A-Level Physics Formula Guide
Cambridge A-Level Physics (9702) | Understanding & Application
All formulas are given in your exam
The goal is not memorization - it's understanding when and how to use each formula
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How to Use Formulas in Exams
- Identify what you're given (known values with units)
- Identify what you need to find (the unknown)
- Choose a formula that connects these quantities
- Rearrange to make the unknown the subject
- Substitute values (with consistent units)
- Calculate and state the answer with units
Physical Constants
Given on page 2 of all papers. You don't need to memorize these.
Kinematics (Uniformly Accelerated Motion)
Use when acceleration is constant. Choose formula based on which variable you DON'T have.
Displacement from initial velocity, acceleration, time
Use when: You don't know final velocity (v)
Variables: s = displacement (m), u = initial velocity (m/s), a = acceleration (m/s²), t = time (s)
Tip: This is a quadratic in t - can have two solutions
Worked Example:
A ball is thrown vertically upward at 20 m/s. How high does it rise? (g = 9.81 m/s²)
At max height, v = 0. Use v² = u² + 2as
0 = 20² + 2(-9.81)s
s = 400 ÷ 19.62
s = 20.4 m
Practice:
A car accelerates from rest at 3.0 m/s² for 5.0 s. Find the distance travelled.
Show Answer
s = ut + ½at² = 0 + ½ × 3.0 × 5.0² = 37.5 m
Final velocity without knowing time
Use when: You don't know time (t)
Rearrangements: s = (v² - u²)/2a, a = (v² - u²)/2s
Tip: Very useful for projectile problems at max height (v = 0)
Worked Example:
An object is dropped from 45 m. Find its speed when it hits the ground.
u = 0 (dropped), s = 45 m, a = 9.81 m/s²
v² = u² + 2as = 0 + 2 × 9.81 × 45 = 882.9
v = 29.7 m/s
Practice:
A car brakes from 25 m/s to rest over 50 m. Find the deceleration.
Show Answer
a = (v² - u²)/2s = (0 - 625)/(2 × 50) = -6.25 m/s²
Pressure & Fluids
For liquids and gases at rest or in equilibrium.
Hydrostatic pressure
Use when: Calculating pressure difference due to depth in a fluid
Variables: Δp = pressure change (Pa), ρ = density (kg/m³), g = 9.81 m/s², Δh = depth change (m)
Note: This gives pressure DIFFERENCE. Total pressure = atmospheric + ρgΔh
Worked Example:
Find the pressure at 10 m depth in seawater (ρ = 1025 kg/m³).
ρ = 1025 kg/m³, g = 9.81 m/s², Δh = 10 m
Δp = ρgΔh = 1025 × 9.81 × 10
Δp = 1.01 × 10⁵ Pa (≈ 1 atm extra)
Practice:
A mercury barometer reads 760 mm. Find the atmospheric pressure. (ρ_Hg = 13600 kg/m³)
Show Answer
p = ρgh = 13600 × 9.81 × 0.760 = 1.01 × 10⁵ Pa
Upthrust (Archimedes' principle)
Use when: Finding buoyancy force on submerged/floating object
Key insight: V is volume of fluid DISPLACED, not volume of object (unless fully submerged)
For floating objects: Upthrust = Weight
Worked Example:
A 0.50 m³ block is fully submerged in water. Find the upthrust.
ρ = 1000 kg/m³, g = 9.81 m/s², V = 0.50 m³
F = ρgV = 1000 × 9.81 × 0.50
F = 4905 N ≈ 4.9 kN
Practice:
A 2.0 kg object floats with 75% of its volume submerged. Find its density.
Show Answer
Upthrust = Weight: ρ_water × 0.75V × g = ρ_object × V × g
ρ_object = 0.75 × 1000 = 750 kg/m³
Circular Motion (Paper 4)
Objects moving in circles with constant speed still accelerate towards the center.
Key insight: Angular velocity is constant for uniform circular motion
Note: v is tangent to the circle, ω is same for all points on rigid body
Worked Example:
A wheel of radius 0.30 m rotates at 5.0 rad/s. Find the speed of a point on the rim.
r = 0.30 m, ω = 5.0 rad/s
v = rω = 0.30 × 5.0
v = 1.5 m/s
Practice:
A satellite orbits at 7500 m/s at radius 6.8 × 10⁶ m. Find its angular velocity.
