To find the magnitude of the magnetic force acting on a proton moving in a magnetic field, we can use the formula for the magnetic force on a charged particle: \[ F = qvB \sin(\theta) \] Where: - \( F \) is the magnetic force, - \( q \) is the charge of the particle, - \( v \) is the velocity of the particle, - \( B \) is the magnetic field strength, - \( \theta \) is the angle between the velocity vector and the magnetic field vector. ### Step-by-Step Calculation 1. **Identify the Given Values**: - Charge of the proton, \( q = 1.6 \times 10^{-19} \, \text{C} \) - Velocity of the proton, \( v = 2 \times 10^6 \, \text{m/s} \) - Magnetic field strength, \( B = 0.5 \, \text{T} \) - Since the proton is moving perpendicular to the magnetic field, the angle \( \theta = 90^\circ \). 2. **Calculate \( \sin(\theta) \)**: - For \( \theta = 90^\circ \), \( \sin(90^\circ) = 1 \). 3. **Substitute the Values into the Formula**: \[ F = (1.6 \times 10^{-19} \, \text{C}) \times (2 \times 10^6 \, \text{m/s}) \times (0.5 \, \text{T}) \times 1 \] 4. **Perform the Multiplication**: - First, calculate \( 1.6 \times 10^{-19} \times 2 \times 10^6 \): \[ 1.6 \times 2 = 3.2 \] \[ 10^{-19} \times 10^6 = 10^{-13} \] Thus, \( 1.6 \times 10^{-19} \times 2 \times 10^6 = 3.2 \times 10^{-13} \, \text{C m/s} \). - Now multiply by \( 0.5 \): \[ F = 3.2 \times 10^{-13} \times 0.5 = 1.6 \times 10^{-13} \, \text{N} \] ### Conclusion The magnitude of the magnetic force acting on the proton is: \[ \boxed{1.6 \times 10^{-13} \, \text{N}} \] ### Explanation of Other Options - **Option A**: [blank] - No value provided, so cannot be considered. - **Option C**: [blank] - No value provided, so cannot be considered. - **Option D**: \( 1.6 \times 10^{-14} \, \text{N} \) - This value is incorrect because it is an order of magnitude lower than the calculated force. The calculations clearly show that the force is \( 1.6 \times 10^{-13} \, \text{N} \), which is ten times greater than this option. ### Revision Summary - The magnetic force on a charged particle is calculated using \( F = qvB \sin(\theta) \). - For a proton moving perpendicular to the magnetic field, \( \sin(90^\circ) = 1 \). - Substitute the values of charge, velocity, and magnetic field strength to find the force. - The correct answer is \( 1.6 \times 10^{-13} \, \text{N} \).