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Q.

The current density varies with radial distance r as $J = {ar}^{2}$ in a cylindrical wire of radius R. The current passing through the wire between raidal distance R/3 and R/2 is

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a

$\frac{65 {πaR}^{4}}{2592}$

b

$\frac{25 π a R^{4}}{72}$

c

$\frac{65 {πa}^{2} R^{3}}{2938}$

d

$\frac{81 {πa}^{2} R^{4}}{144}$

answer is A.

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Detailed Solution

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$Given J = a r^{2} as j = \frac{di}{dA} \Rightarrow di = jdA or i = \int jdA since area of the wire A = π r^{2} dA = 2 π r dr ∴ i = \int jdA = \int_{R / 3}^{R / 2} ar 2 π rdr = 2 π a {[\frac{r^{4}}{4}]}_{R / 3}^{R / 2} = \frac{2 π a}{4} [\frac{R^{4}}{16} - \frac{R^{4}}{81}] = \frac{π a}{2} [\frac{65 R^{4}}{16 \times 81}] = \frac{65 π {aR}^{4}}{2592}$

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