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.