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 \( v = 3 \times 10^6 \, m/s \) - Magnetic field strength \( B = 0.5 \, T \) - 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**: \[ F = (2 \times 10^{-6} \, C) \cdot (3 \times 10^6 \, m/s) \cdot (0.5 \, T) \cdot 1 \] 4. **Perform the Multiplication**: - First, calculate \( (2 \times 10^{-6}) \cdot (3 \times 10^6) \): \[ 2 \times 3 = 6 \quad \text{and} \quad 10^{-6} \times 10^{6} = 10^{0} = 1 \] So, \( (2 \times 10^{-6}) \cdot (3 \times 10^6) = 6 \). - Now, multiply by \( 0.5 \): \[ F = 6 \cdot 0.5 = 3 \, N \] ### Conclusion The magnitude of the magnetic force acting on the charged particle is **3 N**. Therefore, the correct option is **D**. ### Explanation of Other Options - **Option A (0.003 N)**: This value is too small and does not account for the correct multiplication of the charge, velocity, and magnetic field strength. - **Option B (0.03 N)**: This value is also incorrect; it seems to be a miscalculation of the factors involved. - **Option C (0.3 N)**: This value is still too low and does not reflect the correct application of the formula. ### Revision Summary - The magnetic force on a charged particle is calculated using \( F = q \cdot v \cdot B \cdot \sin(\theta) \). - For perpendicular motion, \( \sin(90^\circ) = 1 \). - Ensure to convert units correctly (e.g., microcoulombs to coulombs). - Always check the calculations step-by-step to avoid common pitfalls in multiplication and unit conversion.