Q. 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) $\frac{IR}{2\pi al}$
B) $\frac{IR^{2}}{al}$
C) $\frac{I^{2}R}{al}$
D) $\frac{I^{2}R}{2\pi al}$
Explanation
(d) 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 $S = \frac{EB}{\mu_{0}} = \frac{iR}{\mu_{0}l}.\frac{\mu_{0}i}{2\pi a} = \frac{i^{2}R}{2\pi al}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.