True or false: a static electric field and a static magnetic field together imply energy flow.
Answer: C
Crossed static fields do give a non-zero Poynting vector, and the vector is not zero, but with nothing changing and no dissipation its divergence vanishes and no energy is transferred anywhere. The circulation is a well known curiosity of the formalism.
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True or false: the reactive part of the complex Poynting vector transfers net energy.
Answer: D
Correct answer: False, the reactive part representing energy that merely oscillates to and fro
In the complex Poynting vector \( \tfrac{1}{2}\mathbf{E}\times\mathbf{H}^{*} \), the real part is the time-averaged power actually transported and the imaginary part is stored energy surging back and forth twice per cycle with no net transfer. Near an antenna the reactive part dominates - that is what the near field is - but it is precisely the part that does not radiate.
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True or false: Poynting's theorem is an additional postulate beyond Maxwell's equations.
Answer: A
It follows entirely from the four equations together with the constitutive relations, by a vector identity and a substitution. That energy conservation should emerge without being assumed is one of the theory's strongest recommendations.
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Assertion: energy reaches a resistor through the space around its leads. Reason: the Poynting vector within the conductors points inward.
Answer: B
The inward vector at the conductor surface is precisely the statement that energy crosses into the metal from outside, and its surface integral is the ohmic loss. The two statements are two descriptions of the same field configuration.
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Assertion: the power density from a point source falls as the inverse square of distance. Reason: the field amplitude falls as the inverse first power.
Answer: C
Power density goes as the square of the field, so an amplitude falling as the first power gives a density falling as the square. Both follow from the energy spreading over a sphere of growing area, so the reason is a legitimate explanation.
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Assertion: radiation resistance dissipates no heat. Reason: it is not a real resistance but accounts for power leaving as radiation.
Answer: D
The power it absorbs from the source is carried away by the wave rather than converted to heat in the antenna, though the source cannot tell the difference. An antenna also has a real loss resistance, and the ratio of the two is its efficiency.
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Match the term of Poynting's theorem with what it represents.
1. Surface integral of \( \mathbf{E} \times \mathbf{H} \) 2. Volume integral of \( \sigma E^2 \) 3. Time derivative of the energy density integral
P. Ohmic dissipation Q. Power crossing the boundary R. Rate of change of stored energy
Answer: A
The three terms account respectively for what crosses the boundary, what is turned into heat and what is added to the store, and together they balance exactly. Every energy question in field theory reduces to identifying which of the three is wanted.
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Match the region with the character of the Poynting vector there.
1. Inside a current-carrying wire 2. Dielectric of a coaxial cable 3. Far field of an antenna
P. Axial, carrying the load power Q. Radially inward, equal to the ohmic loss R. Radially outward, falling as the inverse square
Answer: B
Correct answer: 1-Q, 2-P, 3-R
Three regions, three characters. Inside a current-carrying wire the vector points radially inward and its surface integral equals the ohmic loss. In a coaxial cable the vector is axial in the dielectric, carrying the load power along between the conductors rather than inside them. In the far field it is radially outward and falls as the inverse square, since the same power spreads over a growing sphere.
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The steps of deriving Poynting's theorem are listed below. Put them in order.
1. Substitute the two curl equations 2. Expand the divergence of the cross product by the vector identity 3. Integrate over a volume and apply the divergence theorem 4. Recognise the energy density and dissipation terms
Answer: C
The identity comes first because it is what introduces the curls, then the field equations are substituted, then the resulting terms are recognised as energy densities and dissipation, and the integration converts the whole into the familiar volume and surface statement.
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Consider the following statements about the Poynting vector.
1. Its units are watts per square metre 2. It is the cross product of the two field intensities 3. Its imaginary part carries net power
Answer: D
Correct answer: 1 and 2 only are correct
Statements 1 and 2 are correct: the units are watts per square metre, and the vector is \( \mathbf{E}\times\mathbf{H} \). Statement 3 has the two parts the wrong way round - the real part carries the net power and the imaginary part represents energy oscillating back and forth without net transfer.
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