Why the Path Bends into a Circle
A force does work on a particle only if it has a component along (or opposed to) the direction of motion. The magnetic force on a moving charge, $\mathbf{F} = q(\mathbf{v} \times \mathbf{B})$, is always perpendicular to the velocity $\mathbf{v}$. Therefore it does no work, and the speed of the particle never changes — only the direction of its momentum is altered. This is the defining contrast with the electric force $q\mathbf{E}$, which can have a component along the motion and so can transfer energy.
Consider a uniform field $\mathbf{B}$ and take, first, the case where $\mathbf{v}$ is perpendicular to $\mathbf{B}$. The perpendicular force $q\,\mathbf{v}\times\mathbf{B}$ acts exactly like a centripetal force: constant in magnitude, always pointing toward a fixed centre, always at right angles to the velocity. Under such a force the particle traces a closed circle in the plane perpendicular to $\mathbf{B}$.
If, instead, the velocity has a component along $\mathbf{B}$, that parallel component is untouched by the field — the magnetic force on it is zero. The motion then splits into a circle in the perpendicular plane plus a steady drift along the field line, which combine into a helix. We treat the pure circle first and return to the helix below.
The magnetic force supplies the centripetal force toward the centre; the velocity stays tangent to the circle, so its magnitude is constant.
Radius of the Circular Path
For circular motion of radius $r$, the centripetal force required is $mv^2/r$. The magnetic force supplying it has magnitude $qvB$ (for $\mathbf{v}\perp\mathbf{B}$). Equating the two expressions for the centripetal force,
$$\frac{mv^2}{r} = qvB \quad\Longrightarrow\quad r = \frac{mv}{qB}$$
This is NCERT Eq. (4.5). Because the linear momentum is $p = mv$, the same relation reads $r = p/qB$, so the radius is directly proportional to the momentum. The larger the momentum, the larger the radius and the bigger the circle described. A heavier or faster particle of the same charge in the same field sweeps a wider arc; a stronger field $B$ tightens it.
Radius scales with speed, but period does not
Students often assume that if the speed depends on something, so must the time period. Here the two behave oppositely: $r = mv/qB$ rises with $v$ (and with $p$), while $T = 2\pi m/qB$ is completely independent of $v$ and $r$. Double the speed and you double the radius — but the time for one loop is unchanged.
Remember: $r \propto v \propto p$, but $T$ and $\nu$ are speed-independent.
Period and Cyclotron Frequency
If $\omega$ is the angular frequency, then $v = \omega r$. Substituting $r = mv/qB$ gives
$$\omega = 2\pi\nu = \frac{qB}{m}$$
which is NCERT Eq. [4.6(a)], independent of the velocity or energy of the particle. The time taken for one revolution is the period
$$T = \frac{2\pi}{\omega} = \frac{2\pi m}{qB}$$
and the frequency of the uniform circular motion is the cyclotron frequency
$$\nu_c = \frac{qB}{2\pi m}$$
The independence of $\nu_c$ from speed and radius is not a curiosity — it is the single fact on which the cyclotron is built. A faster particle sweeps a larger circle, but it covers that larger circumference proportionally faster, so the time per revolution stays fixed.
An electron ($m = 9\times10^{-31}$ kg, $q = 1.6\times10^{-19}$ C) moves at $v = 3\times10^{7}$ m/s in a field $B = 6\times10^{-4}$ T perpendicular to it. Find the radius and the frequency of revolution.
Radius: $r = \dfrac{mv}{qB} = \dfrac{(9\times10^{-31})(3\times10^{7})}{(1.6\times10^{-19})(6\times10^{-4})} = 28\times10^{-2}$ m $= 28$ cm.
Frequency: $\nu = \dfrac{v}{2\pi r} = 17\times10^{6}$ Hz $= 17$ MHz. (Energy $=\tfrac12 mv^2 \approx 2.5$ keV.)
Helical Motion and Pitch
Now allow the velocity to make an angle with $\mathbf{B}$ that is neither $0^\circ$ nor $90^\circ$. Resolve it into a component $v_\perp$ perpendicular to $\mathbf{B}$ and a component $v_\parallel$ parallel to $\mathbf{B}$. The perpendicular component produces the usual circular motion (now of radius $r = mv_\perp/qB$, the radius of the helix), while the parallel component is unaffected and carries the particle steadily along the field. Superposed, these give a helical path.
The distance moved along the magnetic field in one full rotation is called the pitch $p$. Since one revolution takes a time $T = 2\pi m/qB$ and the particle advances at speed $v_\parallel$ along the field during that time,
$$p = v_\parallel\,T = \frac{2\pi m\,v_\parallel}{qB}$$
This is NCERT Eq. [4.6(b)]. Note that the pitch uses the same period $T$ as the circular motion, so a particle with a larger parallel velocity stretches its helix into a longer pitch without changing the time per loop.
The perpendicular component $v_\perp$ loops the particle around $\mathbf{B}$; the parallel component $v_\parallel$ slides it forward one pitch per revolution.
Use v⊥ in the radius, full v in the angle
For a helix the radius depends only on the perpendicular speed: $r = mv_\perp/qB = m(v\sin\theta)/qB$, where $\theta$ is the angle between $\mathbf{v}$ and $\mathbf{B}$. The pitch uses the parallel speed: $p = (v\cos\theta)T$. Plugging the full speed $v$ into the radius formula is a frequent slip when $\theta \ne 90^\circ$.
