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, \( q = 1.6 \times 10^{-19} \) C - Velocity of the proton, \( v = 2 \times 10^6 \) m/s - Magnetic field strength, \( B = 0.5 \) 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 \( 1.6 \times 10^{-19} \times 2 \times 10^6 \): \[ 1.6 \times 2 = 3.2 \] \[ 10^{-19} \times 10^6 = 10^{-13} \] So, \( 1.6 \times 10^{-19} \times 2 \times 10^6 = 3.2 \times 10^{-13} \). - Now, multiply by \( 0.5 \): \[ 3.2 \times 10^{-13} \times 0.5 = 1.6 \times 10^{-13} \, \text{N} \] 5. **Final Result**: \[ F = 1.6 \times 10^{-13} \, \text{N} \] ### Conclusion The magnitude of the magnetic force acting on the proton is \( 1.6 \times 10^{-13} \) N. Therefore, the correct option is **A**. ### Explanation of Other Options - **Option B: \( 1.6 \times 10^{-14} \) N**: This value is too small. It likely results from an error in the multiplication or misunderstanding of the charge or velocity. - **Option C: \( 1.6 \times 10^{-12} \) N**: This value is too large, possibly due to an incorrect multiplication of the charge and velocity. - **Option D: \( 2.0 \times 10^{-13} \) N**: This value is also incorrect, as it does not match the calculated force and suggests a miscalculation in the application of the formula. ### Revision Summary - Use the formula \( F = qvB \sin(\theta) \) to calculate magnetic force. - Ensure to use the correct values for charge, velocity, and magnetic field strength. - Remember that \( \sin(90^\circ) = 1 \) when the velocity is perpendicular to the magnetic field. - Carefully perform multiplication and pay attention to scientific notation to avoid common pitfalls.