Force on a moving charge

1. Overview

A magnetic field exerts a force on a charged particle only when that particle is in motion relative to the field. This force, often called the magnetic component of the Lorentz force, arises because a moving charge constitutes an infinitesimal electric current. The interaction is fundamental to modern physics, governing the behavior of electrons in television tubes, the paths of ions in mass spectrometers, and the confinement of plasma in fusion reactors. Crucially, this magnetic force is always perpendicular to the velocity of the particle, meaning it changes the particle's direction but never its speed or kinetic energy.

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2. Key Definitions

• Magnetic Flux Density ($B$): The force per unit current per unit length acting on a straight conductor placed perpendicular to the magnetic field. It is a vector quantity measured in Tesla (T).
• Tesla (T): The magnetic flux density that produces a force of one Newton per metre on a conductor carrying a current of one Ampere placed perpendicular to the field ($1\text{ T} = 1\text{ N A}^{-1}\text{ m}^{-1}$).
• Hall Effect: The development of a transverse potential difference (the Hall voltage) across a conductor or semiconductor when a magnetic field is applied perpendicular to the direction of the conventional current.
• Velocity Selection: A configuration of perpendicular (crossed) electric and magnetic fields used to allow only charged particles with a specific velocity to pass through undeflected.
• Number Density ($n$): The number of free charge carriers per unit volume of a material (measured in $\text{m}^{-3}$).

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3. Content

3.1 Direction of the Force

The direction of the magnetic force $F$ on a moving charge is determined using Fleming’s Left-Hand Rule (LHR). Because the rule is based on conventional current (the flow of positive charge), the application depends on the sign of the particle's charge.

• Thumb: Direction of the magnetic Force ($F$).
• First Finger: Direction of the external Magnetic Field ($B$).
• Second Finger: Direction of the Conventional Current ($I$).

Application to Particles:

1. Positive Charges (e.g., Protons, $\alpha$-particles): The second finger points in the same direction as the velocity ($v$).
2. Negative Charges (e.g., Electrons, $\beta$-particles): The second finger points in the opposite direction to the velocity ($v$).

Geometric Relationship: The force $F$ is always mutually perpendicular to both the magnetic field $B$ and the velocity $v$. If $v$ and $B$ are in the plane of the page, $F$ will act into or out of the page.

3.2 Magnitude of the Force

The magnitude of the force $F$ acting on a charge $Q$ moving with velocity $v$ at an angle $\theta$ to a magnetic field of flux density $B$ is given by:

$$F = BQv \sin \theta$$

• Maximum Force: Occurs when the particle moves perpendicular to the field ($\theta = 90^\circ, \sin 90^\circ = 1$). Here, $F = BQv$.
• Zero Force: Occurs when the particle moves parallel or anti-parallel to the field ($\theta = 0^\circ$ or $180^\circ, \sin \theta = 0$).
• Work Done: Since the force is always perpendicular to the direction of motion (displacement), the work done by the magnetic field on the charge is zero ($W = Fd \cos 90^\circ$). Consequently, the kinetic energy and speed of the particle remain constant.

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3.3 The Hall Effect

Origin of the Hall Voltage

1. Deflection: When a current $I$ flows through a conductor of thickness $t$ and width $d$ in a perpendicular magnetic field $B$, the charge carriers (usually electrons) experience a magnetic force $F_B = Bqv$.
2. Charge Accumulation: This force deflects the charge carriers toward one side of the conductor. For electrons, this leaves a net positive charge on the opposite side.
3. Electric Field Creation: The separation of charges creates a transverse Electric Field ($E$) across the width of the conductor.
4. Equilibrium: This electric field exerts an electric force $F_E = qE$ on the carriers in the opposite direction to the magnetic force. Charge continues to accumulate until the electric force perfectly balances the magnetic force: $$F_E = F_B$$
5. Steady State: Once $qE = Bqv$, the carriers move in a straight line. The potential difference maintained across the width at this equilibrium is the Hall Voltage ($V_H$).