If $\theta = 0^\circ$: straight line. If $\theta = 90^\circ$: pure circle. Otherwise: helix.
What Depends on What
The cleanest way to lock this subtopic is to separate the quantity that scales with speed (the radius) from the two that do not (period and frequency). The table below is the single most testable comparison in the topic.
| Quantity | Formula | Depends on speed v? | Depends on B? |
|---|---|---|---|
| Radius $r$ | r = mv/qB = p/qB | Yes — $r \propto v \propto p$ | Yes — $r \propto 1/B$ |
| Time period $T$ | T = 2πm/qB | No — independent of $v$ and $r$ | Yes — $T \propto 1/B$ |
| Cyclotron frequency $\nu_c$ | νc = qB/2πm | No — independent of $v$ and $r$ | Yes — $\nu_c \propto B$ |
| Pitch $p$ (helix) | p = v∥T = 2πm v∥/qB | Yes — $p \propto v_\parallel$ | Yes — $p \propto 1/B$ |
The whole circular orbit rests on one force law. Revise where $q(\mathbf{v}\times\mathbf{B})$ comes from in Magnetic Force & the Lorentz Force.
The Cyclotron
The cyclotron, invented by E.O. Lawrence in 1929, accelerates charged particles such as protons, deuterons, and $\alpha$-particles to high velocities. It consists of two semicircular hollow metallic chambers called dees (after their D-shape), placed in an evacuated chamber with a small gap between them. A magnetic field perpendicular to the plane of the dees is maintained by an electromagnet, and a rapidly oscillating potential difference is applied across the gap by an oscillator, producing an oscillating electric field only in the gap.
Inside a dee there is no electric field, so a particle simply moves in a semicircle of radius $r = mv/qB$ at constant speed. The acceleration happens only in the gap: each time the particle crosses, the electric field gives it a kick. Crucially, while it is inside the metal dees the particle takes a fixed half-period $T/2 = \pi m/qB$ to come around — a time that does not depend on its speed. So if the oscillator reverses the field with exactly this rhythm, the gap voltage is always pointing the right way when the particle arrives.
This matching is the cyclotron resonance condition: the oscillator frequency $\nu_o$ is set equal to the cyclotron frequency,
$$\nu_o = \nu_c = \frac{qB}{2\pi m}$$
On resonance the particle gains energy at every gap crossing, moving into a circle of larger radius each time. The energy and radius increase progressively, the orbit spiralling outward, until the particle reaches the rim of the dees and is extracted through an opening.
The particle gains a kick each time it crosses the gap and spirals into ever-larger semicircles; the oscillator runs at the fixed cyclotron frequency $\nu_c = qB/2\pi m$.
Maximum Energy and Limitations
The largest orbit the particle can occupy is set by the radius $R$ of the dees. At this maximum radius, $r = R$ gives the maximum speed from $R = mv_{\max}/qB$, so
$$v_{\max} = \frac{qBR}{m}$$
The corresponding maximum kinetic energy is $K = \tfrac12 m v_{\max}^2$, which gives the standard cyclotron result
$$K_{\max} = \frac{1}{2}m v_{\max}^2 = \frac{q^2 B^2 R^2}{2m}$$
So the achievable energy grows with the square of both the field strength and the dee radius, and falls with particle mass. A larger machine or a stronger magnet yields a more energetic beam.
The cyclotron has one fundamental limitation. The whole design assumes $\nu_c = qB/2\pi m$ is constant, which holds only while the mass $m$ is constant. As the particle approaches relativistic speeds, its relativistic mass increase makes each revolution take slightly longer, so its frequency drops below the fixed oscillator frequency and the particle drifts out of resonance. The accelerating kicks fall out of step, and energy gain stalls. This is why the simple cyclotron cannot push particles to arbitrarily high energies, and why high-energy machines use frequency-modulated or synchrotron designs.
The cyclotron works because the period is speed-independent
A common conceptual error is to think the oscillator must speed up as the particle gains energy. It must not. Since $T = 2\pi m/qB$ does not change with speed, a single fixed-frequency oscillator stays synchronised on every wider orbit. The mechanism breaks only when relativistic mass growth makes $T$ rise — that is the limitation, not the operating principle.
Resonance: $\nu_o = \nu_c = qB/2\pi m$, fixed. Failure: relativistic $m$ rises, $\nu_c$ falls.
Motion in a magnetic field & the cyclotron — at a glance
- Magnetic force $q(\mathbf{v}\times\mathbf{B})$ is $\perp$ to $\mathbf{v}$, does no work; speed is unchanged, only direction.
- $\mathbf{v}\perp\mathbf{B}$ → circle of radius $r = mv/qB = p/qB$, so $r \propto v \propto p$ and $r \propto 1/B$.
- Period $T = 2\pi m/qB$ and cyclotron frequency $\nu_c = qB/2\pi m$ are both independent of speed and radius.
- Velocity at an angle to $\mathbf{B}$ → helix; pitch $p = v_\parallel T = 2\pi m v_\parallel/qB$; radius uses $v_\perp$.
- Cyclotron: dees + perpendicular $\mathbf{B}$ + oscillator at resonance $\nu_o = \nu_c$; particle spirals outward gaining energy each gap crossing.
- Maximum energy $K_{\max} = q^2B^2R^2/2m$; limited by relativistic mass increase that breaks resonance.