Poynting Theorem Questions and Answers

Answer: A

Correct answer: 80 \( \mu \)W/m²

An isotropic radiator spreads its power evenly over a sphere: \[ S = \frac{P}{4\pi r^2} = \frac{1000}{4\pi (10^3)^2} = \frac{1000}{1.257\times10^7} = 8.0\times10^{-5}\ \mathrm{W/m^2}. \] That is 80 microwatts per square metre, and the inverse square law follows from the same expression - the sphere’s area grows as \( r^2 \) while the power crossing it stays the same.

Answer: B

Multiplying the power density by 377 and taking the square root gives \( \sqrt{7.96 \times 10^{-5} \times 377} = 0.173 \) V/m. There is a useful shortcut: the rms field equals the square root of thirty times the power, divided by the distance.

Answer: C

Correct answer: 614 V/m

Rearranging \( S = E^2/377 \) gives \[ E = \sqrt{377 S} = \sqrt{377\times1000} = \sqrt{377{,}000} = 614\ \mathrm{V/m}. \] Sunlight therefore carries a field of several hundred volts per metre, which is a useful figure to hold: it shows that a field strength which sounds large in circuit terms is entirely ordinary as a wave.

Answer: D

Integrating the Poynting vector over the annular cross-section returns exactly the product of the voltage and the current. That agreement between the field and circuit calculations is the standard demonstration of the theorem.

Answer: A

Correct answer: 50 W

The inward flux of the Poynting vector over the conductor’s surface must equal the power dissipated inside it, which is \[ I^2R = 10^2 \times 0.5 = 50\ \mathrm{W}. \] This is the theorem’s most direct check: the energy dissipated as heat did not travel along the wire but entered radially through its surface from the surrounding field, and the two figures agree exactly.

Answer: B

Correct answer: 73 W

\[ P = I^2 R_{rad} = 1^2 \times 73 = 73\ \mathrm{W}. \] Radiation resistance is defined so that this works: it is the fictitious resistance that would dissipate the power the antenna actually radiates. The value 73 ohms is that of a half-wave dipole, and the current must be rms for the answer to be an average power.

Answer: C

Correct answer: 51.5 mA/m

\[ S = \eta_0 H^2 \quad\Rightarrow\quad H = \sqrt{\frac{S}{377}} = \sqrt{\frac{1}{377}} = 0.0515\ \mathrm{A/m}. \] Note that the impedance divides for \( E \) and multiplies for \( H \), since \( S=EH \) and \( E=\eta_0 H \). Getting that the wrong way round gives 19.4 A/m, which is offered here.

Answer: D

Dividing gives eight kilowatts per square metre, some eight times the intensity of sunlight, and the corresponding field is about 1.7 kilovolts per metre. Standing waves within the cavity make the local value vary widely about that mean.

Answer: A

The received power is the product of the power density and the effective aperture, that is half a milliwatt. The effective aperture is related to the gain and the square of the wavelength, so it falls as the frequency rises for an antenna of fixed gain.

Answer: B

The vector is largest in the space around the conductors, which guide the flow rather than carry it. Within the copper the vector points inward and accounts only for the ohmic loss, not for the power delivered to the load.