Show Answer
ω = v/r = 7500 ÷ (6.8 × 10⁶) = 1.1 × 10⁻³ rad/s
Use when: Finding acceleration needed to maintain circular path
Worked Example:
A car travels at 20 m/s around a bend of radius 50 m. Find the centripetal acceleration.
v = 20 m/s, r = 50 m
a = v²/r = 20² ÷ 50 = 400 ÷ 50
a = 8.0 m/s²
Practice:
An electron orbits at ω = 2.0 × 10⁶ rad/s at r = 5.0 × 10⁻¹¹ m. Find its acceleration.
Show Answer
a = rω² = 5.0 × 10⁻¹¹ × (2.0 × 10⁶)² = 2.0 × 10² m/s²
Examples: Tension, friction, gravity, or normal force can provide this
Worked Example:
A 1200 kg car travels at 15 m/s around a bend of radius 45 m. Find the friction force required.
m = 1200 kg, v = 15 m/s, r = 45 m
F = mv²/r = 1200 × 15² ÷ 45 = 1200 × 225 ÷ 45
F = 6000 N
Practice:
A 0.50 kg ball on a 0.80 m string swings in a horizontal circle at 4.0 m/s. Find the tension.
Show Answer
F = mv²/r = 0.50 × 4.0² ÷ 0.80 = 0.50 × 16 ÷ 0.80 = 10 N
Waves & Doppler Effect
For sound waves when source or observer is moving.
Doppler effect for sound (moving source)
Use when: Sound source moves toward/away from stationary observer
Variables: f₀ = observed frequency (Hz), f_s = source frequency (Hz), v = wave speed (m/s), v_s = source speed (m/s)
Sign convention: Source approaching → use (v - v_s) → f₀ increases | Source receding → use (v + v_s) → f₀ decreases
Worked Example:
An ambulance siren emits 800 Hz and approaches at 30 m/s. Find the observed frequency. (v_sound = 340 m/s)
f_s = 800 Hz, v = 340 m/s, v_s = 30 m/s (approaching → use v - v_s)
f₀ = f_s × v / (v - v_s) = 800 × 340 / (340 - 30)
f₀ = 800 × 340 / 310
f₀ = 877 Hz (higher pitch)
Practice:
The same ambulance now moves away at 30 m/s. Find the observed frequency.
Show Answer
f₀ = 800 × 340 / (340 + 30) = 800 × 340 / 370 = 735 Hz (lower pitch)
Electricity
Current flow and resistance in circuits.
Current from drift velocity
Use when: Relating microscopic electron movement to macroscopic current
Variables: I = current (A), A = cross-sectional area (m²), n = number density (electrons per m³), v = drift velocity (m/s), q = charge per carrier (C)
Tip: Drift velocities are surprisingly slow (often mm/s) despite fast electrical signals
Worked Example:
A copper wire (A = 1.0 × 10⁻⁶ m², n = 8.5 × 10²⁸ m⁻³) carries 2.0 A. Find the drift velocity.
I = 2.0 A, A = 1.0 × 10⁻⁶ m², n = 8.5 × 10²⁸ m⁻³, q = 1.60 × 10⁻¹⁹ C
v = I/(Anq) = 2.0 / (1.0 × 10⁻⁶ × 8.5 × 10²⁸ × 1.60 × 10⁻¹⁹)
v = 1.47 × 10⁻⁴ m/s = 0.15 mm/s
Practice:
If the current doubles, what happens to drift velocity?
Show Answer
v ∝ I (direct proportion), so drift velocity doubles to 0.30 mm/s
Resistors in series
Key principle: Same current through each; voltages add up
Total resistance is always LARGER than any individual resistor
Worked Example:
Three resistors (10Ω, 20Ω, 30Ω) are connected in series. Find the total resistance.
R = R₁ + R₂ + R₃ = 10 + 20 + 30
R = 60 Ω
Resistors in parallel
Key principle: Same voltage across each; currents add up
Total resistance is always SMALLER than the smallest individual resistor
Shortcut for 2 resistors: R = R₁R₂/(R₁+R₂)
Worked Example:
Two resistors (6Ω and 12Ω) are connected in parallel. Find the total resistance.
Using shortcut: R = R₁R₂/(R₁+R₂) = 6 × 12 / (6 + 12)
R = 72 / 18
R = 4 Ω (less than 6Ω, as expected)
Practice:
Three 12Ω resistors are connected in parallel. Find the total resistance.
