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, - \( q \) 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 \( v = 3 \times 10^6 \, m/s \) - Magnetic field strength \( B = 0.5 \, T \) - Since the velocity is 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**: \[ F = (2 \times 10^{-6} \, C) \cdot (3 \times 10^6 \, m/s) \cdot (0.5 \, T) \cdot 1 \] 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 \( 0.5 \): \[ F = 6 \cdot 0.5 = 3 \, N \] 5. **Final Calculation**: - Since we have \( F = 3 \, N \), we need to check if we made any mistakes in the options provided. ### Correct Option The calculated force is \( 3 \, N \), which does not match any of the options provided (A, B, C, D). However, if we consider the possibility of a miscalculation or misinterpretation of the options, we can check the calculations again. ### Re-evaluation of Options - **Option A: 0.003 N** - This is too small compared to our calculated force. - **Option B: 0.009 N** - This is also too small. - **Option C: 0.030 N** - This is still too small. - **Option D: 0.150 N** - This is also too small. ### Conclusion It appears that the options provided do not include the correct answer based on the calculations. The correct answer based on the calculations is \( 3 \, N \). ### Summary - The magnetic force on a charged particle is calculated using \( F = q \cdot v \cdot B \cdot \sin(\theta) \). - For a charge of \( +2 \, \mu C \), velocity of \( 3 \times 10^6 \, m/s \), and magnetic field of \( 0.5 \, T \) with \( \theta = 90^\circ \), the force is \( 3 \, N \). - The options provided do not match the calculated force, indicating a possible error in the question or options. - Always ensure to check the units and calculations carefully to avoid mistakes.