CLASS 12 · CHAPTER 4 · MAGNETISM

Moving Charges and Magnetism

Speed this particle up as much as you like. Its circular orbit grows every time, but the clock timing one full revolution never ticks a nanosecond differently. That stubborn fact is the entire working principle behind the cyclotron.

01 · See it

Watch it happen

Particle speed v: 2.0 × 10⁶ m/s

Drag the slider to speed the particle up, then watch the orbit size change below, and watch the timing readouts to its right stay put.

11.39 mm

Radius r

35.77 ns

Period T

27.95 MHz

Frequency f

2.0 × 10⁶ m/s

Speed v (T unchanged)

Drag the speed slider from one end to the other. The orbit visibly grows, exactly as you’d expect for a faster particle. But watch the readouts: radius r changes every time, while period T and frequency f sit completely still. The particle takes the same real time to complete one lap no matter how fast it’s going, because the dot you see sweeping around the circle moves at a constant angular rate the whole time, only the circle itself changes size.

02 · Derive it

Where the formula comes from

A charge $q$ moving with velocity $v$ through a magnetic field $B$ feels the magnetic (Lorentz) force:

Magnetic force

\vec{F} = q\vec{v}\times\vec{B}

When $v$ is perpendicular to $B$ , this force has magnitude $F=qvB$ and always points perpendicular to the velocity itself. A force that’s always perpendicular to velocity can only change direction, never speed, so it supplies exactly the centripetal force needed for uniform circular motion:

Centripetal condition

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

Solving for the orbit radius:

Radius

r = \dfrac{mv}{qB}

The time for one full revolution is the circumference divided by the speed, $T = 2\pi r/v$ . Substitute $r=mv/(qB)$ and the speed $v$ cancels completely:

Period

T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}

and correspondingly, the frequency of revolution (the cyclotron frequency):

Frequency

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

Neither $T$ nor $f$ contains $v$ or $r$ anywhere, they depend only on the charge, the mass, and the field.

03 · Break it

Where the shortcut stops working

Every instinct built from everyday motion says a bigger loop should take longer to go around. Walk a wider circle at the same pace and it obviously takes longer. That instinct is exactly what this chapter breaks: in cyclotron motion, the particle isn’t walking the bigger loop at the same pace, it’s moving faster around it, by precisely the amount needed to keep the lap time fixed.

Look at the radius formula again: $r=mv/(qB)$ . Radius is directly proportional to speed. Double $v$ and the circle’s radius doubles, so its circumference $2\pi r$ doubles too. But the particle is now also moving at double the speed around that doubled circumference. Time for one lap is distance over speed, and when both the distance and the speed double together, their ratio, the time, is untouched.

Keep the scope precise, though: $T$ and $f$ are independent of speed (and of the radius that speed produces), not independent of everything. Change the field $B$ , or swap in a particle with a different $q$ or $m$ , and the period changes immediately, $T=2\pi m/(qB)$ still depends on all three of those. It’s speed alone that drops out, and that’s precisely the one variable a cyclotron keeps cranking up.

04 · Master it

Apply it under exam conditions

Q1. A proton (q = 1.6×10⁻¹⁹ C, m = 1.67×10⁻²⁷ kg) moves at right angles to a uniform magnetic field B = 0.4 T, with speed v = 3×10⁶ m/s. Find the radius of its circular path.

r = \dfrac{mv}{qB} = \dfrac{(1.67\times10^{-27})(3\times10^{6})}{(1.6\times10^{-19})(0.4)} = \textcolor{#e08a1e}{0.0783\text{ m} \approx 7.83\text{ cm}}

Q2. For that same proton in that same field (B = 0.4 T), find the period and frequency of its circular motion.

T = \dfrac{2\pi m}{qB} = \dfrac{2\pi(1.67\times10^{-27})}{(1.6\times10^{-19})(0.4)} = \textcolor{#e08a1e}{1.64\times10^{-7}\text{ s}}

f = \dfrac{1}{T} \approx \textcolor{#e08a1e}{6.10\times10^{6}\text{ Hz} = 6.10\text{ MHz}}

Notice this calculation never needed the speed from Q1, or the 7.83 cm radius it produced. Had that proton instead been moving at 3×10⁷ m/s (ten times faster, with a radius ten times larger, about 78.3 cm), T and f here would come out identically, since neither v nor r appears anywhere in either formula.

05 · FAQs

Quick answers

If the particle speeds up, doesn't it have further to travel around the now-bigger circle, so shouldn't each revolution take longer?+

No, and this is the chapter's sharpest trap. A faster particle does trace a bigger circle, true, but it also covers that bigger circumference proportionally faster. The two effects cancel exactly: T = 2πm/(qB) has no v in it at all. Double the speed, and both the circumference and the speed around it double together, leaving the time for one lap completely unchanged.

So does the period depend on the radius of the orbit instead?+

Not that either, for the same reason. A bigger orbit (from a faster particle) is covered at a proportionally higher speed, so the time per revolution stays fixed. T depends only on the particle's charge-to-mass ratio and the field strength B, nothing about its motion.

Why does the magnetic force cause circular motion at all, instead of just pushing the particle off to one side?+

Because the magnetic force F = qv×B is always perpendicular to the velocity, by definition of the cross product. A force perpendicular to velocity can never speed a particle up or slow it down, it only ever redirects it. Constant redirection at a constant rate, with no change in speed, is exactly circular motion at a constant speed.

What does any of this have to do with an actual cyclotron?+

Everything, it's the whole design principle. A cyclotron alternates a voltage across a gap at a single fixed frequency to repeatedly kick a particle to higher speeds. Because the particle's revolution period never changes as it speeds up, that fixed-frequency voltage stays perfectly in step with it lap after lap, so one alternating voltage can keep accelerating the particle indefinitely without ever needing to be retuned (this breaks down only at speeds close to light speed, well beyond CBSE scope).

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