Let's solve the problem stepwise, as requested. --- ## Step 1: **Force Vectors on the Particle (Part a)** **Explanation:** - A positively charged particle moving in a magnetic field experiences a force given by the Lorentz force: \[ \vec{F} = q\vec{v} \times \vec{B} \] - On the right side of the circle, the velocity \(\vec{v}\) is into the page (as shown), and \(\vec{B}\) is at an angle \(\theta\) from the vertical (upward and to the right). - Use the right-hand rule: Point your fingers in the direction of \(\vec{v}\) (into the page), curl them toward \(\vec{B}\) (upward/right), and your thumb points toward the force direction. - **Right side:** \(\vec{F}\) will point radially inward and upward (since \(\vec{v}\) is into the page and \(\vec{B}\) is up/right). - **Left side:** The velocity is out of the page, and \(\vec{B}\) is up/left. The force will also point radially inward and upward. **Summary:** On both sides, the magnetic force always points inward (toward the center of the circle) and has a vertical (upward) component due to the angle \(\theta\). --- ## Step 2: **Direction and Magnitude of Average Net Force (Part b)** **Explanation:** - The average net force (over a full circle) is the **centripetal force** needed to keep the particle in a circular path. - This force must point **toward the center of the circle** in the horizontal plane, but since the magnetic field is tilted, the force also has a vertical component. - Net force magnitude: \[ F = qvB\sin\theta \] (since the effective perpendicular component of \(\vec{B}\) to the velocity is \(B\sin\theta\)). - **Direction:** The net force vector (average over the orbit) points radially inward in the x-z plane (toward the center of the circle, plus a small vertical/upward component). --- ## **Final Answer** ### a) - The force vector on the right side points radially inward and upward (into the page, then up/right). - On the left side, it points radially inward and upward (out of the page, then up/left). ### b) - **Direction of average net force:** Radially inward with an upward component (in the x-z plane). - **Magnitude:** \[ F_{\text{net}} = qvB\sin\theta \] - **Vector:** Drawn from the center of the circle, pointing inward and slightly upward at angle \(\theta\) from the horizontal. --- Let me know if you'd like to proceed to parts (c) and (d)!

Let's continue solving the problem through all parts, following the structured approach. --- ## Step 3: **Acceleration of the Particle (Part c)** **Explanation:** - The acceleration \(\vec{a}\) of the particle can be determined using Newton's second law: \[ \vec{F}_{\text{net}} = m\vec{a} \] - Therefore, the magnitude of acceleration is given by: \[ a = \frac{F_{\text{net}}}{m} \] Substituting the expression for the net force: \[ a = \frac{qvB\sin\theta}{m} \] - **Direction:** The direction of \(\vec{a}\) is also toward the center of the circle (horizontal radial component) with a vertical component upward, similar to the net force. --- ## Step 4: **Period of Orbit and Vertical Velocity After One Orbit (Part d)** **Explanation:** 1. **Period \(T\) of the Orbit:** - For a particle in uniform circular motion, the period is given by: \[ T = \frac{2\pi r}{v} \] where \(r\) is the radius of the circular path. 2. **Vertical Velocity After One Orbit:** - The vertical component of velocity \(v_z\) after one orbit can be calculated using the vertical acceleration \(a\) over the time \(T\): \[ v_z = a \cdot T = \left(\frac{qvB\sin\theta}{m}\right) \cdot T \] Substituting \(T\): \[ v_z = \left(\frac{qvB\sin\theta}{m}\right) \cdot \left(\frac{2\pi r}{v}\right) = \frac{2\pi r qB\sin\theta}{m} \] 3. **Net Velocity Vector:** - The horizontal component remains constant as \(v_x = v\). - Therefore, the net velocity vector \(\vec{v}_{\text{net}}\) after one orbit is: \[ \vec{v}_{\text{net}} = \begin{pmatrix} v \\ \frac{2\pi r qB\sin\theta}{m} \end{pmatrix} \] where the first component is the horizontal velocity and the second component is the vertical velocity. --- ## **Final Answers** ### c) - **Acceleration \(a\):** \[ a = \frac{qvB\sin\theta}{m} \] - **Direction:** Toward the center of the circle with an upward component. ### d) 1. **Period \(T\):** \[ T = \frac{2\pi r}{v} \] 2. **Vertical Velocity \(v_z\) after one orbit:** \[ v_z = \frac{2\pi r qB\sin\theta}{m} \] 3. **Net Velocity Vector**: \[ \vec{v}_{\text{net}} = \begin{pmatrix} v \\ \frac{2\pi r qB\sin\theta}{m} \end{pmatrix} \] --- This concludes the solution for all parts of the problem. Let me know if you need further assistance!