Question

A long straight wire resistance R, radius a and length l carries a constant current l. The Poynting vector for the wire will be

• A. $\frac{IR}{2\pi al}$
• B. $\frac{I{R}^2}{al}$
• C. $\frac{I^{2}R}{al}$
• D. $\frac{I^{2}R}{2\pi al}$

Answer 1

The correct option is D $\frac{I^{2}R}{2\pi al}$

Since,

Electric field in wire,

$E=\frac{V}{L}=\frac{IR}{L}$ ($V=IR$, ohm's Law)

and,

$H=\frac{I}{2\pi a}$ (current distributed along surface of wire)

So, magnitude of poynting vector, $\overrightarrow{S}$

$S=EH=\frac{I^{2}R}{2\pi aL}$