Derivation of $V_H = \frac{BI}{ntq}$

1. At equilibrium: $$qE = Bqv \implies E = Bv$$
2. The relationship between electric field $E$, Hall voltage $V_H$, and width $d$ is: $$E = \frac{V_H}{d} \implies V_H = Bvd$$
3. Use the transport equation for current: $$I = nAqv$$ Where $A$ is the cross-sectional area. Since $A = \text{thickness} (t) \times \text{width} (d)$: $$I = n(td)qv$$
4. Rearrange for drift velocity $v$: $$v = \frac{I}{ntdq}$$
5. Substitute $v$ back into the $V_H$ equation: $$V_H = B \left( \frac{I}{ntdq} \right) d$$
6. The width $d$ cancels out, giving the final expression: $$V_H = \frac{BI}{ntq}$$

Use of a Hall Probe

A Hall probe is a small device used to measure magnetic flux density $B$.

• Material: It uses a thin slice of semiconductor rather than metal. This is because semiconductors have a much lower number density ($n$) of charge carriers. Since $V_H \propto 1/n$, a smaller $n$ produces a significantly larger, measurable Hall voltage.
• Orientation: The probe must be held so that the magnetic field lines are perpendicular to the flat face of the semiconductor slice to obtain the maximum $V_H$ reading.
• Calibration: Because $V_H$ is directly proportional to $B$ (provided $I$ is kept constant), the voltmeter attached to the probe can be calibrated to display $B$ directly in Tesla.

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3.4 Motion of Charged Particles in a Uniform Magnetic Field

When a charged particle enters a uniform magnetic field with a velocity perpendicular to the field lines, it undergoes uniform circular motion.

1. Centripetal Force: The magnetic force $F = BQv$ acts as the centripetal force because it is always perpendicular to the velocity. $$BQv = \frac{mv^2}{r}$$
2. Radius of Path: Rearranging for $r$: $$r = \frac{mv}{BQ}$$

Key Observations:

• Mass and Velocity: The radius is proportional to the momentum ($p = mv$). Heavier or faster particles are harder to deflect and follow a path with a larger radius.
• Field and Charge: The radius is inversely proportional to $B$ and $Q$. A stronger field or a larger charge results in a smaller radius (tighter curve).
• Direction of Curvature: Protons and electrons will curve in opposite directions. Due to the much smaller mass of the electron, its radius of curvature is significantly smaller than that of a proton entering the same field at the same speed.

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3.5 Velocity Selection

A velocity selector is a device that uses "crossed" electric and magnetic fields to filter a beam of charged particles.

1. Setup: An electric field $E$ (produced by parallel plates) and a magnetic field $B$ are applied at right angles to each other and to the direction of the incoming particles.
2. Force Balance: For a specific velocity, the upward electric force $F_E = QE$ exactly balances the downward magnetic force $F_B = BQv$ (or vice versa, depending on the field directions).
3. Selection Condition: $$QE = BQv$$ $$v = \frac{E}{B}$$
4. Outcome:
   • Particles with $v = E/B$ experience zero net force and pass through a narrow exit slit undeflected.
   • Particles that are too fast ($v > E/B$) experience a magnetic force greater than the electric force ($BQv > QE$) and are deflected.
   • Particles that are too slow ($v < E/B$) experience an electric force greater than the magnetic force ($QE > BQv$) and are deflected.

• Note: Velocity selection is independent of the mass and the charge of the particle; only the velocity matters.

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Worked Example 1 — Circular Motion of an Alpha Particle

Question: An $\alpha$-particle ($m = 6.64 \times 10^{-27}\text{ kg}$, $Q = +3.2 \times 10^{-19}\text{ C}$) is accelerated to a kinetic energy of $2.4\text{ MeV}$. It then enters a uniform magnetic field of $0.15\text{ T}$ perpendicular to its path. Calculate the radius of the circular path.

Answer:

1. Convert Energy to Joules: $E_k = 2.4 \times 10^6 \times 1.6 \times 10^{-19} = 3.84 \times 10^{-13}\text{ J}$
2. Find Velocity: $E_k = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2E_k}{m}}$ $v = \sqrt{\frac{2 \times 3.84 \times 10^{-13}}{6.64 \times 10^{-27}}} = 1.076 \times 10^7\text{ m s}^{-1}$
3. Calculate Radius: $r = \frac{mv}{BQ}$ $r = \frac{(6.64 \times 10^{-27}) \times (1.076 \times 10^7)}{0.15 \times (3.2 \times 10^{-19})}$ $r = \frac{7.145 \times 10^{-20}}{4.8 \times 10^{-20}} = 1.488... \text{ m}$ Final Answer: $r = 1.5\text{ m}$ (to 2 s.f.)

