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Lorentz Force

CSCA Lorentz Force study guide organized around the publicly available CSCA syllabus. Practice Physics questions on aicsca.com.

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Topic: Electromagnetism > Magnetic Field > Lorentz Force

1. Core Concept: Lorentz Force

**Lorentz Force** is the force exerted by a magnetic field on a **moving charge**. It is the microscopic origin of Ampere's force.

**CSCA Exam Focus**:

* **No Work Done**: The Lorentz force is always perpendicular to the velocity. It changes the **direction** of velocity but **NEVER changes the speed** (Kinetic energy remains constant).

* **Direction Logic**: Distinguishing between positive and negative charges is the most common pitfall.

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2. Core Formulas

(1) Magnitude

$$F = |q| v B \sin\theta$$

* $F$: Lorentz Force (N).

* $q$: Charge (C), use **absolute value** for magnitude calculation.

* $v$: Velocity (m/s).

* $B$: Magnetic Flux Density (T).

* $\theta$: Angle between **Velocity $\vec{v}$** and **Field $\vec{B}$**.

**Special Cases**:

* $\vec{v} \perp \vec{B}$ ($\theta = 90^\circ$): $F = qvB$ (**Maximum**, leads to circular motion).

* $\vec{v} \parallel \vec{B}$ ($\theta = 0^\circ$): $F = 0$ (No force, linear motion).

(2) Direction: Left-Hand Rule

**Use Left Hand for Force**. Pay attention to the charge sign!

* **Step 1**: Open **Left Hand**, let B-field lines **pierce the palm**.

* **Step 2**: Point four fingers based on charge type:

* **Positive (+)**: Fingers point **WITH velocity** ($v$).

* **Negative (-)**: Fingers point **OPPOSITE to velocity** ($-v$).

* **Step 3**: **Thumb** points to **Lorentz Force** ($F$).

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3. Motion of Charged Particles in Uniform Magnetic Field

This is the core model for calculation problems. Usually, gravity is ignored.

Model A: Uniform Circular Motion

When incident velocity $\vec{v} \perp \vec{B}$.

* **Dynamics Equation**: Lorentz force provides centripetal force.

$$qvB = m \frac{v^2}{r} = m \omega^2 r = m v \frac{2\pi}{T}$$

* **Two Key Derivations (Memorize)**:

1. **Radius**:

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

* Inference: Higher speed $v$ $\rightarrow$ Larger radius $r$.

2. **Period**:

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

* Inference: **Period is INDEPENDENT of speed and radius**! It depends only on charge-to-mass ratio ($q/m$) and field ($B$).

* *Basis for the Cyclotron.*

Model B: Helical Motion

When incident velocity $\vec{v}$ is at an angle $\theta$ to $\vec{B}$.

* **Decomposition**:

* $v_{\parallel} = v \cos\theta$: Parallel to B $\rightarrow$ **Uniform Linear Motion**.

* $v_{\perp} = v \sin\theta$: Perpendicular to B $\rightarrow$ **Uniform Circular Motion**.

* **Trajectory**: Helix (Spiral).

* **Pitch**: The axial distance covered in one period.

$$h = v_{\parallel} \cdot T = (v \cos\theta) \cdot \frac{2\pi m}{qB}$$

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4. Typical Examples

**Example 1: The Negative Charge Trap**

**Problem**: A stream of electrons moves horizontally to the right into a uniform magnetic field pointing INTO the page. What is the direction of the Lorentz force?

**Solution**:

1. **Field**: In $\rightarrow$ Pierces left palm (Palm faces out).

2. **Fingers**: Electrons are **Negative**! Velocity is Right, so **fingers must point LEFT**.

3. **Thumb**: Points **Vertically Down**.

**Answer**: Downward.

**Example 2: Radius and Kinetic Energy**

**Problem**: A proton ($H^+$) and an alpha particle ($He^{2+}$, mass 4m, charge 2q) enter the same field perpendicularly with the **same Kinetic Energy**. Find ratio of radii.

**Solution**:

1. **Formula**: Need $r$ in terms of $E_k$.

From $E_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}}$, substitute into $r = \frac{mv}{qB}$:

$$r = \frac{m \sqrt{2E_k/m}}{qB} = \frac{\sqrt{2mE_k}}{qB}$$

2. **Ratio**:

$B, E_k$ are same, so $r \propto \frac{\sqrt{m}}{q}$.

$$\frac{r_H}{r_{\alpha}} = \frac{\sqrt{1}/1}{\sqrt{4}/2} = \frac{1}{1} = 1:1$$

**Answer**: Ratio is 1:1.

**Example 3: Time of Motion**

**Problem**: A particle turns through a center angle of $60^\circ$ in the magnetic field. Find the time $t$.

**Solution**:

Use the Period $T$.

$$t = \frac{\theta}{360^\circ} \times T = \frac{60}{360} T = \frac{1}{6} T$$

Substitute $T = \frac{2\pi m}{qB}$ to solve.