Section 5.2: Lorentz Force
The Lorentz force describes the force exerted on a moving charge in electric and magnetic fields. It is central to understanding particle motion in magnetic fields and the operation of devices like cyclotrons.
The Lorentz force on a particle of charge \(q\) moving with velocity \(\mathbf{v}\) in a magnetic field \(\mathbf{B}\) is given by:
- Force is perpendicular to both \(\mathbf{v}\) and \(\mathbf{B}\).
- Magnitude: \( F = q v B \sin\theta \), where \(\theta\) is the angle between velocity and field.
- Direction: Use the right-hand rule for positive charges.
Example 1
An electron (\(q = -1.6\times10^{-19}\) C, \(m = 9.11\times10^{-31}\) kg) moves at 2×10⁶ m/s perpendicular to a uniform magnetic field of 0.01 T. Find the magnitude of the Lorentz force and radius of its path.
Force: \( F = q v B = 1.6\times10^{-19} \times 2\times10^6 \times 0.01 = 3.2\times10^{-15}\ \text{N} \)
Radius: \( r = \frac{mv}{qB} = \frac{9.11\times10^{-31} \cdot 2\times10^6}{1.6\times10^{-19} \cdot 0.01} \approx 1.14\times10^{-3}\ \text{m} \)
Example 2
A proton moves at 1×10⁵ m/s at 30° to a 0.02 T magnetic field. Find the magnitude of the force on the proton.
\( F = q v B \sin\theta = 1.6\times10^{-19} \cdot 1\times10^5 \cdot 0.02 \cdot \sin30° \)
\( F = 3.2\times10^{-16} \cdot 0.5 = 1.6\times10^{-16}\ \text{N} \)
Practice Problems
- A particle of charge 2 μC moves at 3×10³ m/s perpendicular to a 0.05 T field. Find the Lorentz force.
- An alpha particle (mass 6.64×10⁻²⁷ kg, charge 3.2×10⁻¹⁹ C) moves in a circle in 0.01 T field. Find radius for 1×10⁵ m/s speed.
- A proton moves at 0.01 m/s at 60° to a uniform 0.1 T field. Compute the Lorentz force magnitude.
- Explain qualitatively why the speed of a charged particle in a uniform magnetic field does not change.
- Sketch the path of a negatively charged particle entering a uniform magnetic field perpendicular to velocity.