Motion of a Charged Particle in a Magnetic Field

Three Cases of Motion

When a charge $q$ moves with velocity $\vec v$ in a uniform magnetic field $\vec B$, it experiences a magnetic force of magnitude

$$F = qvB\sin\theta$$

where $\theta$ is the angle between $\vec v$ and $\vec B$. This force is always perpendicular to both $\vec v$ and $\vec B$ (given by the right-hand rule). The particle’s trajectory then depends entirely on $\theta$.

Case 1 — $\vec v$ parallel (or anti-parallel) to $\vec B$

$$F = qvB\sin 0° = 0$$

No force acts, so the particle moves in a straight line at constant velocity. The magnetic field is invisible to a charge moving exactly along its direction.

Case 2 — $\vec v$ perpendicular to $\vec B$

The force $\vec F = q\vec v\times\vec B$ has constant magnitude $qvB$ and is always perpendicular to $\vec v$. This serves as a centripetal force, producing uniform circular motion.

Case 3 — $\vec v$ at an angle $\theta$ to $\vec B$

Decompose $\vec v$ into parallel and perpendicular components:

• $v_\parallel = v\cos\theta$ feels no force → uniform motion along $\vec B$.
• $v_\perp = v\sin\theta$ feels a perpendicular force → circular motion in the plane normal to $\vec B$.

The combination is a helical path along the field direction.

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Case 2 in Detail — Circular Motion

When $\vec v\perp\vec B$, the magnetic force $qvB$ provides the centripetal force:

$$qvB = \frac{mv^2}{r}$$

Solving for the radius of the circle:

$$\boxed{r = \frac{mv}{qB} = \frac{p}{qB}}$$

where $p=mv$ is the particle’s momentum. Larger $p$ → wider circle; stronger $B$ → tighter circle.

Time period, frequency, and angular frequency

The particle travels $2\pi r$ in one revolution at speed $v$, so:

$$T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}\quad(\text{in seconds})$$ $$f = \frac{1}{T} = \frac{qB}{2\pi m}\quad(\text{in Hz})$$ $$\omega = 2\pi f = \frac{qB}{m}\quad(\text{in rad/s})$$

Crucial property: None of $T$, $f$, or $\omega$ depends on the speed $v$ or the circle’s radius $r$. This is the cyclotron property.

[Exam Tip] Remember: $T = 2\pi m/(qB)$ depends only on the mass-to-charge ratio and $B$, not on $v$.

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Case 3 in Detail — Helical Motion

With $v$ at angle $\theta$ to $\vec B$:

$$v_\parallel = v\cos\theta\,,\quad v_\perp = v\sin\theta$$

– Parallel component → no force → straight motion along $\vec B$.
– Perpendicular component → uniform circle with radius

$$\boxed{r = \frac{m\,v_\perp}{qB} = \frac{m\,v\sin\theta}{qB}}$$

Time period is the same as in the pure circular case:

$$T = \frac{2\pi m}{qB}$$

The pitch $p$ (axial distance per revolution) is:

$$\boxed{p = v_\parallel\,T = v\cos\theta\cdot \frac{2\pi m}{qB}}$$

Summary table

Quantity	Circular (Case 2)	Helix (Case 3)
Radius	$r = \frac{mv}{qB}$	$r = \frac{mv\sin\theta}{qB}$
Time period	$T = \frac{2\pi m}{qB}$	$T = \frac{2\pi m}{qB}$
Frequency	$f = \frac{qB}{2\pi m}$	$f = \frac{qB}{2\pi m}$
Angular freq.	$\omega = \frac{qB}{m}$	$\omega = \frac{qB}{m}$
Pitch	–	$p = v\cos\theta\,T$

[Exam Tip] Ratio of pitch to radius is $p/r = 2\pi\cot\theta$. This shortcut often appears in exams.

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The Cyclotron

Invented by E. O. Lawrence in 1932, the cyclotron uses the speed-independence of $T$ to accelerate charged particles to high energies.

Construction

• Two hollow D-shaped electrodes ("dees"), separated by a narrow gap.
• Uniform magnetic field $\vec B$ perpendicular to the plane of the dees.
• High-frequency alternating voltage applied between the dees.
• Ion source at the center.

Working principle

1. A positive ion starts at the center; the AC voltage accelerates it across the gap.
2. Inside a dee, the ion moves in a semicircle of radius $r=mv/(qB)$ under magnetic force alone.
3. Upon crossing the gap again, the AC has reversed, accelerating the ion further into a larger semicircle.
4. This repeats, spiraling outward and increasing the ion’s energy.

