Flashcards · IB Diploma Programme Physics · Topic 4

D. Fields flashcards

35 cards on Topic 4, subtopics D.1 to D.4. Pick a subtopic or study them all.

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D.1 Gravitational fields

Kepler’s three laws: First: orbits are ellipses with the Sun at a focus. Second: the Sun–planet line sweeps out equal areas in equal times. Third: T² ∝ r³.
Newton’s law of gravitation: F = Gm₁m₂/r², attractive, along the line joining the centres (r centre to centre). G = 6.67 × 10⁻¹¹ N m² kg⁻².
Gravitational field strength: g = F/m, the force per unit mass on a small point mass (N kg⁻¹). For a point mass or uniform sphere, g = GM/r².
g at 500 km above a planet of mass 6.0 × 10²³ kg and radius 3.0 × 10⁶ m?: r = 3.5 × 10⁶ m, g = 6.67 × 10⁻¹¹ × 6.0 × 10²³ ÷ (3.5 × 10⁶)² = 3.27 N kg⁻¹
A second moon orbits at twice the radius of a moon with period 2.0 days. Its period?: T ∝ r^(3/2), so T = 2.0 × 2^(3/2) = 5.7 days
When can an extended body be treated as a point mass?: When it is spherically symmetric (such as a uniform sphere), or the separation is much larger than the size of either body.
Gravitational potential at a point: The work done per kilogram to bring a small test mass from infinity to the point: Vg = −GM/r. Zero at infinity, negative everywhere else. (HL only)
Escape speed: The speed at a point that lets a body reach infinity with no kinetic energy left: vesc = √(2GM/r), independent of the body’s mass. (HL only)
Orbital speed and total energy of a satellite in a circular orbit: v = √(GM/r). Ek = GMm/2r, Ep = −GMm/r, so E = −GMm/2r. (HL only)
Atmospheric drag acts on a satellite. Effect on its orbit?: Total energy decreases, so the radius decreases and the speed increases (Ek rises while Ep falls by twice as much). (HL only)

D.2 Electric and magnetic fields

Coulomb’s law: F = kq₁q₂/r², k = 1/(4πε₀) ≈ 8.99 × 10⁹ N m² C⁻². Like charges repel, unlike attract.
Force between 2.0 μC and 3.0 μC point charges 0.10 m apart?: F = 8.99 × 10⁹ × 2.0 × 10⁻⁶ × 3.0 × 10⁻⁶ ÷ 0.10² = 5.4 N (repulsive)
What did Millikan’s experiment show?: Balanced charged oil drops (qE = mg) always carried whole-number multiples of e = 1.60 × 10⁻¹⁹ C, so charge is quantised.
Three ways to charge a body: Friction: electrons transfer, giving equal and opposite charges. Contact: charge is shared, same sign. Induction (with earthing): opposite sign to the charged body.
Electric field strength: E = F/q, the force per unit positive test charge (N C⁻¹ or V m⁻¹). Point charge: E = kQ/r². Parallel plates: E = V/d, uniform.
600 V across plates 1.5 cm apart. Field strength?: E = 600 ÷ 0.015 = 4.0 × 10⁴ V m⁻¹
Field inside and outside a charged conducting sphere: Charge sits on the surface; E = 0 inside; outside, the field is radial, as if the charge were a point at the centre.
Electric field lines vs magnetic field lines: Electric lines start on positive and end on negative charges. Magnetic lines form closed loops (outside a magnet from N to S).
Electric potential, and its value for a point charge: The work done per unit charge to bring a small positive test charge from infinity to the point. Ve = kQ/r, a scalar with the sign of Q. (HL only)
Electric potential energy of two point charges, and field from potential: Ep = kq₁q₂/r (negative for unlike charges). E = −ΔVe/Δr: the field points towards falling potential. (HL only)
Why is no work done moving a charge along an equipotential?: ΔVe = 0, so W = qΔVe = 0. Field lines cross equipotentials at 90°. (HL only)

D.3 Motion in electromagnetic fields

Force and acceleration of a charge in an electric field: F = qE, a = qE/m. Along E for a positive charge, against it for a negative one. Crossing a uniform field at right angles gives a parabolic path.
Magnetic force on a moving charge and on a conductor: F = qvB sin θ F = BIL sin θ θ is the angle between v (or I) and B.
Why does a magnetic field not change a charged particle’s speed?: The force is always at right angles to the velocity, so it does no work. Only the direction changes.
Radius and period of a charged particle’s circular path in a magnetic field: qvB = mv²/r gives r = mv/(qB). T = 2πm/(qB), independent of speed.
Velocity selector: Crossed electric and magnetic fields: a particle goes straight through when qE = qvB, i.e. v = E/B, whatever its charge or mass.
Force between two parallel current-carrying wires: F/L = μ₀I₁I₂ / (2πr), μ₀ = 4π × 10⁻⁷ T m A⁻¹. Same-direction currents attract; opposite currents repel.
10 A and 5.0 A in the same direction, 2.0 cm apart. Force per metre?: F/L = 4π × 10⁻⁷ × 10 × 5.0 ÷ (2π × 0.020) = 5.0 × 10⁻⁴ N m⁻¹, attractive

D.4 Induction (HL only)

Magnetic flux: Φ = BA cos θ, in Wb (T m²), θ the angle between B and the normal to the area. (HL only)
Faraday’s law of induction: ε = −N ΔΦ/Δt: the induced emf equals the rate of change of flux linkage. (HL only)
Emf induced in a straight conductor moving through a field: ε = BvL, with B, v and the conductor mutually perpendicular. (HL only)
Lenz’s law: The induced current opposes the change in flux that produces it, as conservation of energy requires (work must be done to produce the electrical energy). (HL only)
A 60-turn coil’s flux rises uniformly from 0 to 4.0 × 10⁻³ Wb in 0.50 s. Emf?: 60 × 4.0 × 10⁻³ ÷ 0.50 = 0.48 V (zero while the flux then stays constant). (HL only)
Emf of a coil rotating in a uniform field: ε = NBAω sin ωt, peak NBAω. Zero when the coil’s plane is perpendicular to B (flux maximum); peak when parallel. (HL only)
A generator’s rotation frequency is increased by a factor k. Effect on its output?: Peak emf × k and period ÷ k: a taller, narrower sine wave. (HL only)