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: \[ F = q \cdot v \cdot B \cdot \sin(\theta) \] Where: - \( F \) is the magnetic force (in Newtons, N) - \( q \) is the charge of the particle (in Coulombs, C) - \( v \) is the velocity of the particle (in meters per second, m/s) - \( B \) is the magnetic field strength (in Teslas, T) - \( \theta \) 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 \( v = 5 \, m/s \) - Magnetic field strength \( B = 0.3 \, T \) - 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**: \[ F = (2 \times 10^{-6} \, C) \cdot (5 \, m/s) \cdot (0.3 \, T) \cdot 1 \] 4. **Perform the Calculation**: \[ F = 2 \times 10^{-6} \cdot 5 \cdot 0.3 \] \[ F = 2 \times 5 \times 0.3 \times 10^{-6} \] \[ F = 3 \times 10^{-6} \, N \] \[ F = 0.000003 \, N = 0.003 \, N \] ### Conclusion The magnitude of the magnetic force acting on the particle is **0.003 N**. Therefore, the correct option is **A**. ### Explanation of Other Options - **Option B (0.03 N)**: This value is ten times larger than the correct answer. It may arise from a miscalculation, possibly by incorrectly multiplying the charge or velocity. - **Option C (0.3 N)**: This value is also incorrect and is significantly larger than the correct answer. It could result from misunderstanding the relationship between the variables or misapplying the formula. - **Option D (0.6 N)**: This is the largest option and is incorrect. It likely results from a calculation error or misunderstanding of the sine function in the context of the angle. ### Common Pitfalls - **Misunderstanding the Angle**: Remember that the angle \( \theta \) is crucial. If the particle is not moving perpendicular to the magnetic field, you must use the correct sine value. - **Unit Conversion**: Ensure that all units are consistent, especially when dealing with microcoulombs and converting them to coulombs. - **Forgetting to Use the Sine Function**: If the angle is not \( 90^\circ \), you must calculate \( \sin(\theta) \) correctly. ### Revision Summary - Use the formula \( F = q \cdot v \cdot B \cdot \sin(\theta) \) for magnetic force. - Ensure all units are in SI units (C, m/s, T). - Remember that \( \sin(90^\circ) = 1 \) when the velocity is perpendicular to the magnetic field. - Double-check calculations to avoid common errors in multiplication and unit conversion.