Worked Example 2 — Hall Effect in a Semiconductor

Question: A semiconductor slice has a thickness of $0.20\text{ mm}$ and a charge carrier density of $5.0 \times 10^{22}\text{ m}^{-3}$. When a current of $15\text{ mA}$ flows through it in a magnetic field of $0.40\text{ T}$, calculate the Hall voltage produced. (Charge on carrier $q = 1.6 \times 10^{-19}\text{ C}$)

Answer:

1. Identify and Convert Units: $B = 0.40\text{ T}$ $I = 15 \times 10^{-3}\text{ A}$ $n = 5.0 \times 10^{22}\text{ m}^{-3}$ $t = 0.20 \times 10^{-3}\text{ m}$ $q = 1.6 \times 10^{-19}\text{ C}$
2. Apply Equation: $V_H = \frac{BI}{ntq}$ $V_H = \frac{0.40 \times 15 \times 10^{-3}}{(5.0 \times 10^{22}) \times (0.20 \times 10^{-3}) \times (1.6 \times 10^{-19})}$
3. Calculation: $V_H = \frac{0.006}{1.6 \times 10^0} = 0.00375\text{ V}$ Final Answer: $V_H = 3.8\text{ mV}$ (to 2 s.f.)

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4. Key Equations

Equation	Symbols	Data Sheet?
$F = BQv \sin \theta$	$F$: Force, $B$: Flux Density, $Q$: Charge, $v$: Velocity	Yes
$V_H = \frac{BI}{ntq}$	$V_H$: Hall voltage, $n$: Number density, $t$: Thickness	Yes
$r = \frac{mv}{BQ}$	$r$: Radius, $m$: Mass, $v$: Velocity, $B$: Flux Density	No (Derive)
$v = \frac{E}{B}$	$v$: Selected velocity, $E$: Electric field, $B$: Magnetic field	No (Derive)
$E_k = \frac{p^2}{2m}$	$E_k$: Kinetic energy, $p$: Momentum ($mv$)	No

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5. Common Mistakes to Avoid

• ❌ Wrong: Using the direction of electron travel for the second finger in Fleming's Left-Hand Rule.
• ✅ Right: For electrons, the second finger must point opposite to the direction of motion (Conventional Current).
• ❌ Wrong: Assuming the magnetic force does work or increases the speed of the particle.
• ✅ Right: The magnetic force is always perpendicular to velocity; it only changes the direction. Speed and Kinetic Energy are constant.
• ❌ Wrong: Using the width ($d$) instead of thickness ($t$) in the Hall voltage formula.
• ✅ Right: $t$ is the dimension parallel to the magnetic field lines.
• ❌ Wrong: Forgetting to convert non-SI units like $\text{MeV}$ to $\text{J}$, $\text{mm}$ to $\text{m}$, or $\text{mT}$ to $\text{T}$.
• ✅ Right: Perform all unit conversions before substituting values into the equations.

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6. Exam Tips

1. Describing the Hall Effect: When asked to explain the origin of $V_H$, use the "Three-Step Logic":
   • (1) Magnetic force $F_B$ causes charge carriers to drift to one side.
   • (2) This separation of charge creates an electric field $E$ and an electric force $F_E$.
   • (3) Equilibrium is reached when $F_E = F_B$, resulting in a constant Hall voltage.
2. Drawing Paths: If a particle enters a field perpendicularly, draw a circular arc. If it enters at an angle, the path is a helix (though 9702 usually focuses on perpendicular entry). If the particle leaves the field, the exit path must be a straight line tangent to the circle at the exit point.
3. The "n" Factor: Be prepared to explain why metals are poor Hall probes. Metals have a very high $n$ ($\approx 10^{28}\text{ m}^{-3}$), making $V_H$ too small to measure accurately. Semiconductors have a lower $n$, making $V_H$ larger and easier to detect.
4. Derivations: You are frequently asked to derive $r = mv/BQ$ or $v = E/B$. Always start by stating which forces are being equated (e.g., "Magnetic force provides centripetal force" or "Electric force balances magnetic force").