Circular Motion and Cyclotrons Revision Notes | A Level AQA Physics

1. Why Does a Magnetic Field Curve a Particle's Path?

Picture a proton fired horizontally into a region where a uniform magnetic field points vertically upward. The proton is moving, it carries charge, and there is a magnetic field — so it experiences a force. That force is not in the direction of motion, and not in the direction of the field. It acts at right angles to both.

This is the key insight: because the magnetic force is always perpendicular to the velocity, it never does work on the particle and never changes its speed. What it does change, continuously, is the direction of the velocity. A force that always points perpendicular to motion is exactly the definition of a centripetal force — and a centripetal force produces circular motion.

Fleming's left-hand rule: point your first finger in the direction of the field ($B$), your second finger in the direction of conventional current (i.e. the direction a positive charge moves), and your thumb points in the direction of the force. For a negative charge such as an electron, reverse the thumb direction.

Because the speed stays constant and the force magnitude $F = Bqv$ stays constant, the radius of the circular path also stays constant — the particle traces out a perfect circle in a plane perpendicular to the field.

2. Deriving the Radius of Circular Motion

To find the radius, set the magnetic force equal to the centripetal force. You already know both expressions from earlier in the course: centripetal force is $F = \frac{mv^2}{r}$ and the magnetic force is $F = Bqv$. Equating them gives the radius directly.

Notice what this tells you: a faster particle curves less (larger $r$), a stronger field curves it more (smaller $r$), and a more massive particle curves less (larger $r$). These relationships are worth internalising because they underpin everything that follows about the cyclotron.

Worked example: Calculating the Radius of a Proton's Circular Orbit

A proton travels at $2.0 \times 10^6$ m/s perpendicular to a uniform magnetic field of flux density 0.50 T. Calculate the radius of its circular orbit. (Proton mass $m = 1.67 \times 10^{-27}$ kg, proton charge $q = 1.60 \times 10^{-19}$ C) [3 marks]

1. $r = \frac{mv}{Bq}$ — [M1] Equating magnetic force to centripetal force and rearranging gives this expression. State it explicitly for the method mark.
2. $r = \frac{(1.67 \times 10^{-27}\ \text{kg}) \times (2.0 \times 10^6\ \text{m/s})}{(0.50\ \text{T}) \times (1.60 \times 10^{-19}\ \text{C})}$ — [M1] All values are already in SI units, so no conversion is needed. Carry every unit through the substitution.
3. $\text{Numerator} = 1.67 \times 10^{-27} \times 2.0 \times 10^6 = 3.34 \times 10^{-21}\ \text{kg m/s}$ — $1.67 \times 2.0 = 3.34$ and $10^{-27} \times 10^6 = 10^{-21}$. This is the particle's momentum.
4. $\text{Denominator} = 0.50 \times 1.60 \times 10^{-19} = 8.0 \times 10^{-20}\ \text{T C} = 8.0 \times 10^{-20}\ \text{kg/s}$ — $0.50 \times 1.60 = 0.80$, so $0.80 \times 10^{-19} = 8.0 \times 10^{-20}$.
5. $r = \frac{3.34 \times 10^{-21}}{8.0 \times 10^{-20}} = \frac{3.34}{8.0} \times 10^{-21-(-20)} = 0.4175 \times 10^{-1}\ \text{m} \approx 0.042\ \text{m}$ — [A1] $3.34 \div 8.0 = 0.4175$ and $10^{-21} \div 10^{-20} = 10^{-1}$, giving $0.04175$ m. Rounded to 2 significant figures: $r \approx 0.042$ m.

Answer: $r \approx 0.042$ m (4.2 cm)

3. The Cyclotron: Structure and Operating Principle

Now that you can see how a magnetic field curves a charged particle into a circle, the cyclotron becomes a natural next step. The question it answers is: how do you accelerate a particle to very high energies without needing an enormously long straight accelerator? The answer is to keep bending the particle back on itself and give it a small energy boost each time it crosses a gap.

A cyclotron consists of two hollow, semicircular metal electrodes called dees (because they look like the letter D), placed face-to-face with a small gap between them. The whole assembly sits inside a uniform magnetic field directed perpendicular to the plane of the dees. A source of charged particles — typically protons — sits at the centre.

• The magnetic field (perpendicular to the dees) keeps the particle on a semicircular arc inside each dee.
• The alternating electric field across the gap between the dees accelerates the particle each time it crosses.
• The particle spirals outward because each acceleration increases its speed, and a faster particle has a larger orbital radius ($r = mv/Bq$).
• The particle is extracted when it reaches the outer edge of the dees, at maximum radius and maximum kinetic energy.

Here is the clever part: the time for a particle to complete a semicircle inside a dee is $t = \frac{\pi r}{v} = \frac{\pi m}{Bq}$. This is independent of both $r$ and $v$. So no matter how fast the particle gets, it always takes the same time to cross each dee — which means the alternating voltage can be set to a fixed frequency and will always be in the right phase to accelerate the particle when it arrives at the gap.

This frequency independence is the cyclotron's defining feature. It means the accelerating voltage does not need to be reprogrammed as the particle speeds up — the geometry of the physics does the synchronisation automatically.

4. Energy Gained in a Cyclotron

Each time the particle crosses the gap, the electric field does work on it and its kinetic energy increases. After many crossings, the particle reaches the outer edge of the dees at radius $R$ (the maximum radius, set by the physical size of the dees) and is deflected out by a separate electric field towards a target.

At maximum radius $R$, the particle's speed can be found from $r = \frac{mv}{Bq}$ rearranged to $v = \frac{BqR}{m}$. Substituting into the kinetic energy formula gives the maximum kinetic energy the cyclotron can deliver.

