Question

The current density is a solid cylindrical wire of radius R, as a function of radial distance r is given by $J\left(r\right)=J_{0}\left(1-\frac{r}{R}\right).$ The total current in the radial regon r = 0 to $r=\frac{R}{4}$ will be :

• A. $\frac{5J_{0}\pi R^{2}}{32}$
• B. $\frac{5J_{0}\pi R^{2}}{96}$
• C. $\frac{3J_{0}\pi R^{2}}{64}$
• D. $\frac{J_{0}\pi R^{2}}{128}$

Answer 1

Answer: (B)

$di=JdA=J_{0}\left(1-\frac{r}{R}\right)2\pi rdr\Rightarrow \qquad\quad i=\int^{^{^{r \frac{R}{4}}}}_{_{_{r-0}}}J_{0}\left(1-\frac{r}{R}\right)2\pi rdr\frac{5J_{0}\pi R^{2}}{96}$