When current flows down a wire, work is done, which shows up as Joule heating of the wire (Eq. 7.7).
Question 8.1: When current flows down a wire, work is done, which shows up...
P=VI=I^2R. (7.7)
Though there are certainly easier ways to do it, the energy per unit time delivered to the wire can be calculated using the Poynting vector. Assuming it’s uniform, the electric field parallel to the wire is
E=\frac{V}{L},where V is the potential difference between the ends and L is the length of the wire (Fig. 8.1). The magnetic field is “circumferential”; at the surface (radius a) it has the value
B=\frac{\mu_0I}{2\pi a}.Accordingly, the magnitude of the Poynting vector is
S=\frac{1}{\mu_0}\frac{V}{L}\frac{\mu_0I}{2\pi a}=\frac{VI}{2\pi aL},and it points radially inward. The energy per unit time passing in through the surface of the wire is therefore
\int{S.da}=S(2\pi aL)=VI ,which is exactly what we concluded, on much more direct grounds, in Sect. 7.1.1.^4
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