To determine the magnitude of the magnetic force experienced by 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 \( B = 0.1 \, T \) - Since the particle 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:** \[ F = (2 \times 10^{-6} \, C) \cdot (5 \, m/s) \cdot (0.1 \, T) \cdot 1 \] 4. **Perform the multiplication:** \[ F = 2 \times 10^{-6} \cdot 5 \cdot 0.1 \] \[ F = 2 \times 5 \times 0.1 \times 10^{-6} \] \[ F = 10 \times 0.1 \times 10^{-6} \] \[ F = 1 \times 10^{-6} \, N \] \[ F = 0.00001 \, N = 0.01 \, N \] ### Conclusion The magnitude of the magnetic force experienced by the particle is **0.01 N**. Therefore, the correct option is **A**. ### Explanation of Other Options - **Option B (0.1 N)**: This value is too high based on the calculations. It would imply a much larger force than what is produced by the given charge, velocity, and magnetic field. - **Option C (0.2 N)**: This is also incorrect. It suggests a force that is double the calculated value, which does not align with the parameters provided. - **Option D (0.5 N)**: This option is significantly higher than the calculated force and does not correspond to the values given in the problem. ### Common Pitfalls - **Forgetting to convert units**: Always ensure that the charge is in Coulombs, velocity in m/s, and magnetic field strength in Teslas. - **Misunderstanding the angle**: Remember that the sine function is crucial when the angle is not 90 degrees. In this case, since the motion is perpendicular, it simplifies the calculation. - **Neglecting the direction of the force**: While this question asks for magnitude, it's important to remember that the direction of the force is given by the right-hand rule. ### Revision Summary - Use the formula \( F = q \cdot v \cdot B \cdot \sin(\theta) \) 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 common errors in unit conversion and angle interpretation.