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

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) (I R/2 π a l) (B) (I R2/a l) (C) (I2 R/a l) (D) (I2 R/2 π a l)

Tardigrade Admin · Accepted Answer

mid Poynting vector mid=S=(E B/μ0) Since E=(V/l)=(I R/l) B=(μ0 I/2 π a) ⇒ S=((I R/l))((μ0 I/2 π a)) (1/μ0)=(I2 R/2 π a l)

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

$\frac{I R}{2 πa l}$

$\frac{I R ^{2}}{a l}$

$\frac{I ^{2} R}{a l}$

$\frac{I ^{2} R}{2 πa l}$

$∣$ Poynting vector $∣= S = \frac{EB}{μ _{0}}$
Since $E = \frac{V}{l} = \frac{I R}{l}$
$B = \frac{μ _{0} I}{2 πa} \Rightarrow S = (\frac{I R}{l}) (\frac{μ _{0} I}{2 πa}) \frac{1}{μ _{0}} = \frac{I ^{2} R}{2 πa l}$

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

2πalIR​

alIR2​

alI2R​

2πalI2R​