recall and use F = BQv sin θ

Fundamental Equation

When a charge \$Q\$ moves with velocity \$v\$ through a magnetic field \$B\$ at an angle \$\theta\$, the magnetic force is

\$F = BQv \sin\theta\$

For a straight wire carrying current \$I\$ of length \$L\$, the same idea gives

\$F = BIL \sin\theta\$

⚠️ Remember: \$F\$ is a vector; its direction is given by the right‑hand rule.

Variables Explained

• \$B\$ – Magnetic field strength (tesla, T)
• \$Q\$ – Charge moving through the field (coulombs, C)
• \$v\$ – Speed of the charge (metres per second, m/s)
• \$\theta\$ – Angle between \$v\$ and \$B\$ (degrees or radians)
• \$I\$ – Current in the conductor (amperes, A)
• \$L\$ – Length of conductor in the field (metres, m)

🔌 Analogy: Think of the conductor as a train moving through a magnetic “track” – the magnetic field pushes on the moving charges, creating a force that can move the whole train.

Worked Example

Suppose a 0.5 m long copper wire moves at 20 m/s through a uniform magnetic field of 0.5 T. The wire carries a current of 2 A. The field is perpendicular to the wire.

Since \$\sin 90^\circ = 1\$, the force is

\$F = BIL = 0.5 \times 2 \times 0.5 = 0.5\ \text{N}\$

🧲 The force pushes the wire perpendicular to both the field and its motion.

Exam Tips & Tricks

1. Always check the units – Tesla × Coulomb × m/s gives newtons.
2. Remember that \$\sin 90^\circ = 1\$; this often simplifies calculations.
3. Use the right‑hand rule to determine the direction of the force.
4. When given a current instead of charge, convert using \$Q = I\Delta t\$ or use \$F = BIL\$ directly.
5. Draw a quick sketch of the field, wire, and force vectors – visualising helps avoid mistakes.
6. Practice with different angles: \$\theta = 0^\circ\$ gives no force, \$\theta = 45^\circ\$ gives \$F = BQv/\sqrt{2}\$.

📚 Remember: The key to success is understanding the relationship between motion, magnetic fields, and the resulting force.