Question

The currect density $\overline{J}$ inside a long, solid, cylindrical wire of radius $a=12mm$ is in the direction of the central axis, and its magnitude varies linearly with radial distance $r$ from the axis according to $J=\frac{J_{0}r}{a}$ where $J_0=\frac{{10}^5}{4\pi }A/m^2.$ Find the magnitude of the magnetic field at $r=\frac{a}{2}$ in $\mu T.$

Answer 1

$a=12mm=12\times {10}^{-3}m$

$J=\frac{J_{0}Y}{a},$ $J_0=\frac{{10}^5}{\pi }A/m^2$

$B=\frac{{\mu }_0{i}_{ind}}{2\pi r}....\left(i\right)$

$i_{ind}=\underset{0}{\overset{r}{\int }}J\left(r\right)dA$

$=\underset{0}{\overset{r}{\int }}\left(\frac{J_{0}r}{a}\right)\left(2\pi r\right)$

$=\left[\frac{2\pi J_0}{a}\right]\underset{0}{\overset{r}{\int }}{r}^2$

$=\frac{2\pi J_0}{a}\frac{r^3}{3}.....\left(ii\right)$

Substitute equation (ii) in (i)

$B=\frac{{\mu }_0}{2\pi r}\left[\left(\frac{2\pi J_0}{a}\right)\left(\frac{r^3}{3}\right)\right]$

$B=\frac{{\mu }_0{J}_0{r}^2}{3a}....\left(iii\right)$

$=\frac{4\pi \times {10}^{-7}\times {10}^{+5}\times 9^2}{4\times 3\times a}$

$=\frac{\pi \times {10}^{-7}\times {10}^5\times 12\times {10}^{-3}}{3}$

$B=1.25\times {10}^{-4}T$