Show Answer
1/R = 1/12 + 1/12 + 1/12 = 3/12 = 1/4, so R = 4 Ω
Gravitational Fields (Paper 4)
For orbital motion and gravity beyond uniform field approximation.
Gravitational potential
Use when: Finding potential energy per unit mass at distance r from center of mass M
Variables: φ = potential (J kg⁻¹), G = 6.67 × 10⁻¹¹ N m² kg⁻², M = mass of body (kg), r = distance from center (m)
Potential is NEGATIVE - zero at infinity, decreases (more negative) closer to mass
Worked Example:
Find the gravitational potential at Earth's surface. (M = 6.0 × 10²⁴ kg, r = 6.4 × 10⁶ m)
G = 6.67 × 10⁻¹¹ N m² kg⁻², M = 6.0 × 10²⁴ kg, r = 6.4 × 10⁶ m
φ = -GM/r = -(6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (6.4 × 10⁶)
φ = -6.25 × 10⁷ J kg⁻¹
Practice:
Find the potential at height 400 km above Earth's surface (ISS orbit).
Show Answer
r = 6.4 × 10⁶ + 4 × 10⁵ = 6.8 × 10⁶ m
φ = -(6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (6.8 × 10⁶) = -5.88 × 10⁷ J kg⁻¹
Gravitational potential energy
Use when: Finding energy in orbital/escape problems
Relationship: E_P = mφ (potential energy = mass × potential)
For escape velocity: KE + PE = 0 → ½mv² = GMm/r → v = √(2GM/r)
Worked Example:
Find the escape velocity from Earth's surface.
For escape: KE = -PE, so ½mv² = GMm/r
v = √(2GM/r) = √(2 × 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 6.4 × 10⁶)
v = √(1.25 × 10⁸)
v = 1.12 × 10⁴ m/s = 11.2 km/s
Practice:
How much energy is needed to move a 1000 kg satellite from Earth's surface to 400 km altitude?
Show Answer
ΔE = mΔφ = 1000 × (-5.88 × 10⁷ - (-6.25 × 10⁷)) = 1000 × 3.7 × 10⁶ = 3.7 × 10⁹ J
Ideal Gases (Paper 4)
Kinetic theory linking microscopic and macroscopic properties.
Worked Example:
2.0 mol of gas at 300 K occupies 0.050 m³. Find the pressure.
n = 2.0 mol, R = 8.31 J mol⁻¹ K⁻¹, T = 300 K, V = 0.050 m³
p = nRT/V = 2.0 × 8.31 × 300 ÷ 0.050
p = 9.97 × 10⁴ Pa ≈ 100 kPa
Practice:
0.50 mol of gas at 2.0 × 10⁵ Pa and 400 K. Find the volume.
Show Answer
V = nRT/p = 0.50 × 8.31 × 400 ÷ (2.0 × 10⁵) = 8.3 × 10⁻³ m³
Relationship: N = nNₐ and R = kNₐ (where Nₐ = Avogadro constant)
Alternative form: p = ⅓ρ⟨c²⟩ where ρ = Nm/V = density
Note: ⟨c²⟩ ≠ ⟨c⟩² (mean square speed ≠ square of mean speed)
Worked Example:
Find the mean KE of a gas molecule at 300 K.
k = 1.38 × 10⁻²³ J K⁻¹, T = 300 K
KE = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 300
KE = 6.21 × 10⁻²¹ J
Practice:
Find the root mean square speed of N₂ (m = 4.65 × 10⁻²⁶ kg) at 300 K.
Show Answer
⟨c²⟩ = 3kT/m = 3 × 1.38 × 10⁻²³ × 300 ÷ 4.65 × 10⁻²⁶ = 2.67 × 10⁵
c_rms = √⟨c²⟩ = 517 m/s
Thermodynamics (Paper 4)
Energy transfer in heat engines and the first law.
First law of thermodynamics
Use when: Analyzing energy changes in a gas system
Variables: ΔU = change in internal energy (J), q = heat added TO system (J), W = work done ON system (J)
Sign convention: +ve when energy enters system | For ideal gas: ΔU ∝ ΔT (isothermal → ΔU = 0)
Worked Example:
A gas receives 500 J of heat and does 200 J of work on its surroundings. Find ΔU.
q = +500 J (heat added), W = -200 J (work done BY gas = -ve W ON gas)
ΔU = q + W = 500 + (-200)
ΔU = +300 J (internal energy increases)
Practice:
In an isothermal expansion, 800 J of work is done by the gas. How much heat enters?
