Specification: AQA A-level Physics (7408) 3.12.2.3 Turning points option | Paper: 3 Section B | Data and formulae booklet: c = 1/√(μ₀ε₀) and the values of μ₀ and ε₀ are given in the exam; the speed in a material, v = 1/√(μ₀ε₀εr), is not, so learn it
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Maxwell’s prediction: an electromagnetic wave is a transverse wave of oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel, and in phase. Maxwell calculated their speed in a vacuum as c = 1/√(μ₀ε₀); this matched the measured speed of light, so he concluded that light is an electromagnetic wave.
ε₀ and μ₀: the permittivity of free space ε₀ relates to the electric field strength due to a charged object in free space; the permeability of free space μ₀ relates to the magnetic flux density due to a current-carrying wire in free space. In a non-magnetic material of relative permittivity εr the speed falls to v = 1/√(μ₀ε₀εr).
Hertz’s discovery of radio waves: Hertz produced radio waves with a spark gap and detected them with a loop of wire with a small gap. He reflected them from a metal sheet to set up stationary waves; adjacent nodes are λ/2 apart, and with the known frequency, c = fλ gave the speed of light. He also showed they could be polarised, so they are transverse.
The equation
speed of electromagnetic waves in a vacuum = 1 ÷ √(permeability of free space × permittivity of free space) c = 1/√(μ₀ε₀)
| Quantity | Symbol | Unit |
|---|---|---|
| speed of electromagnetic waves in a vacuum | c | metres per second (m s⁻¹) |
| permeability of free space, 4π × 10−7 | μ₀ | henries per metre (H m⁻¹) |
| permittivity of free space, 8.85 × 10−12 | ε₀ | farads per metre (F m⁻¹) |
| relative permittivity of a material (non-magnetic) | εr | no unit (a ratio) |
| speed in the material | v | metres per second (m s⁻¹) |
Rearranged: ε₀ = 1 ÷ (μ₀c²) | μ₀ = 1 ÷ (ε₀c²) | in a material: v = 1/√(μ₀ε₀εr) = c ÷ √εr, so εr = (c ÷ v)² | Square c when you rearrange: forgetting the square gives ε₀ 3 × 108 times too big.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) Use the data booklet values μ₀ = 4π × 10−7 H m⁻¹ and ε₀ = 8.85 × 10−12 F m⁻¹ to calculate the speed of electromagnetic waves in a vacuum, to 3 significant figures.
Show solution
F: c = 1/√(μ₀ε₀)
I: c = 1/√(4π × 10−7 × 8.85 × 10−12) = 1/√(1.112… × 10−17)
F: c is already the subject
A: 1 ÷ 3.334… × 10−9 = 2.998… × 108 m s⁻¹ = 3.00 × 108 m s⁻¹ (3 s.f.)
Q2 (Core) A transparent non-magnetic material has a relative permittivity of 2.25 at the frequency used, so the permittivity of the material is 2.25ε₀. Calculate the speed of electromagnetic waves in the material, to 3 significant figures. (μ₀ = 4π × 10−7 H m⁻¹, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
F: v = 1/√(μ₀ε₀εr)
I: v = 1/√(4π × 10−7 × 8.85 × 10−12 × 2.25)
F: v is already the subject
A: 1.999… × 108 m s⁻¹ = 2.00 × 108 m s⁻¹ (3 s.f.), which is c ÷ √2.25 = c ÷ 1.5
Q3 (Core) The speed of light in a vacuum is 3.00 × 108 m s⁻¹ and μ₀ = 4π × 10−7 H m⁻¹. Calculate the permittivity of free space ε₀, to 3 significant figures.
Show solution
F: c = 1/√(μ₀ε₀)
I: 3.00 × 108 = 1/√(4π × 10−7 × ε₀)
F: square both sides: c² = 1 ÷ (μ₀ε₀), so ε₀ = 1 ÷ (μ₀c²) = 1 ÷ (4π × 10−7 × (3.00 × 108)²) = 1 ÷ 1.130… × 1011
A: 8.841… × 10−12 = 8.84 × 10−12 F m⁻¹ (3 s.f.)
Watch out: square c before dividing. Writing ε₀ = 1 ÷ (μ₀c) is a common slip and gives a value 3 × 108 times too big.
