Magnetic force on charges and the velocity selector

Answer

**F = qvB** when the charge moves at right angles to the field B (given as F = qvB sinθ). It is **zero** for a stationary charge.

Answer

**Perpendicular** to the velocity v (and to B). Because it is always sideways, it changes the charge's **direction** but never its **speed**.

Answer

The force F = qvB is always perpendicular to v, so it acts as a **centripetal force**, curving the path into a **circle** of radius r = mv/(qB).

Answer

$r = \dfrac{mv}{qB}$ — a heavier or faster particle curves in a bigger circle; a stronger field or bigger charge curves it tighter.

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A device with **crossed** electric and magnetic fields (E and B at right angles). Only charges of one speed pass straight through; the rest are deflected.

Answer

The electric and magnetic forces **balance**: **qE = qvB**. The net force is then zero, so the charge is undeflected.

Answer

$v = \dfrac{E}{B}$ — from qE = qvB, the charge q cancels.

Answer

**No** — q cancels in qE = qvB, so every undeflected particle has the same speed v = E ÷ B, whatever its charge or mass.

Answer

The magnetic force qvB is smaller, so the **electric force qE wins** and the charge is deflected the way qE points.

Answer

The magnetic force qvB is larger, so the **magnetic force wins** and the charge is deflected the other way.

Answer

v = E ÷ B = (2.0 × 10⁴) ÷ 0.10 = 2.0 × 10⁵ m s⁻¹.

Answer

The force is perpendicular to the motion, so it does **no work** on the charge — only its direction changes, not its speed.