Show Answer
Isothermal → ΔU = 0, W = -800 J (BY gas)
0 = q + (-800), so q = +800 J (heat must enter to maintain temperature)
Work done BY gas at constant pressure
Use when: Gas expands or compresses at constant pressure (isobaric process)
Variables: W = work done BY gas (J), p = pressure (Pa), ΔV = volume change (m³)
On p-V diagram, work = area under curve (for any process, not just isobaric)
Worked Example:
A gas at 2.0 × 10⁵ Pa expands from 0.010 m³ to 0.025 m³. Find the work done by the gas.
p = 2.0 × 10⁵ Pa, ΔV = 0.025 - 0.010 = 0.015 m³
W = pΔV = 2.0 × 10⁵ × 0.015
W = 3000 J (done BY the gas)
Practice:
A gas is compressed from 0.030 m³ to 0.010 m³ at 1.5 × 10⁵ Pa. What work is done on the gas?
Show Answer
W_BY = pΔV = 1.5 × 10⁵ × (-0.020) = -3000 J
Work done ON gas = +3000 J
Simple Harmonic Motion (Paper 4)
Oscillations where restoring force ∝ displacement from equilibrium.
Defining equation of SHM
Key insight: Acceleration proportional to displacement, directed towards equilibrium
Variables: a = acceleration (m/s²), ω = angular frequency (rad s⁻¹) = 2πf = 2π/T, x = displacement (m)
The negative sign indicates acceleration is opposite to displacement direction
Worked Example:
A pendulum oscillates at 0.50 Hz. Find the acceleration when displaced 0.10 m from equilibrium.
f = 0.50 Hz, x = 0.10 m
ω = 2πf = 2π × 0.50 = π rad/s
a = -ω²x = -π² × 0.10
a = -0.99 m/s² (towards equilibrium)
Practice:
A mass-spring system has T = 0.40 s and amplitude 0.15 m. Find the maximum acceleration.
Show Answer
ω = 2π/T = 2π/0.40 = 5π rad/s
a_max = ω²x₀ = (5π)² × 0.15 = 37 m/s² (at extremes)
Velocity as function of displacement
Use when: You know position and amplitude, need velocity (very common!)
Key points: v = 0 at x = ±x₀ (extremes), v = ±ωx₀ at x = 0 (equilibrium)
Maximum speed: v_max = ωx₀ = 2πfx₀ (occurs at equilibrium)
Worked Example:
A mass oscillates with amplitude 0.20 m and f = 2.0 Hz. Find the speed when x = 0.12 m.
x₀ = 0.20 m, f = 2.0 Hz, x = 0.12 m
ω = 2πf = 2π × 2.0 = 4π rad/s
v = ω√(x₀² - x²) = 4π × √(0.20² - 0.12²) = 4π × √(0.04 - 0.0144)
v = 4π × √0.0256 = 4π × 0.16
v = 2.0 m/s
Practice:
The same oscillator: find the maximum speed.
Show Answer
v_max = ωx₀ = 4π × 0.20 = 2.5 m/s (at equilibrium, x = 0)
Displacement as function of time
Choice depends on starting point: sin if starting from equilibrium, cos if starting from extreme
Variables: x₀ = amplitude (m), ω = 2πf = 2π/T (rad/s), t = time (s)
Worked Example:
An oscillator starts at equilibrium, amplitude 0.08 m, period 0.50 s. Find displacement at t = 0.125 s.
Starting from equilibrium → use sin. x₀ = 0.08 m, T = 0.50 s
ω = 2π/T = 2π/0.50 = 4π rad/s
x = x₀sin(ωt) = 0.08 × sin(4π × 0.125) = 0.08 × sin(π/2)
x = 0.08 m (at maximum displacement)
Electric Fields (Paper 4)
Point charges and radial electric fields.
Electric potential around a point charge
Use when: Finding potential at distance r from a point charge
Variables: V = potential (V or J C⁻¹), Q = charge (C), ε₀ = 8.85 × 10⁻¹² F/m, r = distance (m)
Unlike gravity, V can be +ve (positive Q) or -ve (negative Q)
Worked Example:
Find the electric potential 0.30 m from a +5.0 μC charge.
