A) \[\frac{IR}{2\pi al}\]
B) \[\frac{I{{R}^{2}}}{al}\]
C) \[\frac{{{I}^{2}}R}{al}\]
D) \[\frac{{{I}^{2}}R}{2\pi al}\]
Correct Answer: D
Solution :
Electric field \[E=\frac{v}{l}=\frac{iR}{l}\] (R = Resistance of wire) Magnetic field at the surface of wire \[B=\frac{{{\mu }_{0}}i}{2\pi a}\] (a = radius of wire) Hence poynting vector, directed radially inward is given by \[S=\frac{EB}{{{\mu }_{0}}}=\frac{iR}{{{\mu }_{0}}l}.\frac{{{\mu }_{0}}i}{2\pi a}=\frac{{{i}^{2}}R}{2\pi al}\]
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