Magnetic Poles:

• North and South (like charges, but no magnetic monopoles!)
• Like poles repel, opposite poles attract
• Field lines: N → S outside magnet

Force on Moving Charge

$\vec{F} = q\vec{v} \times \vec{B}$

Magnitude: $F = qvB \sin\theta$

where $\theta$ is angle between $\vec{v}$ and $\vec{B}$.

Direction: Right-Hand Rule #1

1. Point fingers along $\vec{v}$ (velocity)
2. Curl fingers toward $\vec{B}$ (field)
3. Thumb points along $\vec{F}$ (force) for positive charge

For negative charge: Force is opposite!

Key Points:

• $\theta = 90°$: Maximum force ($F = qvB$)
• $\theta = 0°$ or $180°$: No force (parallel or anti-parallel)
• Force is perpendicular to both $\vec{v}$ and $\vec{B}$
• No work done (F ⊥ v, so F·v = 0)

Circular Motion in B Field

If $\vec{v} \perp \vec{B}$, charge moves in circle!

Magnetic force provides centripetal force: $qvB = \frac{mv^2}{r}$

Radius of orbit: $r = \frac{mv}{qB}$

Period: $T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$

Frequency (cyclotron frequency): $f = \frac{1}{T} = \frac{qB}{2\pi m}$

💡 Note: Period and frequency are independent of v and r!

Applications: Mass Spectrometer

Separates ions by mass:

1. Ions accelerated through V
2. Enter perpendicular B field
3. Radius depends on m/q ratio
4. Heavier ions have larger radius

$r = \frac{1}{B}\sqrt{\frac{2mV}{q}}$

Force on Current-Carrying Wire

Current = moving charges, so wire in B field experiences force!

$\vec{F} = I\vec{L} \times \vec{B}$

Magnitude: $F = BIL \sin\theta$

where:

• I = current (A)
• L = length of wire in field (m)
• $\theta$ = angle between wire and B field

Direction: Right-Hand Rule #2

1. Point fingers along current direction
2. Curl toward $\vec{B}$
3. Thumb = force direction

Maximum force: Wire perpendicular to B ($F = BIL$) No force: Wire parallel to B

Force Between Parallel Wires

Two parallel wires carrying currents:

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$

where:

• $\mu_0 = 4\pi \times 10^{-7}$ T·m/A (permeability of free space)
• d = distance between wires

Same direction currents: Attract Opposite direction currents: Repel

💡 This defines the Ampere! 1 A = current that produces 2×10⁻⁷ N/m force between parallel wires 1 m apart.

Torque on Current Loop

Rectangular loop (sides a and b) in B field:

$\tau = NIAB \sin\theta$

where:

• N = number of turns
• I = current
• A = area of loop
• $\theta$ = angle between normal to loop and B field

Magnetic dipole moment: $\mu = NIA$ (A·m²)

$\tau = \mu B \sin\theta$

Maximum torque: Loop perpendicular to B ($\theta = 90°$) Equilibrium: Loop parallel to B ($\theta = 0°$)

Applications: Electric Motor

1. Current through loop in B field
2. Torque rotates loop
3. Commutator reverses current every half-turn
4. Continuous rotation!

Magnetic Field of Long Straight Wire

$B = \frac{\mu_0 I}{2\pi r}$

Direction: Right-Hand Rule #3 (Grip Rule)

• Thumb along current
• Fingers curl around wire → B field direction

Field forms circles around wire.

Magnetic Field of Solenoid

Long coil of wire (N turns, length L):

Inside (uniform field): $B = \mu_0 nI = \mu_0 \frac{N}{L} I$

where n = N/L = turns per unit length

Outside: B ≈ 0

Solenoid is like bar magnet with N and S poles!

Problem-Solving Strategy

1. Identify: Charge or current in B field?
2. Find angle $\theta$ between v (or I) and B
3. Calculate magnitude: F = qvB sinθ or F = BIL sinθ
4. Find direction: Right-hand rule
5. For circular motion: Use r = mv/qB

Common Mistakes

❌ Forgetting sin θ term ❌ Wrong right-hand rule (for negative charges) ❌ Confusing E and B field formulas ❌ Saying magnetic force does work (it doesn't! F ⊥ v) ❌ Using v when parallel to B (F = 0) ❌ Wrong units (convert G to T, cm to m)

Part (a): Magnetic force

$F = |q|vB \sin\theta = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.50)(1)$ $F = 1.6 \times 10^{-13} \text{ N}$

(Direction: Use right-hand rule, then reverse for negative charge)

Part (b): Radius of circular path

$r = \frac{mv}{|q|B} = \frac{(9.11 \times 10^{-31})(2.0 \times 10^6)}{(1.6 \times 10^{-19})(0.50)}$ $r = \frac{1.82 \times 10^{-24}}{8.0 \times 10^{-20}} = 2.3 \times 10^{-5} \text{ m} = 0.023 \text{ mm}$

Very small radius!

Answer:

• (a) F = 1.6 × 10⁻¹³ N
• (b) r = 0.023 mm