The current density in a solid cylindrical wi | vedclass

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(B) The current $i$ through a cross-sectional area element $dA$ is given by $di = J(r) dA$. For a cylindrical wire,the area element at a radial distan

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$\frac{5 J_{0} \pi R^{2}}{96}$

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(B) The current $i$ through a cross-sectional area element $dA$ is given by $di = J(r) dA$. For a cylindrical wire,the area element at a radial distance $r$ with thickness $dr$ is $dA = 2 \pi r dr$. Substituting the given current density $J(r) = J_{0}(1 - \frac{r}{R})$,we get: $di = J_{0}(1 - \frac{r}{R}) (2 \pi r dr) = 2 \pi J_{0} (r - \frac{r^{2}}{R}) dr$. To find the total current $i$ from $r = 0$ to $r = \frac{R}{4}$,we integrate: $i = \int_{0}^{R/4} 2 \pi J_{0} (r - \frac{r^{2}}{R}) dr = 2 \pi J_{0} [\frac{r^{2}}{2} - \frac{r^{3}}{3R}]_{0}^{R/4}$. Substituting the limits: $i = 2 \pi J_{0} [\frac{(R/4)^{2}}{2} - \frac{(R/4)^{3}}{3R}] = 2 \pi J_{0} [\frac{R^{2}}{32} - \frac{R^{3}}{192R}] = 2 \pi J_{0} [\frac{R^{2}}{32} - \frac{R^{2}}{192}]$. Finding a common denominator: $i = 2 \pi J_{0} [\frac{6R^{2} - R^{2}}{192}] = 2 \pi J_{0} [\frac{5R^{2}}{192}] = \frac{5 J_{0} \pi R^{2}}{96}$.

Rod	Length,Diameter,Potential Difference
$(A)$	$L, 3d, V$
$(B)$	$2L, d, 2V$
$(C)$	$3L, 2d, 2V$

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

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