speed $v$ on a circle of radius $r$ : $qvB = \dfrac{mv^2}{r}$ . This is a single algebraic statement — magnetic push on the left, mechanical demand on the right — valid at every instant since both sides are constant.

Step 5. Cancel one factor of $v$ from both sides (allowed since $v \neq 0$ ) and rearrange for the radius: $qB = \dfrac{mv}{r}$ , so $r = \dfrac{mv}{qB}$ . This is the radius of the circular path.

Step 6. Extend to the period by using circumference over speed: $T = \dfrac{2\pi r}{v} = \dfrac{2\pi}{v}\cdot\dfrac{mv}{qB} = \dfrac{2\pi m}{qB}$ . The speed $v$ cancels completely, leaving $T$ dependent only on $q$ , $m$ and $B$ — the central, exam-favourite conclusion of this derivation.

Worked examples

Board level - radius of an electron's circular path

An electron ( $m = 9.1\times10^{-31}\ \text{kg}$ , $q = 1.6\times10^{-19}\ \text{C}$ ) moves with speed $v = 2\times10^{6}\ \text{m/s}$ perpendicular to a uniform magnetic field of $B = 0.01\ \text{T}$ . Find the radius of its circular path, its period, and its cyclotron frequency.

Step 1. Write the radius formula and substitute the given values: $r = \dfrac{mv}{qB} = \dfrac{(9.1\times10^{-31})(2\times10^{6})}{(1.6\times10^{-19})(0.01)}$ .

Step 2. Evaluate the numerator and denominator separately. Numerator: $9.1\times10^{-31}\times2\times10^{6} = 1.82\times10^{-24}$ . Denominator: $1.6\times10^{-19}\times0.01 = 1.6\times10^{-21}$ .

Step 3. Divide: $r = \dfrac{1.82\times10^{-24}}{1.6\times10^{-21}} = 1.1375\times10^{-3}\ \text{m} \approx 1.14\ \text{mm}$ .

Step 4. Find the period using $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi(9.1\times10^{-31})}{(1.6\times10^{-19})(0.01)} = \dfrac{5.72\times10^{-30}}{1.6\times10^{-21}} \approx 3.57\times10^{-9}\ \text{s}$ .

Step 5. Find the cyclotron frequency: $f = 1/T \approx 1/(3.57\times10^{-9}) \approx 2.80\times10^{8}\ \text{Hz}$ , about $280\ \text{MHz}$ . Answer: the electron circles with radius about $1.14\ \text{mm}$ , period about $3.57\ \text{ns}$ , and frequency about $280\ \text{MHz}$ .

Competitive level - proton speed, period and comparison with an alpha particle

A proton moves in a circular path of radius $0.2\ \text{m}$ in a magnetic field of $0.25\ \text{T}$ . Take $m_p = 1.67\times10^{-27}\ \text{kg}$ and $q = 1.6\times10^{-19}\ \text{C}$ . Find (a) the proton's speed, (b) its period of revolution, and (c) state, without recalculating in full, how the period would compare for an alpha particle (charge $2q$ , mass about $4m_p$ ) in the same field.

Step 1. Rearrange $r = mv/(qB)$ for speed: $v = \dfrac{qBr}{m} = \dfrac{(1.6\times10^{-19})(0.25)(0.2)}{1.67\times10^{-27}}$ .

Step 2. Evaluate the numerator: $1.6\times10^{-19}\times0.25\times0.2 = 8\times10^{-21}$ .

Step 3. Divide: $v = \dfrac{8\times10^{-21}}{1.67\times10^{-27}} \approx 4.79\times10^{6}\ \text{m/s}$ .

Step 4. Find the period from $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi(1.67\times10^{-27})}{(1.6\times10^{-19})(0.25)} = \dfrac{1.049\times10^{-26}}{4\times10^{-20}} \approx 2.62\times10^{-7}\ \text{s}$ , about $0.26\ \mu\text{s}$ .

Step 5. For part (c), use the ratio form of the period formula directly rather than recomputing from scratch: since $T = 2\pi m/(qB)$ , the period scales as $m/q$ . An alpha particle has mass about $4m_p$ and charge $2q$ , so its $m/q$ ratio is $4m_p/(2q) = 2(m_p/q)$ , twice the proton's — its period in the same field would be about twice as long, roughly $5.2\times10^{-7}\ \text{s}$ , without touching a calculator again. Answer: the proton's speed is about $4.79\times10^{6}\ \text{m/s}$ , its period is about $2.62\times10^{-7}\ \text{s}$ , and an alpha particle in the same field would take about twice as long per revolution. Timing note: under two minutes. Once $r = mv/(qB)$ is rearranged for the unknown, every part is one substitution; the alpha-particle comparison in part (c) is a scaling argument, not a fresh calculation, and reaching for $T \propto m/q$ instead of recomputing from zero is exactly the kind of shortcut that saves time under exam pressure.