Q = 5.0 × 10⁻⁶ C, r = 0.30 m, k = 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻²
V = kQ/r = 9.0 × 10⁹ × 5.0 × 10⁻⁶ / 0.30
V = 1.5 × 10⁵ V = 150 kV
Practice:
Find the potential at 0.20 m from a -3.0 μC charge.
Show Answer
V = kQ/r = 9.0 × 10⁹ × (-3.0 × 10⁻⁶) / 0.20 = -1.35 × 10⁵ V (negative)
Electric potential energy
Use when: Finding work done moving charge q near charge Q
Sign: +ve for like charges (repulsion - work needed to bring together), -ve for unlike charges (attraction)
E_P = qV (similar to gravitational: E_P = mφ)
Worked Example:
Two protons are 2.0 × 10⁻¹⁰ m apart. Find their electric potential energy.
Q = q = 1.6 × 10⁻¹⁹ C, r = 2.0 × 10⁻¹⁰ m
E_P = kQq/r = 9.0 × 10⁹ × (1.6 × 10⁻¹⁹)² / (2.0 × 10⁻¹⁰)
E_P = 9.0 × 10⁹ × 2.56 × 10⁻³⁸ / 2.0 × 10⁻¹⁰
E_P = 1.15 × 10⁻¹⁸ J (or 7.2 eV)
Practice:
A proton and electron are 5.0 × 10⁻¹¹ m apart. Find their potential energy.
Show Answer
E_P = k × (+e) × (-e) / r = 9.0 × 10⁹ × (-(1.6 × 10⁻¹⁹)²) / 5.0 × 10⁻¹¹
E_P = -4.6 × 10⁻¹⁸ J (negative = bound system)
Capacitors (Paper 4)
Combinations and discharge through resistors.
Capacitors in series
Key principle: Same charge on each capacitor; voltages add
Note: Opposite of resistors! Series capacitance is SMALLER than smallest individual
For 2 capacitors: C = C₁C₂/(C₁+C₂)
Worked Example:
Two capacitors (4.0 μF and 6.0 μF) are connected in series. Find the total capacitance.
Using shortcut: C = C₁C₂/(C₁+C₂) = 4.0 × 6.0 / (4.0 + 6.0)
C = 24 / 10
C = 2.4 μF (smaller than 4.0 μF)
Capacitors in parallel
Key principle: Same voltage across each; charges add
Note: Parallel capacitance is LARGER (opposite of resistors)
Worked Example:
The same capacitors (4.0 μF and 6.0 μF) are connected in parallel. Find the total capacitance.
C = C₁ + C₂ = 4.0 + 6.0
C = 10 μF (larger than either)
Practice:
Three 12 μF capacitors in parallel, then in series with 6 μF. Find total C.
Show Answer
Parallel: 12 + 12 + 12 = 36 μF
Then series: C = 36 × 6 / (36 + 6) = 216/42 = 5.1 μF
Capacitor discharge (exponential decay)
Use when: Finding Q, V, or I during discharge
Variables: x = charge, voltage, or current at time t; x₀ = initial value; τ = RC = time constant (s)
At t = RC: value = x₀/e ≈ 0.37x₀ (37% remaining). At t = 5RC: essentially discharged (0.7%)
Worked Example:
A 100 μF capacitor charged to 12 V discharges through 50 kΩ. Find V after 3.0 s.
V₀ = 12 V, R = 50 × 10³ Ω, C = 100 × 10⁻⁶ F, t = 3.0 s
RC = 50 × 10³ × 100 × 10⁻⁶ = 5.0 s
V = V₀e⁻ᵗ/ᴿᶜ = 12 × e⁻³·⁰/⁵·⁰ = 12 × e⁻⁰·⁶ = 12 × 0.549
V = 6.6 V
Practice:
How long until the voltage drops to 3.0 V?
Show Answer
3.0 = 12 × e⁻ᵗ/⁵
e⁻ᵗ/⁵ = 0.25
-t/5 = ln(0.25) = -1.39
t = 6.9 s
Magnetic Fields (Paper 4)
Forces on currents and moving charges in magnetic fields.
Force on current-carrying conductor
Use when: Wire carrying current in a magnetic field
Variables: B = magnetic flux density (T), I = current (A), L = length in field (m), θ = angle between I and B
Direction: Fleming's left-hand rule (First = B, seCond = I, thuMb = Motion). Max force when θ = 90°
Worked Example:
A 0.20 m wire carries 5.0 A perpendicular to a 0.30 T field. Find the force.