Resonance condition

To stay in sync with the AC, the revolution frequency must match:

$$\boxed{f_{AC} = f_{cyc} = \frac{qB}{2\pi m}}$$

Maximum kinetic energy

At the outer radius $R$ of a dee,

$$v_{\max} = \frac{qBR}{m},\qquad K_{\max} = \tfrac12mv_{\max}^2 = \frac{q^2B^2R^2}{2m}$$

Limitations

1. Relativistic effect: As $v\to0.1c$, $m$ increases, $f_{cyc}$ drops, and the ion falls out of resonance.
2. Electrons: Reach relativistic speeds at low KE, so standard cyclotrons cannot accelerate them effectively.

[Exam Tip] Be ready to derive $K_{\max}=q^2B^2R^2/(2m)$ and explain why electrons fall out of sync due to relativistic mass increase.

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Memory Capsule

Three motion cases

Case	Condition	Trajectory
1	$\vec v\parallel\vec B$	Straight line
2	$\vec v\perp\vec B$	Uniform circle
3	$0<\theta<90^°,\;\vec v\angle\vec B$	Helix

Circular motion (Case 2)

$$r = \frac{mv}{qB},\quad T = \frac{2\pi m}{qB},\quad f = \frac{qB}{2\pi m},\quad \omega = \frac{qB}{m}$$

Helical motion (Case 3)

$$r = \frac{mv\sin\theta}{qB},\quad p = v\cos\theta\cdot T = \frac{2\pi m v\cos\theta}{qB}$$

Key ratios

$$\frac{p}{r} = 2\pi\cot\theta$$

Cyclotron

– Resonance: $f_{AC}=qB/(2\pi m)$. – $K_{\max}=q^2B^2R^2/(2m)$. – Electrons cannot be accelerated (relativistic mass increase).

Solved Examples

Example 1. Radius of a proton’s circular path

A proton ($m=1.67\times10^{-27}\,$kg, $q=1.6\times10^{-19}\,$C) enters $B=0.5\,$T with $v=4\times10^6\,$m/s perpendicular to $B$. Find $r$.

Solution:
Use $r=mv/(qB)$:

$$r=\frac{(1.67\times10^{-27})(4\times10^6)}{(1.6\times10^{-19})(0.5)}\approx8.35\times10^{-2}\,\text{m}=8.35\,\text{cm}$$

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Example 2. Time period of an electron in a magnetic field

An electron ($m_e=9.1\times10^{-31}\,$kg) enters $B=1.5\times10^{-4}\,$T perpendicular to $v$. Find $T$.

Solution:
$$T=\frac{2\pi m_e}{qB}=\frac{2\pi(9.1\times10^{-31})}{(1.6\times10^{-19})(1.5\times10^{-4})}\approx2.4\times10^{-7}\,\text{s}$$

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Example 3. Pitch of a helix

A particle with $q/m=10^7\,$C/kg enters $B=0.2\,$T at $v=10^5\,$m/s and $\theta=60°$. Find (a) $r$ and (b) $p$.

Solution:

(a) $r=\dfrac{v\sin60°}{(q/m)B}=\dfrac{10^5(\sqrt3/2)}{10^7\times0.2}\approx4.33\times10^{-2}\,\text{m}$.
(b) $T=2\pi/( (q/m)B)=\dfrac{2\pi}{10^7\times0.2}\approx3.14\times10^{-6}\,\text{s}$,
$$p=v\cos60°\,T=10^5\times0.5\times3.14\times10^{-6}\approx0.157\,\text{m}$$

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Example 4. Cyclotron — maximum KE

A cyclotron with $R=0.5\,$m and $B=1.2\,$T accelerates protons. Find $K_{\max}$ in MeV.

Solution: $$K_{\max}=\frac{q^2B^2R^2}{2m}=\frac{(1.6\times10^{-19})^2(1.2)^2(0.5)^2}{2(1.67\times10^{-27})}\approx2.76\times10^{-12}\,\text{J}$$ $$\approx\frac{2.76\times10^{-12}}{1.6\times10^{-13}}\approx17.3\,\text{MeV}$$

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Example 5: Cyclotron frequency

A cyclotron is to accelerate protons in a magnetic field of $0.8$ T. What should be the frequency of the alternating voltage applied to the dees?