Common traps

Thinking the magnetic force changes the particle's speed. It changes only the direction of motion; speed and kinetic energy are strictly constant throughout circular motion in a uniform field, because the force is always perpendicular to velocity and so does zero work.
Expecting a faster particle to complete a revolution more quickly. The period $T = 2\pi m/(qB)$ has no $v$ in it at all — a faster particle simply moves on a proportionally larger circle, and the two effects cancel exactly, so $T$ stays the same for every particle of the same $q/m$ in the same field.
Using the plain circular-motion radius formula $r = mv/(qB)$ when the velocity is not purely perpendicular to $\vec B$ . If there is a component along the field, only the perpendicular component $v_\perp$ enters the circle radius, and the actual path is a helix, not a flat circle.
Getting the sense of curving backwards. Positive and negative charges moving the same way in the same field curve in opposite directions, because the force $q\vec v\times\vec B$ reverses when the sign of $q$ reverses — always re-check with the right-hand rule (and a sign flip for negative charge) rather than guessing.
Forgetting to convert units before substituting, especially field in gauss instead of tesla or mass in atomic mass units instead of kilograms — a very common silent error in numericals with protons and alpha particles.

You should be able to…

Derive $r = mv/(qB)$ by equating the magnetic force to the centripetal force requirement for circular motion.
Compute the radius, period, and cyclotron frequency of a charged particle moving perpendicular to a uniform magnetic field.
Explain, in terms of work done by a perpendicular force, why the particle's speed stays constant throughout the motion.
Explain why the period of revolution is independent of the particle's speed, using $T = 2\pi m/(qB)$ .
Describe qualitatively how a helical path arises when the velocity has a component along the field, and identify which velocity component determines the circle's radius.
Predict and justify the sense of curving (clockwise or anticlockwise) for positive versus negative charges in a given field direction.

Ask the tutor: Want to see why a helical path spirals at a fixed pitch, or how this same radius formula becomes the design equation for a cyclotron's dees — ask the AI tutor to walk through either extension from this derivation.

Why doesn't the magnetic field just pull the charged particle straight into the centre of the circle

Because the force is always perpendicular to the particle's current velocity, not fixed toward one point in space. At the instant the particle enters, the force points toward what will become the centre, but as the particle's velocity direction changes moment by moment, the force direction rotates right along with it, always staying perpendicular to whatever the current velocity is. That continuously turning inward push is exactly what curves the path into a closed circle instead of pulling the particle straight to a point, the same way a taut string constantly redirects a whirled stone without ever reeling it in.

If the period does not depend on speed, does that mean a very fast particle and a very slow particle finish a lap at literally the same moment

Yes, provided they have the same charge-to-mass ratio $q/m$ and are in the same field $B$ — that is exactly what $T = 2\pi m/(qB)$ says, since $v$ has cancelled out of the formula entirely. If you release a fast proton and a slow proton together at the start line of the circle, moving in the same field, both complete one full revolution in the same time $T$ , even though the fast one is tracing a much bigger circle to do it. This surprising independence from speed is precisely the property that lets a cyclotron-type device use a fixed-frequency alternating field to keep re-accelerating particles as they spiral outward.

What actually happens at the exact entry point if the velocity is not perfectly perpendicular to the field

Split the entry velocity into two independent parts: the component along $\vec B$ , called $v_\parallel$ , and the component perpendicular to it, called $v_\perp$ . The parallel part feels absolutely no magnetic force, since a vector crossed with itself (or anything parallel to it) is zero, so it just carries the particle steadily forward along the field direction at constant speed. Meanwhile the perpendicular part $v_\perp$ behaves exactly as in the pure perpendicular case, curving into a circle of radius $r = mv_\perp/(qB)$ . Add the two motions together and you get a helix — a circular spiral advancing steadily along the field, like a spring being drawn out.

Is the radius formula r equals mv over qB something I need to derive on the board, or can I just quote it

CBSE expects you to be able to derive it, since it is built from the syllabus's own magnetic force law and ordinary centripetal-force mechanics — a question phrased as "show that" or "derive" wants the full argument from $\vec F = q\vec v\times\vec B$ through to $r = mv/(qB)$ , in the same steps as the derivation above. In a numerical problem, however, you can go straight to using $r = mv/(qB)$ and $T = 2\pi m/(qB)$ once you have justified, in one line, that the force supplies the required centripetal force — full re-derivation is not expected inside a numerical, only inside an explicit "derive" question.