Question

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

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

Answer 1

Answer: (D)

$\mid$ Poynting vector $\mid=S=\frac{E B}{\mu_{0}}$
Since $E=\frac{V}{l}=\frac{I R}{l}$
$B=\frac{\mu_{0} I}{2 \pi a} \Rightarrow S=\left(\frac{I R}{l}\right)\left(\frac{\mu_{0} I}{2 \pi a}\right) \frac{1}{\mu_{0}}=\frac{I^{2} R}{2 \pi a l}$