Let's break down part (b) of the problem into two steps with explanations and provide the final answer. --- ### **Step 1: Calculate the Magnetic Force Acting on the Proton** The magnetic force (\( F \)) on a moving charge in a magnetic field is given by: \[ F = qvB \sin \theta \] Since the proton's velocity is perpendicular to the magnetic field (\(\theta = 90^\circ\)), \(\sin \theta = 1\): - \(q = 1.60 \times 10^{-19}\) C - \(v = 1.00 \times 10^4\) m/s - \(B = 1.00 \times 10^{-3}\) T \[ F = (1.60 \times 10^{-19} \text{ C})(1.00 \times 10^4 \text{ m/s})(1.00 \times 10^{-3} \text{ T}) \] \[ F = 1.60 \times 10^{-18} \text{ N} \] **Explanation:** The force is perpendicular to both the velocity and the magnetic field. By the right-hand rule, for a positive charge, the force will point toward the center of the circular path (providing the centripetal force). --- ### **Step 2: Calculate the Acceleration of the Proton** From Newton's Second Law: \[ F = ma \implies a = \frac{F}{m} \] The mass of a proton (\(m\)) is \(1.67 \times 10^{-27}\) kg. \[ a = \frac{1.60 \times 10^{-18} \text{ N}}{1.67 \times 10^{-27} \text{ kg}} \] \[ a \approx 9.58 \times 10^8 \text{ m/s}^2 \] **Explanation:** This acceleration is the centripetal acceleration, always directed toward the center of the semicircular path. --- ### **Final Answers** - **Magnitude of force:** \(1.60 \times 10^{-18}\) N - **Magnitude of acceleration:** \(9.58 \times 10^8\) m/s² - **Direction:** Both the force and acceleration point toward the center of the semicircular path (perpendicular to the velocity at each point).