Discussion
A conductor of resistance 0.5 \( \Omega \) carries 10 A. The inward flux of the Poynting vector over its surface is
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.
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