To determine the angle between the velocity \( v \) of a charge and the magnetic field \( B \) when the force on the charge is half of the maximum force, we need to understand how the magnetic force on a moving charge is calculated. ### Step 1: Understanding the Magnetic Force The magnetic force \( F \) on a charge \( q \) moving with velocity \( v \) in a magnetic field \( B \) is given by the formula: \[ F = qvB \sin(\theta) \] where: - \( F \) is the magnetic force, - \( q \) is the charge, - \( v \) is the velocity of the charge, - \( B \) is the magnetic field strength, - \( \theta \) is the angle between the velocity vector \( v \) and the magnetic field vector \( B \). ### Step 2: Maximum Force Condition The maximum force occurs when \( \sin(\theta) = 1 \), which happens at \( \theta = 90^\circ \). In this case, the formula simplifies to: \[ F_{\text{max}} = qvB \] ### Step 3: Finding Half of the Maximum Force We are looking for the angle \( \theta \) where the force \( F \) is half of the maximum force: \[ F = \frac{1}{2} F_{\text{max}} = \frac{1}{2} (qvB) \] Substituting the expression for \( F \): \[ qvB \sin(\theta) = \frac{1}{2} (qvB) \] ### Step 4: Simplifying the Equation We can cancel \( qvB \) from both sides (assuming \( q \), \( v \), and \( B \) are not zero): \[ \sin(\theta) = \frac{1}{2} \] ### Step 5: Solving for \( \theta \) The angle \( \theta \) that satisfies \( \sin(\theta) = \frac{1}{2} \) is: \[ \theta = 30^\circ \] ### Conclusion: Correct Option Thus, the angle between the velocity \( v \) and the magnetic field \( B \) when the force is half of the maximum force is: **Correct Option: C. 30°** ### Step 6: Analyzing Other Options - **Option A: 90°** - This is the angle for maximum force, not half. At this angle, \( \sin(90^\circ) = 1 \), so the force is \( F_{\text{max}} \). - **Option B: 45°** - At this angle, \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707 \). The force would be \( F = qvB \cdot 0.707 \), which is greater than half of the maximum force. - **Option D: 0°** - At this angle, \( \sin(0^\circ) = 0 \), meaning there is no force acting on the charge. This is not relevant to our question. ### Revision Summary - The magnetic force on a charge is given by \( F = qvB \sin(\theta) \). - Maximum force occurs at \( \theta = 90^\circ \) where \( F_{\text{max}} = qvB \). - To find when the force is half of the maximum, set \( \sin(\theta) = \frac{1}{2} \), leading to \( \theta = 30^\circ \). - Other angles (0°, 45°, 90°) do not yield half the maximum force.