Final answer:
The magnitude of the magnetic force on an electron with a charge of -1.60×10⁻¹⁹ C, moving at 2.20×10⁵ m/s through a 1.75 T magnetic field at an angle of 40.0° to the field, is approximately 7.83×10⁻¹⁵ N.
Step-by-step explanation:
The magnitude of the magnetic force on an electron moving through a magnetic field can be determined using the formula F = |q|vBsin(θ), where F is the magnetic force, |q| is the magnitude of the charge, v is the velocity of the electron, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector. For the given values in the student's question, we have an electron with charge -1.60×10⁻¹⁹ C, velocity 2.20×10⁵ m/s, and a magnetic field of 1.75 T. As the particle's velocity lies in the xy-plane directed at 50.0° to the x-axis, it makes an angle of 90° - 50.0° = 40.0° with the magnetic field directed in the y-direction.
Using these values, we calculate the magnetic force:
- F = |q|vBsin(θ)
- F = (1.60×10⁻¹⁹ C)(2.20×10⁵ m/s)(1.75 T)sin(40.0°)
- F = 1.60×10⁻¹⁹ C × 2.20×10⁵ m/s × 1.75 T × 0.6428
- F = 7.83×10⁻¹⁵ N
Therefore, the magnitude of the magnetic force on the electron is approximately 7.83×10⁻¹⁵ N.