Motion in electromagnetic fields

A wire carrying a **current** in a **magnetic field** feels a **force** (a sideways push) — the principle behind electric motors.

$F = BIL\sin\theta$ — force = field strength × current × length × sin(angle between current and field). Given in the data booklet.

The **angle between the current and the magnetic field**. When the wire is perpendicular to the field, θ = 90° and sin θ = 1, so F = BIL.

The **tesla (T)**.

When the current is **at right angles** to the field (θ = 90°, sin θ = 1).

When the current runs **along (parallel to)** the field (θ = 0°, sin 0° = 0).

On the **left** hand at right angles: **F**irst finger = **F**ield, se**C**ond finger = **C**urrent, thu**M**b = force/**M**otion.

All three are **mutually perpendicular** (at right angles to one another).

The **force reverses** direction. (Reversing the field does the same.)

The force **doubles** — F = BIL, so F is proportional to I.

F = BIL = 0.50 × 2.0 × 0.10 = 0.10 N.

**F = BIL sin θ**: largest at 90°, **zero** when the current runs along the field.

The gradient is **BL**, so B = gradient ÷ L (or B = ΔF/(LΔI)).

They all point **outwards** (or all inwards if the current reverses): the loop is stretched or squeezed, with no overall force.

$F = qE$ — the charge times the field strength. In the direction of the field for a **positive** charge, opposite it for a **negative** charge.

Two steps: force $F = qE$, then Newton's second law $a = \dfrac{F}{m} = \dfrac{qE}{m}$.

Because $a = \dfrac{qE}{m}$ and the electron's **mass m is tiny** (9.1 × 10⁻³¹ kg), so even a modest force gives an acceleration of order 10¹⁴ m s⁻².

A **parabola** — like a projectile. Constant velocity along the plates, constant acceleration across them.

A **positive** charge accelerates **along** the field; an **electron** (negative) accelerates **opposite** to the field.

**Constant velocity** — there is no force along the plates, so the horizontal speed never changes.

$s = \tfrac{1}{2}at^{2}$ (starting from rest sideways) — NOT s = vt, because the sideways motion is accelerated.

Yes — the booklet gives $E = \dfrac{F}{q}$; rearranged that is F = qE.

F = qE = (1.6 × 10⁻¹⁹)(2.0 × 10⁴) = 3.2 × 10⁻¹⁵ N.

a = F ÷ m = (8.0 × 10⁻¹⁶) ÷ (9.1 × 10⁻³¹) ≈ 8.8 × 10¹⁴ m s⁻².

Because the sideways motion is **accelerated** (constant force qE), not at constant velocity. Use s = ½at² instead.

**eV = ½mv²**, so v = √(2eV/m): the work done by the field becomes kinetic energy.

**No.** Its path is a straight line (from rest or along the field) or a **parabola** (fired across it).

For electrons, protons and ions: **mg is tiny next to qE**. Not for oil drops: there the net force is qE − mg.

**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.

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

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).

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

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.

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

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

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

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

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

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

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

From r = mv/(qB): **q/m = v/(Br)**. Measure r at a known v and B.

The **lighter** one: r = mv/(qB) ∝ m, so the heavier ion moves on the **bigger** circle.

They curve **opposite ways** round their circles; the radius also depends on each one's m/q and speed.