B = 0.30 T, I = 5.0 A, L = 0.20 m, θ = 90° (perpendicular)
F = BIL sinθ = 0.30 × 5.0 × 0.20 × sin90° = 0.30 × 5.0 × 0.20 × 1
F = 0.30 N
Practice:
The same wire makes a 30° angle with the field. What is the new force?
Show Answer
F = 0.30 × 5.0 × 0.20 × sin30° = 0.30 × 0.5 = 0.15 N
Force on moving charge
Use when: Charged particle moving through a magnetic field
Variables: q = charge (C), v = velocity (m/s), θ = angle between v and B
Force ⊥ velocity → no work done, only direction changes → circular motion. Radius: r = mv/(Bq)
Worked Example:
An electron (m = 9.11 × 10⁻³¹ kg) moves at 2.0 × 10⁷ m/s perpendicular to a 0.50 T field. Find (a) the force and (b) the radius of its circular path.
q = 1.60 × 10⁻¹⁹ C, v = 2.0 × 10⁷ m/s, B = 0.50 T
(a) F = Bqv = 0.50 × 1.60 × 10⁻¹⁹ × 2.0 × 10⁷
F = 1.6 × 10⁻¹² N
(b) r = mv/(Bq) = 9.11 × 10⁻³¹ × 2.0 × 10⁷ / (0.50 × 1.60 × 10⁻¹⁹)
r = 2.3 × 10⁻⁴ m = 0.23 mm
Practice:
A proton (m = 1.67 × 10⁻²⁷ kg) enters the same field at the same speed. Find its radius.
Show Answer
r = mv/(Bq) = 1.67 × 10⁻²⁷ × 2.0 × 10⁷ / (0.50 × 1.60 × 10⁻¹⁹) = 0.42 m
(Much larger due to greater mass!)
Electromagnetic Induction (Paper 4)
Faraday's and Lenz's laws - generating EMF from changing magnetic flux.
Units: 1 Wb = 1 T m² = 1 V s
Lenz: The negative sign - EMF opposes the change producing it
Worked Example:
A coil of 200 turns has flux through it changing from 5.0 mWb to 2.0 mWb in 0.10 s. Find the induced EMF.
N = 200, ΔΦ = (2.0 - 5.0) × 10⁻³ = -3.0 × 10⁻³ Wb, Δt = 0.10 s
ε = -N(ΔΦ/Δt) = -200 × (-3.0 × 10⁻³) ÷ 0.10
ε = 6.0 V
Practice:
A 500-turn coil in a 0.40 T field has area 0.020 m². It rotates 90° in 0.050 s. Find the average EMF.
Show Answer
ΔΦ = BA(cos90° - cos0°) = 0.40 × 0.020 × (0-1) = -8.0 × 10⁻³ Wb
ε = -500 × (-8.0 × 10⁻³) ÷ 0.050 = 80 V
Derivation: From ε = -dΦ/dt where dΦ = B × L × (v dt)
Worked Example:
A 0.50 m rod moves at 4.0 m/s perpendicular to a 0.30 T field. Find the induced EMF.
B = 0.30 T, L = 0.50 m, v = 4.0 m/s
ε = BLv = 0.30 × 0.50 × 4.0
ε = 0.60 V
Practice:
An aircraft with 30 m wingspan flies at 250 m/s in Earth's 5.0 × 10⁻⁵ T field. Find the EMF between wingtips.
Show Answer
ε = BLv = 5.0 × 10⁻⁵ × 30 × 250 = 0.375 V
Hall Effect (Paper 4)
Voltage developed across a conductor in a magnetic field.
Hall voltage
Use when: Finding B-field using Hall probe, or determining charge carrier type
Variables: V_H = Hall voltage (V), B = flux density (T), I = current (A), n = number density (m⁻³), t = thickness (m), q = charge (C)
V_H is larger for semiconductors (small n) than metals (large n) - that's why semiconductor Hall probes are used
Worked Example:
A Hall probe (t = 0.50 mm, n = 2.0 × 10²⁵ m⁻³) carries 20 mA in a 0.40 T field. Find V_H.
B = 0.40 T, I = 0.020 A, n = 2.0 × 10²⁵ m⁻³, t = 5.0 × 10⁻⁴ m, q = 1.60 × 10⁻¹⁹ C
V_H = BI/(ntq) = 0.40 × 0.020 / (2.0 × 10²⁵ × 5.0 × 10⁻⁴ × 1.60 × 10⁻¹⁹)
V_H = 8.0 × 10⁻³ / (1.6 × 10³)
V_H = 5.0 × 10⁻⁶ V = 5.0 μV
Practice:
The Hall voltage is 12 μV in a 0.30 T field. Find B when V_H = 20 μV (same current).
