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

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a

$\frac{65 π a R^{4}}{2592}$

b

$\frac{25 π a R^{4}}{72}$

c

$\frac{65 π a^{2} R^{3}}{2938}$

d

$\frac{81 π a^{2} R^{4}}{144}$

answer is A.

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$\begin{array}{l} dI = jdA = {ar}^{2} \times 2 πrdr \\ = 2 {πar}^{3} dr \\ I = 2 πa \int_{R / 3}^{R / 2} r^{3} dr \\ = 2 πa \times \frac{1}{4} {[r^{4}]}_{R / 3}^{R / 2} \\ = \frac{πa}{2} [\frac{R^{4}}{16} - \frac{R^{4}}{81}] \\ I = \frac{{πaR}^{4}}{2} [\frac{81 - 16}{16 \times 81}] = \frac{65 {πaR}^{4}}{2592} \end{array}$

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