Solution. Resonance condition: $f_{AC} = qB/(2\pi m)$.

$$f = \dfrac{(1.6\times 10^{-19})(0.8)}{2\pi (1.67\times 10^{-27})} = \dfrac{1.28\times 10^{-19}}{1.049\times 10^{-26}}$$

$$f \approx 1.22\times 10^7\;\text{Hz} = 12.2\;\text{MHz}$$

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Example 6: Doubling the field

A charged particle moves in a circle of radius $r$ in a magnetic field $B$. If $B$ is doubled (keeping speed and charge the same), what is the new radius?

Solution. Since $r = mv/(qB)$, $r \propto 1/B$. So doubling $B$ halves the radius:

$$r' = r/2$$

Time period $T = 2\pi m/(qB)$ also halves: $T' = T/2$. Speed is unchanged (no work done).

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Example 7: Two particles in the same field

A proton and an alpha particle ($q_\alpha = 2e$, $m_\alpha = 4m_p$) enter the same magnetic field with the same speed, perpendicular to the field. Compare (a) their radii (b) their time periods.

Solution.

(a) $r = mv/(qB)$. With same $v$, $B$:

$$\dfrac{r_\alpha}{r_p} = \dfrac{m_\alpha/q_\alpha}{m_p/q_p} = \dfrac{4m_p/(2e)}{m_p/e} = 2$$

The alpha particle traces a circle with twice the radius of the proton.

(b) $T = 2\pi m/(qB)$:

$$\dfrac{T_\alpha}{T_p} = \dfrac{m_\alpha/q_\alpha}{m_p/q_p} = \dfrac{4/2}{1/1} = 2$$

The alpha takes twice as long to complete one revolution.

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Example 8: Why electrons cannot be accelerated in a standard cyclotron

An electron is to be accelerated in a hypothetical cyclotron. Estimate its speed classically after it gains a kinetic energy of $0.5$ MeV, and explain why this result indicates a breakdown of the classical framework.

Solution. Using the non-relativistic formula for kinetic energy, $K = \frac{1}{2}m_e v^2 \Rightarrow v = \sqrt{\dfrac{2K}{m_e}}$.

Converting kinetic energy from MeV to Joules: $$K = 0.5 \times 10^6 \times 1.6 \times 10^{-19}\;\text{J} = 8.0 \times 10^{-14}\;\text{J}$$

Substituting the values: $$v = \sqrt{\dfrac{2 \times 8.0 \times 10^{-14}}{9.1 \times 10^{-31}}} = \sqrt{\dfrac{1.6 \times 10^{-13}}{9.1 \times 10^{-31}}}$$

$$v \approx \sqrt{1.76 \times 10^{17}} = \sqrt{17.6 \times 10^{16}} \approx 4.19 \times 10^8\;\text{m/s}$$

This calculated value exceeds the universal speed limit—the speed of light ($c = 3 \times 10^8\;\text{m/s}$)—which is physically impossible.

Conclusion: The classical estimate breaks down completely at this energy level. In reality, as the electron's speed approaches $c$, its relativistic mass ($m$) increases significantly. Since the cyclotron frequency depends inversely on mass ($f_{cyc} = \frac{qB}{2\pi m}$), the frequency drops, causing the electron to quickly fall out of step with the fixed frequency of the oscillating electric field. This is why cyclotrons are used for heavy particles like protons or alpha particles, but not electrons.

Example 9: Helical motion — full picture

A proton with KE $= 5\times 10^{-13}$ J enters a region of $\vec{B} = 0.1$ T at $30°$ to $\vec{B}$. ($m_p = 1.67\times 10^{-27}$ kg, $q = 1.6\times 10^{-19}$ C.) Find (a) speed (b) radius of helix (c) pitch.

Solution.

(a) From $K = \frac{1}{2}mv^2$:

$$v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2(5\times 10^{-13})}{1.67\times 10^{-27}}} = \sqrt{5.99\times 10^{14}} \approx 2.45\times 10^7\;\text{m/s}$$

(b) Radius:

$$r = \dfrac{m v\sin 30°}{qB} = \dfrac{(1.67\times 10^{-27})(2.45\times 10^7)(0.5)}{(1.6\times 10^{-19})(0.1)}$$

$$r = \dfrac{2.045\times 10^{-20}}{1.6\times 10^{-20}} \approx 1.28\;\text{m}$$

(c) Pitch: $p = v\cos 30°\cdot T$, with $T = 2\pi m/(qB) = 2\pi(1.67\times 10^{-27})/[(1.6\times 10^{-19})(0.1)] \approx 6.55\times 10^{-7}$ s.

$$p = (2.45\times 10^7)(0.866)(6.55\times 10^{-7}) \approx 13.9\;\text{m}$$