# Problem 1: Magnetic Forces on Charges and Currents --- ## (a) **Why a Magnetic Field Exerts No Force on a Stationary Charge (Lorentz Force Law)** **Lorentz Force Law:** \[ \vec{F} = q (\vec{E} + \vec{v} \times \vec{B}) \] - \( \vec{F} \): total force on the charge - \( q \): charge - \( \vec{E} \): electric field - \( \vec{v} \): velocity of the charge - \( \vec{B} \): magnetic field **Explanation:** - For a stationary charge, \( \vec{v} = \). - The magnetic part of the force is \( q (\vec{v} \times \vec{B}) \). - Thus, \( q ( \times \vec{B}) = \). - **Conclusion:** **A magnetic field exerts no force on a stationary charge because the magnetic force depends on the velocity of the charge. If velocity is zero, the force is zero.** --- ## (b) **Magnetic Force Perpendicularity (Mathematical Proof, Trajectory & Force Diagram)** **Mathematical Proof:** \[ \vec{F}_B = q (\vec{v} \times \vec{B}) \] - The cross product \( \vec{v} \times \vec{B} \) is always perpendicular to both \( \vec{v} \) and \( \vec{B} \). **Diagram:** - If \( \vec{v} \) is to the right and \( \vec{B} \) is into the page, \( \vec{F}_B \) is upward (right-hand rule). **Trajectory:** - Since the force is always perpendicular to velocity, it causes the particle to move in a **circle** (or helix if there is a velocity component parallel to \( \vec{B} \)). **Summary:** - **The magnetic force is always perpendicular to the velocity of the charged particle.** --- ## (c) **Magnetic Forces Do No Work** **Work Done:** \[ W = \int \vec{F} \cdot d\vec{s} \] For magnetic force: \[ \vec{F}_B = q (\vec{v} \times \vec{B}) \] \[ \vec{F}_B \cdot \vec{v} = q (\vec{v} \times \vec{B}) \cdot \vec{v} = \] (because the cross product is perpendicular to \( \vec{v} \)) **Conclusion:** - **Magnetic forces do no work** because the force is always perpendicular to the displacement of the particle. --- ## (d) **Trajectories in Fields** ### 1. **Uniform Magnetic Field (\( \vec{B} \) constant)** - **Trajectories:** - If \( \vec{v} \perp \vec{B} \): **Circle** - If \( \vec{v} \) has component along \( \vec{B} \): **Helix** (circular motion + constant velocity along \( \vec{B} \)) **Diagram:** -  - *[Alt: Circular and helical paths of charged particles in a magnetic field]* ### 2. **Combined Electric (\( \vec{E} \)) and Magnetic (\( \vec{B} \)) Fields** - If \( \vec{E} \) and \( \vec{B} \) are perpendicular: - **E × B drift:** Particle drifts at velocity \( \vec{v}_d = \frac{\vec{E} \times \vec{B}}{B^2} \) - The trajectory is a **cycloid**. **Diagram:** -  - *[Alt: Cycloidal trajectory of a charged particle in crossed E and B fields]* --- ## (e) **Force Between Parallel Current-Carrying Conductors** **Physical Reason:** - Each wire produces a magnetic field that exerts a force on the other wire. **Force Derivation:** - Magnetic field at distance \( r \) from a long straight wire: \[ B = \frac{\mu_ I_1}{2\pi r} \] - Force per unit length on wire 2 due to wire 1: \[ F_{21}/L = I_2 B = I_2 \frac{\mu_ I_1}{2\pi r} \] **Interpretation:** - **If currents are in the same direction:** wires attract. - **If currents are in opposite directions:** wires repel. **Diagram:** -  - *[Alt: Parallel wires attracting or repelling depending on direction of current]* --- ## **Summary Table** | Subpart | Main Point | Equation(s) | Diagram/Result | |---------|------------|-------------|---------------| | (a) | No force on stationary charge | \( \vec{F} = q (\vec{v} \times \vec{B}) \) | None | | (b) | Force ⟂ velocity | Cross-product | Circular/Helical Trajectory | | (c) | No work by \( \vec{F}_B \) | \( \vec{F}_B \cdot \vec{v} = \) | None | | (d) | Trajectories in fields | | Circles, helices, cycloids | | (e) | Parallel wires attract/repel | \( F/L = \frac{\mu_ I_1 I_2}{2\pi r} \) | Attraction/Repulsion | --- **Final Answer:** Each part above provides the step-by-step solution, mathematical expressions, and physical interpretation as required. Let me know if you need more detailed diagrams or further explanations!