Show Answer
V_H ∝ B (everything else constant)
B_new = B × (V_H_new/V_H) = 0.30 × (20/12) = 0.50 T
Alternating Current (Paper 4)
Sinusoidal variation of current and voltage.
AC voltage or current
Use when: Finding instantaneous value of AC signal at time t
Variables: x = instantaneous value (V or A), x₀ = peak value, ω = 2πf (rad/s), f = frequency (Hz)
UK mains: V₀ ≈ 325 V (peak), f = 50 Hz, so V = 325sin(100πt)
Worked Example:
An AC supply has peak voltage 340 V and frequency 50 Hz. Find the voltage at t = 2.5 ms.
V₀ = 340 V, f = 50 Hz, t = 2.5 × 10⁻³ s
ω = 2πf = 2π × 50 = 100π rad/s
V = V₀sin(ωt) = 340 × sin(100π × 2.5 × 10⁻³) = 340 × sin(π/4)
V = 340 × 0.707 = 240 V
RMS (root mean square) value
Use when: Calculating power, or comparing AC to equivalent DC
Key insight: RMS value gives same power dissipation as DC of same value
UK mains: V_rms = 230 V, so V₀ = 230 × √2 ≈ 325 V peak
Worked Example:
Find the average power dissipated in a 100 Ω resistor connected to 230 V rms supply.
V_rms = 230 V, R = 100 Ω
P = V_rms²/R = 230²/100 = 52900/100
P = 529 W
Practice:
An AC current has peak value 5.0 A. Find the RMS current.
Show Answer
I_rms = I₀/√2 = 5.0/√2 = 3.5 A
Quantum Physics (Paper 4)
Wave-particle duality and the photoelectric effect.
Worked Example:
Find the energy of a photon with wavelength 500 nm (green light).
h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s, λ = 500 × 10⁻⁹ m
E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (500 × 10⁻⁹)
E = 3.98 × 10⁻¹⁹ J (or 2.49 eV)
Practice:
Find the frequency of a 4.0 eV photon. (1 eV = 1.60 × 10⁻¹⁹ J)
Show Answer
E = 4.0 × 1.60 × 10⁻¹⁹ = 6.4 × 10⁻¹⁹ J
f = E/h = 6.4 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 9.7 × 10¹⁴ Hz
Worked Example:
UV light (λ = 200 nm) hits a metal with work function 4.5 eV. Find the max KE of photoelectrons.
E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (200 × 10⁻⁹) = 9.95 × 10⁻¹⁹ J = 6.22 eV
KEmax = hf - Φ = 6.22 - 4.5
KEmax = 1.72 eV (or 2.75 × 10⁻¹⁹ J)
Practice:
A metal has threshold wavelength 450 nm. Find the work function in eV.
Show Answer
Φ = hc/λ₀ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (450 × 10⁻⁹) = 4.42 × 10⁻¹⁹ J
Φ = 4.42 × 10⁻¹⁹ ÷ 1.60 × 10⁻¹⁹ = 2.76 eV
Use when: Electron diffraction, electron microscopy
Worked Example:
Find the de Broglie wavelength of an electron accelerated through 100 V.
KE = eV = 1.60 × 10⁻¹⁹ × 100 = 1.60 × 10⁻¹⁷ J
v = √(2KE/m) = √(2 × 1.60 × 10⁻¹⁷ ÷ 9.11 × 10⁻³¹) = 5.93 × 10⁶ m/s
λ = h/(mv) = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 5.93 × 10⁶)
λ = 1.23 × 10⁻¹⁰ m = 0.123 nm
Practice:
A neutron (m = 1.67 × 10⁻²⁷ kg) has wavelength 0.10 nm. Find its speed.
Show Answer
v = h/(mλ) = 6.63 × 10⁻³⁴ ÷ (1.67 × 10⁻²⁷ × 0.10 × 10⁻⁹) = 3970 m/s
Emission: Electron drops to lower level, photon released
Nuclear Physics (Paper 4)
Mass-energy equivalence and radioactive decay.
Shortcut: 1 u of mass ≈ 931 MeV of energy
Worked Example:
A nuclear reaction releases 0.0035 u of mass. Find the energy released in MeV.