Q4 (Core) A short radio pulse takes 250 ns to travel along 50.0 m of coaxial cable. The insulator in the cable is non-magnetic. Calculate the relative permittivity εr of the insulator, to 3 significant figures. (μ₀ = 4π × 10−7 H m⁻¹, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 250 ns = 250 × 10−9 s; speed v = 50.0 ÷ (250 × 10−9) = 2.00 × 108 m s⁻¹
F: v = 1/√(μ₀ε₀εr)
I: 2.00 × 108 = 1/√(4π × 10−7 × 8.85 × 10−12 × εr)
F: εr = 1 ÷ (μ₀ε₀v²) = 1 ÷ (1.112… × 10−17 × 4.00 × 1016) = 1 ÷ 0.4448…
A: 2.247… = 2.25 (3 s.f., no unit). This is the value for polyethylene.
Q5 (Exam) In a repeat of Hertz’s experiment, radio waves of frequency 100 MHz are reflected from a large metal sheet so that stationary waves form. A detector is moved away from the sheet and finds that the distance from the 1st node to the 6th node is 7.45 m. (a) Calculate the speed of the radio waves, to 3 significant figures. (b) Calculate the percentage difference between this and Maxwell’s value, 1/√(μ₀ε₀), to 1 significant figure. (μ₀ = 4π × 10−7 H m⁻¹, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Step 1 – wavelength
from the 1st to the 6th node is 5 node spacings, so λ/2 = 7.45 ÷ 5 = 1.49 m and λ = 2.98 m
Step 2 – (a) speed
F: c = fλ
I: c = 100 × 106 × 2.98
A: 2.98 × 108 m s⁻¹ (3 s.f.)
Step 3 – (b) compare with Maxwell
1/√(μ₀ε₀) = 2.998… × 108 m s⁻¹; difference = (2.998… − 2.98) ÷ 2.998… × 100 = 0.621…%
A: 0.6% (1 s.f.), so the radio waves travel at the speed of light, as Maxwell predicted
Watch out: 6 nodes mark out 5 gaps, not 6, and each gap is half a wavelength. Using 7.45 ÷ 6 or forgetting to double gives the wrong λ.
Q6 (Exam) At radio frequencies, water is non-magnetic and has a relative permittivity of 80. A radio signal of frequency 100 MHz passes from air into water. Calculate (a) the speed of the radio waves in water and (b) their wavelength in water, each to 3 significant figures. (μ₀ = 4π × 10−7 H m⁻¹, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Step 1 – (a) speed in water
F: v = 1/√(μ₀ε₀εr) = (1/√(μ₀ε₀)) ÷ √εr
I: v = 2.998… × 108 ÷ √80 = 2.998… × 108 ÷ 8.944…
A: 3.352… × 107 m s⁻¹ = 3.35 × 107 m s⁻¹ (3 s.f.)
Step 2 – (b) wavelength in water
the frequency does not change when the wave enters the water, so λ = v ÷ f = 3.352… × 107 ÷ (100 × 106)
A: 0.335 m (3 s.f.), compared with about 3.00 m in air
Watch out: the frequency is set by the source and stays the same; it is the speed and wavelength that fall.
Q7 (Exam) Show that the unit of 1/√(μ₀ε₀) is m s⁻¹. You may use 1 H = 1 V s A⁻¹ and 1 F = 1 C V⁻¹.
Show solution
F: c = 1/√(μ₀ε₀), so find the unit of μ₀ε₀ first
I: unit of μ₀ε₀ = H m⁻¹ × F m⁻¹ = (V s A⁻¹) × (C V⁻¹) × m⁻²
F: the volts cancel and 1 C = 1 A s, so this is s × A s × A⁻¹ × m⁻² = s² m⁻²
A: 1/√(s² m⁻²) = 1 ÷ (s m⁻¹) = m s⁻¹, the unit of speed
Watch out: in a “show that” question every step must be written down. Just stating the answer m s⁻¹ scores nothing.
