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Question-64730




Question Number 64730 by Tanmay chaudhury last updated on 20/Jul/19
Answered by ajfour last updated on 20/Jul/19
E=ρj  ⇒ E=((ρI)/(2πr^2 ))  △V =∫Eds = ((ρI)/(2π))((1/D)−(1/(D+w)))  i=((△V)/R)= ((ρIw)/(2πRD(D+w)))      =((250)/(51)) ≈ 4.9A
$${E}=\rho{j} \\ $$$$\Rightarrow\:{E}=\frac{\rho{I}}{\mathrm{2}\pi{r}^{\mathrm{2}} } \\ $$$$\bigtriangleup{V}\:=\int{Eds}\:=\:\frac{\rho{I}}{\mathrm{2}\pi}\left(\frac{\mathrm{1}}{{D}}−\frac{\mathrm{1}}{{D}+{w}}\right) \\ $$$${i}=\frac{\bigtriangleup{V}}{{R}}=\:\frac{\rho{Iw}}{\mathrm{2}\pi{RD}\left({D}+{w}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{250}}{\mathrm{51}}\:\approx\:\mathrm{4}.\mathrm{9}{A}\: \\ $$
Commented by Tanmay chaudhury last updated on 21/Jul/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by Tanmay chaudhury last updated on 21/Jul/19
Commented by Tanmay chaudhury last updated on 21/Jul/19

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