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 Incoming links: Intensity of an isotropic radiator

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 15D / e / Solution Created 2026-09-24 Updated 2026-10-03

For isotropic radiation, the average power is spread over a sphere. At

r = 0.10 m

, the intensity of an isotropic radiator is

⟨ ∣ S ∣ ⟩ = \frac{P}{4 π r ^{2}} = \frac{1}{4 π ( 0.10 ) ^{2}} ≃ 7.96 W m^{- 2} .

Using the result of part (d),

∣ E_{0} ∣ = \frac{2 ⟨ ∣ S ∣ ⟩}{ϵ _{0} c} ≃ 77 V m^{- 1} .

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Past exam of the physics 2 paper of the University of Oxford / 2018 / 4 / Solution 2026-10-25

The intensity of an isotropic radiator at distance

r

is the radiated power divided by spherical area:

I = P / (4 π r^{2})

. For a locally plane far-field electromagnetic wave in air, its mean Poynting flux is

I = E_{rms}^{2} / Z_{0} = E_{peak}^{2} / (2 Z_{0})

, where the vacuum wave impedance is

Z_{0} ≃ 376.7 Ω

. Thus

I = \frac{1 0 ^{5}}{4 π ( 1 0 ^{5} ) ^{2}} ≃ 7.96 \times 1 0^{- 7} W m^{- 2},

and

E_{rms} ≃ 1.73 \times 1 0^{- 2} V m^{- 1}, E_{peak} ≃ 2.45 \times 1 0^{- 2} V m^{- 1} .

Both conventions are given because “field strength” alone does not specify RMS or peak. This estimate treats the stated transmitter power as radiated power and ignores ground reflection and absorption.

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