Magnetic Fields and the Lorentz Force

The mathematical expression for the Lorentz force is:

$$\mathbf{F} = q\mathbf{E} + q(\mathbf{v} \times \mathbf{B})$$

Where:

• $q$ is the electric charge of the particle.
• $\mathbf{E}$ is the electric field vector.
• $\mathbf{v}$ is the velocity vector of the particle.
• $\mathbf{B}$ is the magnetic field vector.
• $\times$ denotes the vector cross product.

Breaking Down the Components

1. The Electric Component ($q\mathbf{E}$): This force acts in the direction of the electric field (for a positive charge) or opposite to it (for a negative charge). Crucially, this force does not depend on the velocity of the particle. Even if the particle is standing still, the electric field will push it.

2. The Magnetic Component ($q(\mathbf{v} \times \mathbf{B})$): This part of the force is more complex. Because it involves a cross product, the force is always perpendicular to both the velocity of the particle and the direction of the magnetic field.

Key Insight: Because the magnetic force is always perpendicular to the velocity ($\mathbf{F}_B \perp \mathbf{v}$), the magnetic field can change the direction of a particle's motion, but it can never change its speed or kinetic energy.

Determining Direction: The Right-Hand Rule

Since the magnetic force is a cross product, students often struggle to determine which way the particle will be pushed. The standard tool for this is the Right-Hand Rule (RHR):

1. Point your fingers (or thumb, depending on the specific version used in your textbook) in the direction of the velocity $\mathbf{v}$.
2. Curl your fingers toward the direction of the magnetic field $\mathbf{B}$.
3. Your thumb will then point in the direction of the force $\mathbf{F}$ for a positive charge.

Warning: If the particle is an electron (which has a negative charge), the force will act in the exact opposite direction of what your right hand indicates.

Worked Example: Particle in a Uniform Magnetic Field

Let's apply these concepts to a practical scenario.

Problem: A proton (charge $q = +1.6 \times 10^{-19}$ C) enters a uniform magnetic field of $0.5$ T. The proton is traveling at a velocity of $2.0 \times 10^6$ m/s, and its velocity is perpendicular to the magnetic field lines. Calculate the magnitude of the magnetic force acting on the proton.

Solution: Since the velocity is perpendicular to the magnetic field, the angle $\theta$ between $\mathbf{v}$ and $\mathbf{B}$ is $90^\circ$. The magnitude of the magnetic force is given by:

$$F_B = |q|vB\sin(\theta)$$

Substituting the given values:

$$\begin{aligned} F_B &= (1.6 \times 10^{-19}\text{ C}) \cdot (2.0 \times 10^6\text{ m/s}) \cdot (0.5\text{ T}) \cdot \sin(90^\circ) \\ F_B &= (1.6 \times 10^{-19}) \cdot (1.0 \times 10^6) \cdot 1 \\ F_B &= 1.6 \times 10^{-13}\text{ N} \end{aligned}$$

The magnitude of the force is $1.6 \times 10^{-13}$ Newtons.

Summary Table: Electric vs. Magnetic Forces