Optics Question 176
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
Options:
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} $
Show Answer
Answer:
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} $
BETA