Q8 (Exam) Light travels through diamond at 1.24 × 108 m s⁻¹. Assume diamond is non-magnetic. (a) Calculate the relative permittivity of diamond at optical frequencies, to 3 significant figures. (b) Calculate the refractive index of diamond, to 3 significant figures, using c = 3.00 × 108 m s⁻¹, and comment on your answers. (μ₀ = 4π × 10−7 H m⁻¹, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Step 1 – (a) relative permittivity
F: v = 1/√(μ₀ε₀εr), so εr = 1 ÷ (μ₀ε₀v²)
I: εr = 1 ÷ (1.112… × 10−17 × (1.24 × 108)²) = 1 ÷ 0.1710…
A: 5.847… = 5.85 (3 s.f.)
Step 2 – (b) refractive index
n = c ÷ cs = 3.00 × 108 ÷ 1.24 × 108 = 2.419…
A: 2.42 (3 s.f.). Since v = c ÷ √εr and v = c ÷ n, n = √εr: here 2.419…² = 5.85, so the two answers agree.
Q9 (Stretch) In Fizeau’s 1849 experiment, light passed through a gap between the teeth of a rotating wheel with 720 teeth (and 720 equal gaps), travelled to a mirror 8633 m away and back, and was first blocked completely when the wheel turned at 12.6 revolutions per second. At this speed the returning light meets the tooth next to the gap it left through. (a) Calculate Fizeau’s value for the speed of light, to 3 significant figures. (b) Calculate the percentage difference from Maxwell’s value 1/√(μ₀ε₀), to 2 significant figures, and state what the agreement suggested. (μ₀ = 4π × 10−7 H m⁻¹, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Step 1 – time for a gap to be replaced by the next tooth
one revolution takes 1 ÷ 12.6 s and passes 720 gaps and 720 teeth, so gap to tooth takes 1 ÷ (2 × 720 × 12.6) = 1 ÷ 18 144 = 5.511… × 10−5 s
Step 2 – (a) speed of light
F: speed = distance ÷ time, with distance = 2 × 8633 = 17 266 m (there and back)
I: c = 17 266 ÷ 5.511… × 10−5
A: 3.132… × 108 m s⁻¹ = 3.13 × 108 m s⁻¹ (3 s.f.)
Step 3 – (b) compare with Maxwell
(3.132… − 2.998…) ÷ 2.998… × 100 = 4.47…%
A: 4.5% (2 s.f.). Agreement this close to Maxwell’s theoretical value suggested that light is an electromagnetic wave.
Watch out: the light makes a round trip, so use 2 × 8633 m. Using the one-way distance halves the answer.
Q10 (Stretch) A parallel-plate capacitor has plates of area 0.0400 m² separated by 2.00 mm of a non-magnetic plastic, and its capacitance is measured as 4.25 × 10−10 F. (a) Calculate the relative permittivity of the plastic, to 3 significant figures. (b) Calculate the speed of electromagnetic waves in the plastic, to 3 significant figures. (c) Microwaves of frequency 2.45 GHz set up a stationary wave inside a long block of the plastic. Calculate the distance between adjacent nodes, to 3 significant figures. (μ₀ = 4π × 10−7 H m⁻¹, ε₀ = 8.85 × 10−12 F m⁻¹)
Show solution
Convert first: 2.00 mm = 2.00 × 10−3 m; 2.45 GHz = 2.45 × 109 Hz
Step 1 – (a) relative permittivity from C = Aε₀εr ÷ d
εr = Cd ÷ (ε₀A) = (4.25 × 10−10 × 2.00 × 10−3) ÷ (8.85 × 10−12 × 0.0400) = 8.50 × 10−13 ÷ 3.54 × 10−13
A: 2.401… = 2.40 (3 s.f.)
Step 2 – (b) speed in the plastic
F: v = 1/√(μ₀ε₀εr)
I: v = 1/√(1.112… × 10−17 × 2.401…)
A: 1.935… × 108 m s⁻¹ = 1.94 × 108 m s⁻¹ (3 s.f.)
Step 3 – (c) node spacing
λ = v ÷ f = 1.935… × 108 ÷ 2.45 × 109 = 0.07898… m; adjacent nodes are λ/2 apart = 0.03949… m
A: 3.95 cm (3 s.f.)
Watch out: keep εr unrounded into step 2, and remember nodes are half a wavelength apart, not a whole wavelength.
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