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(A) The current density $J$ is given as $J(r) = \beta(r + r_0)^2$. Consider a small elemental area $dA$ at a radial distance $r$ with thickness $dr$ a

Vedclass · Accepted Answer

$\pi \beta \left[ \frac{R^4}{12} + \frac{r_0^2 R^2}{6} + \frac{2 r_0 R^3}{9} \right]$

Vedclass · Answer

(A) The current density $J$ is given as $J(r) = \beta(r + r_0)^2$. Consider a small elemental area $dA$ at a radial distance $r$ with thickness $dr$ and angular width $d	heta$. The area element is $dA = r dr d	heta$. The current $di$ through this element is $di = J(r) dA = \beta(r + r_0)^2 r dr d	heta$. The shaded region consists of two sectors,each with an angular width of $\pi/6$. The total angular width is $\Delta 	heta = \pi/6 + \pi/6 = \pi/3$. The total current $I$ through the shaded region is obtained by integrating $di$ over $r$ from $0$ to $R$ and $	heta$ over the total angular width $\Delta 	heta$: $I = \int_0^{\pi/3} d	heta \int_0^R \beta(r^2 + 2rr_0 + r_0^2) r dr$ $I = \frac{\pi}{3} \beta \int_0^R (r^3 + 2r_0r^2 + r_0^2r) dr$ $I = \frac{\pi \beta}{3} \left[ \frac{r^4}{4} + \frac{2r_0r^3}{3} + \frac{r_0^2r^2}{2} ight]_0^R$ $I = \frac{\pi \beta}{3} \left[ \frac{R^4}{4} + \frac{2r_0R^3}{3} + \frac{r_0^2R^2}{2} ight]$ $I = \pi \beta \left[ \frac{R^4}{12} + \frac{2r_0R^3}{9} + \frac{r_0^2R^2}{6} ight]$

Current density in a cylindrical wire of radius $R$ varies with radial distance as $J(r) = \beta(r + r_0)^2$. The current through the shaded section of the wire shown in the figure is:

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