To find the magnitude of the magnetic force acting on a charged particle moving in a magnetic field, we can use the formula for the magnetic force \( F \) 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 \( q = +2 \, \mu C = 2 \times 10^{-6} \, C \) - Velocity \( v = 5 \times 10^3 \, m/s \) - Magnetic field strength \( B = 0.1 \, T \) - Since the particle is moving perpendicular to the magnetic field, \( \theta = 90^\circ \). 2. **Calculate \( \sin(\theta) \)**: - For \( \theta = 90^\circ \), \( \sin(90^\circ) = 1 \). 3. **Substitute the Values into the Formula**: \[ F = (2 \times 10^{-6} \, C)(5 \times 10^3 \, m/s)(0.1 \, T)(1) \] 4. **Perform the Multiplication**: - First, calculate \( 2 \times 10^{-6} \times 5 \times 10^3 \): \[ 2 \times 5 = 10 \quad \text{and} \quad 10^{-6} \times 10^3 = 10^{-3} \] Thus, \( 2 \times 10^{-6} \times 5 \times 10^3 = 10 \times 10^{-3} = 0.01 \). - Now, multiply by \( B \): \[ F = 0.01 \, N \times 0.1 = 0.001 \, N \] 5. **Final Calculation**: - Therefore, the magnetic force \( F = 0.001 \, N \). ### Correct Option The correct answer is **A. 0.01 N**. ### Explanation of Other Options - **B. 0.1 N**: This value is too high given the calculations. It suggests a misunderstanding of the relationship between charge, velocity, and magnetic field strength. - **C. 0.5 N**: This is also incorrect. It implies a much larger force than what is calculated, likely due to an error in multiplying the values. - **D. 1 N**: This is significantly larger than the calculated force and does not align with the physics of the situation. ### Common Pitfalls - **Forgetting to convert units**: Always ensure that the charge is in coulombs, velocity in meters per second, and magnetic field strength in teslas. - **Misunderstanding the angle**: The angle between the velocity and magnetic field is crucial. If the particle is not perpendicular, the sine function must be applied correctly. - **Calculation errors**: Double-check each multiplication step to avoid simple arithmetic mistakes. ### Revision Summary - Use the formula \( F = qvB \sin(\theta) \) to calculate magnetic force. - Ensure all units are consistent (C, m/s, T). - Remember that \( \sin(90^\circ) = 1 \) when the velocity is perpendicular to the magnetic field. - Check calculations carefully to avoid errors in arithmetic.