To increase the maximum energy, you can increase $B$, increase $R$ (build bigger dees), or use a particle with higher charge. In practice, building larger dees becomes very expensive, which is one reason more advanced accelerators such as synchrotrons were developed.

Worked example: Calculating Maximum Kinetic Energy and Speed in a Cyclotron

A cyclotron has dees of radius $R = 0.60$ m and operates with a magnetic flux density of $B = 1.2$ T. It accelerates protons (mass $m = 1.67 \times 10^{-27}$ kg, charge $q = 1.60 \times 10^{-19}$ C). Calculate (a) the maximum speed of the protons and (b) their maximum kinetic energy. [4 marks]

1. $v_{\text{max}} = \frac{BqR}{m} = \frac{1.2 \times 1.60 \times 10^{-19} \times 0.60}{1.67 \times 10^{-27}}$ — [M1] Rearranging $r = mv/Bq$ for $v$ and substituting $r = R = 0.60$ m.
2. $1.2 \times 1.60 \times 10^{-19} \times 0.60 = 1.2 \times 0.60 \times 1.60 \times 10^{-19} = 0.72 \times 1.60 \times 10^{-19} = 1.152 \times 10^{-19}$ — $1.2 \times 0.60 = 0.72$; $0.72 \times 1.60 = 1.152$. Units: T $\cdot$ C $\cdot$ m = kg m/s.
3. $v_{\text{max}} = \frac{1.152 \times 10^{-19}}{1.67 \times 10^{-27}} = \frac{1.152}{1.67} \times 10^{-19-(-27)} = 0.6898 \times 10^{8}\ \text{m/s} \approx 6.9 \times 10^7\ \text{m/s}$ — [A1] $1.152 \div 1.67 \approx 0.690$; $10^{-19} \div 10^{-27} = 10^{8}$. So $v_{\text{max}} \approx 6.9 \times 10^7$ m/s. This is about 23% of the speed of light — a useful sanity check that relativistic effects are starting to become relevant.
4. $E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 1.67 \times 10^{-27} \times (6.9 \times 10^7)^2$ — [M1] Using $E_k = \frac{1}{2}mv^2$ with the speed found in step 3. Expand the square carefully.
5. $(6.9 \times 10^7)^2 = 6.9 \times 6.9 \times 10^{14} = 47.61 \times 10^{14} = 4.761 \times 10^{15}\ \text{m}^2\text{/s}^2$ — $6.9 \times 6.9$: $6 \times 6.9 = 41.4$ and $0.9 \times 6.9 = 6.21$, so $41.4 + 6.21 = 47.61$. Then $10^7 \times 10^7 = 10^{14}$.
6. $E_k = \frac{1}{2} \times 1.67 \times 10^{-27} \times 4.761 \times 10^{15} = 0.5 \times 1.67 \times 4.761 \times 10^{-12}$ — $1.67 \times 4.761 = 7.951$; $0.5 \times 7.951 = 3.976$. So $E_k \approx 4.0 \times 10^{-12}$ J.

Answer: (a) $v_{\text{max}} \approx 6.9 \times 10^7$ m/s; (b) $E_k \approx 4.0 \times 10^{-12}$ J

5. How the Cyclotron Accelerates Particles Step by Step

It helps to trace one complete cycle to make sure the mechanism is clear. Start with a proton at rest near the centre of the gap between the two dees.

• Step 1 — First acceleration: The alternating voltage makes dee A positive and dee B negative (or vice versa). The proton is accelerated across the gap from the positive dee towards the negative dee, gaining kinetic energy $\Delta E_k = qV$ where $V$ is the accelerating voltage.
• Step 2 — First semicircle: Inside dee B, the magnetic field (perpendicular to the plane) exerts a force $F = Bqv$ on the proton, curving it into a semicircle of radius $r_1 = mv_1 / Bq$. The electric field does not penetrate the metal dee, so the proton's speed is constant inside.
• Step 3 — Second acceleration: The alternating voltage has reversed polarity by the time the proton returns to the gap (because the transit time equals exactly half the period of the alternating voltage). The proton is accelerated again, gaining another $\Delta E_k = qV$.
• Step 4 — Larger semicircle: Now moving faster, the proton curves through a larger semicircle of radius $r_2 = mv_2 / Bq > r_1$ inside dee A.
• Step 5 — Repeat: This process repeats, with the proton spiralling outward until it reaches the rim of the dees, where it is extracted.

After $n$ gap crossings, the total kinetic energy gained is $E_k = nqV$. This is a useful cross-check: if you know the accelerating voltage and the number of crossings, you can verify the final kinetic energy independently of the $r = mv/Bq$ route.

6. Limitations of the Cyclotron

The cyclotron is elegant, but it has a fundamental limitation rooted in special relativity. As the particle approaches a significant fraction of the speed of light, its relativistic mass increases. From $r = mv/Bq$, a larger effective mass means a larger radius — but it also means the transit time $t = \pi m / Bq$ increases. The particle starts arriving at the gap slightly late, out of phase with the alternating voltage, and the acceleration becomes inefficient.

In the worked example above, the proton reached about 23% of the speed of light. At this speed, relativistic effects are small but not negligible. For electrons, which are much lighter and reach relativistic speeds far sooner, a standard cyclotron is essentially unusable — the relativistic mass increase becomes significant at much lower energies.

More advanced machines such as the synchrocyclotron (which varies the frequency of the alternating voltage to compensate) and the synchrotron (which varies both the frequency and the magnetic field strength to keep particles on a fixed-radius ring) were developed to overcome this limitation. The Large Hadron Collider at CERN is a synchrotron.

7. Exam Technique and Common Mistakes

Circular motion in a magnetic field and the cyclotron are reliable sources of marks in AQA A Level papers. The questions tend to follow predictable patterns, and knowing where students typically drop marks will help you avoid the same traps.