The current density in a solid cylindrical wi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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: $i = 2 \pi J_{0} [\frac{6R^{2} - R^{2}}{192}] = 2 \pi J_{0} [\frac{5R^{2}}{192}] = \frac{5 J_{0} \pi R^{2}}{96}$.

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:

Similar Questions

$A$ cylindrical wire of radius $0.5 \, mm$ and conductivity $5 \times 10^{7} \, S/m$ is subjected to an electric field of $10 \, mV/m$. The expected value of current in the wire will be $x^{3} \, \pi \, mA$. The value of $x$ is ......

For which of the following dependencies of drift velocity $v_d$ on electric field $E$ is Ohm's law obeyed?

There are $0.8 \times 10^{23}$ free electrons $/ cm^3$ in copper. If a $0.2 \, A$ current is flowing in a copper wire, then the drift velocity of electrons will be, given the cross-sectional area of the wire is $0.01 \, cm^2$.

$A$ wire of diameter $0.02\,m$ contains $10^{28}$ free electrons per cubic meter. For an electrical current of $100\,A$,the drift velocity of the free electrons in the wire is nearly:

Vedclass Test Series

Exam Paper Generator

Online Exam Module