A long straight wire of resistance R, radius | vedclass

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Question

(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

Vedclass · Accepted Answer

$\frac{{{I^2}R}}{{2\pi al}}$

Vedclass · Answer

(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 $S = \frac{{EB}}{{{\mu _0}}} = \frac{{iR}}{{{\mu _0}l}}.\frac{{{\mu _0}i}}{{2\pi a}} = \frac{{{i^2}R}}{{2\pi al}}$

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

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