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A loThe current density across a cylinder conductor of radius R

Kaddsy , 7 Years ago
Grade 12th pass
anser 1 Answers
Susmita
You have to use polar coordonate (r,\theta).
An elemental area dS=rdrd\theta.
Now elementary current is  
dI=J.dS
We have to integrate r from 0 to R,\theta from 0 to 2\pi to obtain total current I.
\int_{0}^{R} Jo (1-\frac{r}{R})rdr\int_{0}^{2\pi}d\theta
 Please carry it out yourself. If you face problem then only ask me.
The ans will be (\pi Jo R^2)/3
The next problem will be in the same manner with the value of J as given in it.
Last Activity: 7 Years ago
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