Δm = 0.0035 u, 1 u = 931 MeV
E = Δmc² = 0.0035 × 931
E = 3.26 MeV
Practice:
In U-235 fission, 200 MeV is released per nucleus. Find the mass converted to energy per fission.
Show Answer
Δm = E/c² = 200 ÷ 931 = 0.215 u (or 3.57 × 10⁻²⁸ kg)
Binding energy: Energy needed to completely separate all nucleons
Worked Example:
A sample contains 2.0 × 10²⁰ atoms with half-life 5.0 days. Find the initial activity.
N = 2.0 × 10²⁰, t½ = 5.0 × 24 × 3600 = 4.32 × 10⁵ s
λ = ln2/t½ = 0.693 ÷ 4.32 × 10⁵ = 1.60 × 10⁻⁶ s⁻¹
A = λN = 1.60 × 10⁻⁶ × 2.0 × 10²⁰
A = 3.2 × 10¹⁴ Bq
Practice:
A source has activity 8.0 × 10⁸ Bq and decay constant 2.0 × 10⁻⁶ s⁻¹. How many nuclei remain?
Show Answer
N = A/λ = 8.0 × 10⁸ ÷ 2.0 × 10⁻⁶ = 4.0 × 10¹⁴ nuclei
Use when: Finding remaining amount/activity after time t
Note: 0.693 = ln(2)
Medical Physics - Ultrasound (Paper 4)
Reflection at tissue boundaries.
Intensity reflection coefficient
Use when: Calculating reflection at tissue boundaries in ultrasound imaging
Variables: I_R/I₀ = fraction of intensity reflected, Z = acoustic impedance = ρc (kg m⁻² s⁻¹)
If Z₁ = Z₂: no reflection (matched impedance). If Z₁ >> Z₂ or Z₁ << Z₂: nearly total reflection
Worked Example:
Find the fraction of ultrasound reflected at a muscle-bone boundary. (Z_muscle = 1.7 × 10⁶, Z_bone = 6.4 × 10⁶ kg m⁻² s⁻¹)
Z₁ = 1.7 × 10⁶, Z₂ = 6.4 × 10⁶
I_R/I₀ = (Z₁-Z₂)²/(Z₁+Z₂)² = (1.7-6.4)²/(1.7+6.4)² × 10¹²/10¹²
I_R/I₀ = (-4.7)²/(8.1)² = 22.09/65.61
I_R/I₀ = 0.34 (34% reflected)
Practice:
Why is gel used between the probe and skin? (Z_air ≈ 400, Z_skin = 1.7 × 10⁶)
Show Answer
Without gel: I_R/I₀ = (400-1.7×10⁶)²/(400+1.7×10⁶)² ≈ 0.999 (99.9% reflected!)
Air causes almost total reflection. Gel matches impedance with skin.
Astrophysics & Cosmology (Paper 4)
Stars, radiation, and the expanding universe.
Use when: Finding star's power output from size and temperature
Use when: Finding peak wavelength from temperature, or vice versa
Worked Example:
A star has surface temperature 6000 K. Find the wavelength at which it emits most intensely.
T = 6000 K, Wien's constant = 2.898 × 10⁻³ m K
λmax = 2.898 × 10⁻³ ÷ 6000
λmax = 4.83 × 10⁻⁷ m = 483 nm (blue-green)
Practice:
A red giant has peak emission at 700 nm. Find its surface temperature.
Show Answer
T = 2.898 × 10⁻³ ÷ (700 × 10⁻⁹) = 4140 K
Sign: Redshift (+ve Δλ) = moving away, Blueshift (-ve Δλ) = moving towards
Key insight: Universe is expanding - more distant galaxies recede faster
Worked Example:
A galaxy is 50 Mpc away. Find its recession velocity. (H₀ = 70 km s⁻¹ Mpc⁻¹)
d = 50 Mpc, H₀ = 70 km s⁻¹ Mpc⁻¹
v = H₀d = 70 × 50
v = 3500 km/s = 3.5 × 10⁶ m/s
Practice:
A galaxy recedes at 21,000 km/s. How far away is it? (H₀ = 70 km s⁻¹ Mpc⁻¹)
Show Answer
d = v/H₀ = 21000 ÷ 70 = 300 Mpc
More A-Level Resources
LumiExams.com | Cambridge A-Level Physics (9702) | Syllabus 2025-2027